Polynomials and Factoring

This deck covers polynomial vocabulary, adding and subtracting polynomials, every way to multiply them, and the special products, then lays out a complete factoring strategy that runs from the GCF through grouping to the cubes. It targets the dropped subtraction sign, the missing middle term when a binomial is squared, trying to factor a sum of squares, and skipping the GCF and then calling what is left unfactorable.

Subject: College Algebra · 134 slides · symbolic lesson

Open the interactive version of this deck · Homework for this lesson

What this lesson covers

The lesson, slide by slide

1. Polynomials and Factoring

Title

College Algebra - Deck 02

Multiply it out. Then learn to run the whole thing backwards.

2. What you will be able to do

Objectives

By the end of this deck you can:

  1. Name the degree, the leading coefficient, and the standard form of any polynomial.
  2. Add, subtract, and multiply polynomials without losing a sign.
  1. Recognize and use the special products on sight.
  2. Factor with a strategy: greatest common factor first, then count the terms.
  1. Factor trinomials whether or not the leading coefficient is one.
  2. Decide when a polynomial is prime, and check every answer by multiplying it back out.

3. What survived from Real Numbers, Exponents, and Radicals?

Warm-up

Discussion prompt

Before we open Polynomials and Factoring: without looking back, what was the main idea of Real Numbers, Exponents, and Radicals, and what could you do by the end of it that you could not do before?

Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.

Answer:

The foundation deck for College Algebra. It covers the real number sets, absolute value as distance, interval and set-builder notation, the order of operations, and the sign rules, then gives every integer exponent rule together with the reason behind it. From there it moves to scientific notation, simplifying and combining radicals, rational exponents, and rationalizing denominators. It targets the four errors that follow students all semester: distributing an exponent across a sum, reading a negative exponent as a negative number, dropping the absolute value out of an even root, and adding unlike radicals.

4. The Language of Polynomials

Section

Part 1

5. What a polynomial is

Concept

A polynomial is a sum of terms, and every term is a number multiplied by a variable raised to a whole-number power.

\[ 4x^{3} - 7x^{2} + x - 9 \]

The whole-number-power rule is the whole definition. If an exponent is negative or fractional, or the variable is stuck under a radical or in a denominator, it is not a polynomial.

\[ \text{not polynomials:}\quad 5x^{-2}, \quad \sqrt{x} + 1, \quad \frac{3}{x} - 8 \]

6. Break it if you can: What a polynomial is

Counterexample

Discussion prompt

A polynomial is a sum of terms, and every term is a number multiplied by a variable raised to a whole-number power.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Answer:

The whole-number-power rule is the whole definition. If an exponent is negative or fractional, or the variable is stuck under a radical or in a denominator, it is not a polynomial.

7. Always take the GCF first

Picture it

Animation

Shows: Always take the GCF first — a rendered Manim animation.

Rendered with Manim.

Takeaway: Skipping this step makes every later step harder.

8. Term, coefficient, degree

Concept

term — One piece of the polynomial, joined to the others by plus or minus. Always read the sign in front of a term as part of that term.

coefficient — The number multiplying the variable part of a term, sign included.

degree of a term — The exponent on the variable in that term. A plain number has degree zero.

The degree of the polynomial is the largest degree among its terms.

9. Take the definitions apart: term vs degree of a term

Definition probe

Sort into buckets

Every line below is part of the definition of term or of degree of a term — one or the other, never both. Put each where it belongs.

term
One piece of the polynomial, joined to the others by plus or minus.; Always read the sign in front of a term as part of that term.
degree of a term
The exponent on the variable in that term.; A plain number has degree zero.
b1
One piece of the polynomial, joined to the others by plus or minus. Always read the sign in front of a term as part of that term.
b2
The exponent on the variable in that term. A plain number has degree zero.

10. Reading one polynomial apart

Concept

\[ 4x^{3} - 7x^{2} + x - 9 \]

termcoefficientdegree
4x^343
-7x^2-72
x11
-9-90

Notice two habits worth building: a bare variable has an invisible coefficient of one, and the minus signs belong to the terms that follow them.

11. Watch it run: Reading one polynomial apart

Pattern

Step through it

Step through Reading one polynomial apart one row at a time. What is driving the change, and what would the row after the last one be?

  1. Step 1: term is 4x^3
  2. Step 2: term is -7x^2
  3. Step 3: term is x
  4. Step 4: term is -9

12. Degree is the size of the biggest engine

Intuition

Think of each term as an engine and the exponent as its horsepower. When the input gets big, the highest-powered term drowns out everything else.

That is why degree gets top billing. It predicts how the polynomial behaves far from zero, how many solutions its equation can have, and how much work factoring will be.

xx^3x^2sum x^3 + x^2
1112
1010001001100
1001000000100001010000

By the last row the cubic term is doing ninety-nine percent of the work. The degree told you that in advance.

13. Watch it run: Degree is the size of the biggest engine

Pattern

Step through it

Step through Degree is the size of the biggest engine one row at a time. What is driving the change, and what would the row after the last one be?

  1. Step 1: x is 1
  2. Step 2: x is 10
  3. Step 3: x is 100

14. Standard form and the leading coefficient

Concept

Standard form means writing the terms from the highest exponent down to the lowest.

\[ 7x^{3} - 4x^{5} + 2 - x \quad \longrightarrow \quad -4x^{5} + 7x^{3} - x + 2 \]

leading coefficient — The coefficient of the highest-degree term once the polynomial is in standard form. Sign included, always.

Here the leading coefficient is negative four, not seven. Rearranging first is what makes that obvious.

15. Factored form hands you the zeros

Picture it

Animation

Shows: Factored form hands you the zeros — a rendered Manim animation.

Rendered with Manim.

Takeaway: The graph crosses exactly where a factor is zero.

16. The names you are expected to use

Concept

Polynomials get named two ways at once: by how many terms they have, and by their degree.

number of termsnameexample
1monomial5x^3
2binomial2x - 7
3trinomialx^2 + 5x + 6
4 or morejust polynomialx^3 + x^2 - x + 1
degreenameexample
0constant12
1linear3x - 1
2quadraticx^2 - 9
3cubicx^3 + 8
4quarticx^4 - 16

17. Watch it run: The names you are expected to use

Pattern

Step through it

Step through The names you are expected to use one row at a time. What is driving the change, and what would the row after the last one be?

  1. Step 1: degree is 0
  2. Step 2: degree is 1
  3. Step 3: degree is 2
  4. Step 4: degree is 3
  5. Step 5: degree is 4

18. What has to happen first: Worked example: describe a polynomial completely

Ranking

Put in order

Put the moves of Worked example: describe a polynomial completely into the order they have to happen.

  1. Order the terms from the largest exponent down
  2. Read the degree off the first term
  3. Read the leading coefficient off that same term
  4. Verify by re-reading the original, unsorted version

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Standard form is a sorting job, nothing more.

19. Worked example: describe a polynomial completely

Worked example

Write in standard form, then give the degree, the leading coefficient, and both names.

\[ 6x^{2} - 9 + 2x^{4} \]

Order the terms from the largest exponent down

Why: Standard form is a sorting job, nothing more. Each term travels with the sign in front of it.

\[ 2x^{4} + 6x^{2} - 9 \]

Read the degree off the first term

Why: In standard form the first term is the highest-degree term by construction, so its exponent is the degree of the whole polynomial.

Read the leading coefficient off that same term

Why: The leading coefficient is the number attached to the highest-degree term, which is now sitting in front.

Three terms means trinomial; degree four means quartic.

\[ \text{degree } 4, \quad \text{leading coefficient } 2, \quad \text{quartic trinomial} \]

Verify by re-reading the original, unsorted version

Why: The original had a fourth-power term with a coefficient of two and exactly three terms. Sorting changed the order, not the content, so the answers hold.

20. Degree controls the shape

Picture it

Animation

Shows: Degree controls the shape — a rendered Manim animation.

Rendered with Manim.

Takeaway: A cubic can turn twice; a quadratic only once.

21. Like terms are the only things you may combine

Concept

like terms — Terms with the same variable raised to the same exponent. Only the coefficients differ.

\[ 3x^{2} \text{ and } -8x^{2} \text{ are like.} \qquad 3x^{2} \text{ and } 3x^{3} \text{ are not.} \]

To combine like terms you add the coefficients and keep the variable part untouched. The exponent never changes.

22. Term to definition: Polynomials and Factoring

Matching

Match the pairs

Match each term to the definition this lesson gave it — not the one you would guess from the word.

  • t1. term
  • t2. coefficient
  • t3. degree of a term
  • t4. leading coefficient
  • t5. like terms
  • d1. One piece of the polynomial, joined to the others by plus or minus. Always read the sign in front of a term as part of that term.
  • d2. The number multiplying the variable part of a term, sign included.
  • d3. The exponent on the variable in that term. A plain number has degree zero.
  • d4. The coefficient of the highest-degree term once the polynomial is in standard form. Sign included, always.
  • d5. Terms with the same variable raised to the same exponent. Only the coefficients differ.

Why: These are the working definitions of term, coefficient, degree of a term, leading coefficient, like terms as Polynomials and Factoring uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.

23. Think units, not letters

Intuition

Three apples plus five apples is eight apples. Three apples plus five oranges is just three apples and five oranges. You cannot merge them, and nobody is upset about it.

A squared term and a cubed term are different units of fruit. Adding polynomials is only ever the apples-and-oranges sort, done carefully.

This is also why the exponent stays put when you combine. Eight apples, not eight apples squared.

24. By analogy: Think units, not letters

Analogy

Discussion prompt

Explain Think units, not letters by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Answer:

Three apples plus five apples is eight apples. Three apples plus five oranges is just three apples and five oranges. You cannot merge them, and nobody is upset about it.

25. Plan first: Worked example: adding two polynomials

Step zero

Discussion prompt

Worked example: adding two polynomials — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Drop the parentheses

Answer:

  1. Drop the parentheses
  2. Group the like terms
  3. Add the coefficients in each group
  4. Verify by substituting the value one into the originals and into the answer

26. Worked example: adding two polynomials

Worked example

\[ \left(3x^{2} - 5x + 7\right) + \left(x^{2} + 8x - 2\right) \]

Drop the parentheses

Why: A plus sign in front of a group changes nothing inside it, so every term keeps the sign it already had.

\[ 3x^{2} - 5x + 7 + x^{2} + 8x - 2 \]

Group the like terms

Why: Squared with squared, first-power with first-power, constants with constants. This is the apples-and-oranges sort.

\[ \left(3x^{2} + x^{2}\right) + \left(-5x + 8x\right) + \left(7 - 2\right) \]

Add the coefficients in each group

Why: Three plus one is four; negative five plus eight is three; seven minus two is five. The variable parts ride along unchanged.

\[ 4x^{2} + 3x + 5 \]

Verify by substituting the value one into the originals and into the answer

Why: The first polynomial becomes three minus five plus seven, which is five. The second becomes one plus eight minus two, which is seven. Their sum is twelve. The answer becomes four plus three plus five, which is also twelve, so the two expressions agree.

27. adding two polynomials — line by line

Picture it

Animation

Shows: Each line of the worked example "adding two polynomials", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The first polynomial becomes three minus five plus seven, which is five. The second becomes one plus eight minus two, which is seven. Their sum is twelve. The answer becomes four plus three plus five, which is also twelve, so the two expressions agree.

28. Subtracting means adding the opposite

Concept

A minus sign in front of a group is a negative one waiting to be distributed. It flips the sign of every single term inside.

\[ -\left(2x^{2} + 6x - 9\right) = -2x^{2} - 6x + 9 \]

All three signs changed, including the one on the last term that was already negative. Write this line out every time until it is automatic.

29. Teach it back: Subtracting means adding the opposite

Explain it

Discussion prompt

Explain Subtracting means adding the opposite to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Answer:

A minus sign in front of a group is a negative one waiting to be distributed. It flips the sign of every single term inside.

30. Something is wrong here: the minus only reaches the first term

Anomaly

Predict first

A student writes this, and it looks reasonable:

Change the sign on the first term only and copy the rest

It is wrong. Say what breaks — and say it before you turn the page.

Correct: The minus feels like it belongs to the term standing right next to it, so the six and the nine get copied down unchanged.

The minus feels like it belongs to the term standing right next to it, so the six and the nine get copied down unchanged.

Why: The minus feels like it belongs to the term standing right next to it, so the six and the nine get copied down unchanged.

31. Trap: the minus only reaches the first term

Trap

The trap

The problem:

\[ \left(5x^{2} - 3x + 4\right) - \left(2x^{2} + 6x - 9\right) \]

Change the sign on the first term only and copy the rest

Why: The minus feels like it belongs to the term standing right next to it, so the six and the nine get copied down unchanged.

\[ 5x^{2} - 3x + 4 - 2x^{2} + 6x - 9 \]

\[ 3x^{2} + 3x - 5 \quad \text{(wrong)} \]

The fix

The same problem:

\[ \left(5x^{2} - 3x + 4\right) - \left(2x^{2} + 6x - 9\right) \]

Distribute the negative one across all three terms

Why: Subtracting a group means subtracting everything in it. Every sign inside flips, including the minus nine, which becomes plus nine.

\[ 5x^{2} - 3x + 4 - 2x^{2} - 6x + 9 \]

\[ 3x^{2} - 9x + 13 \quad \text{(correct)} \]

Check by adding the answer back to the polynomial that was subtracted

Why: The correct answer plus the second polynomial rebuilds the first one exactly, so the correct answer is the true difference; the wrong version does not rebuild it.

32. Decode the notation: Trap: the minus only reaches the first term

Notation

Annotate

From Trap: the minus only reaches the first term — read this one piece at a time. What is each part doing?

On: \( \left(5x^{2} - 3x + 4\right) - \left(2x^{2} + 6x - 9\right) \)

  • The minus feels like it belongs to the term standing right next to it, so the six and the nine get copied down unchanged.
  • Subtracting a group means subtracting everything in it. Every sign inside flips, including the minus nine, which becomes plus nine.
  • The correct answer plus the second polynomial rebuilds the first one exactly, so the correct answer is the true difference; the wrong version does not rebuild it.

33. Worked example: subtracting two polynomials

Worked example

\[ \left(5x^{2} - 3x + 4\right) - \left(2x^{2} + 6x - 9\right) \]

Rewrite the subtraction as adding the opposite polynomial

Why: This single rewrite is what stops the dropped-sign error. Do it on its own line before you combine anything.

\[ 5x^{2} - 3x + 4 + \left(-2x^{2} - 6x + 9\right) \]

Combine the squared terms

Why: Five minus two is three, and the squared part is unchanged.

Combine the first-power terms and then the constants

Why: Negative three minus six is negative nine. Four plus nine is thirteen.

\[ 3x^{2} - 9x + 13 \]

Verify by adding the answer to the second polynomial

Why: Three plus two gives five squared-terms; negative nine plus six gives negative three; thirteen minus nine gives four. That rebuilds the first polynomial exactly, so the subtraction was done correctly.

34. subtracting two polynomials — line by line

Picture it

Animation

Shows: Each line of the worked example "subtracting two polynomials", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Three plus two gives five squared-terms; negative nine plus six gives negative three; thirteen minus nine gives four. That rebuilds the first polynomial exactly, so the subtraction was done correctly.

35. Pattern: adding and subtracting polynomials

Pattern

  1. Rewrite any subtraction as adding the opposite, flipping every sign inside the second group.
  2. Drop the parentheses.
  1. Group like terms - same variable, same exponent.
  2. Add the coefficients only; never touch the exponents.
  1. Write the result in standard form, highest degree first.
  2. Check by adding your answer back to what you subtracted.

36. Rule out three: Check yourself: subtraction

Elimination

Eliminate the wrong options

Simplify (4x^2 - 7x + 1) - (x^2 - 3x - 6).

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

  • A. 3x^2 - 4x + 7
  • B. 3x^2 - 10x - 5
  • C. 3x^2 - 4x - 5
  • D. 5x^2 - 10x - 5

Survives elimination: A

Why: Flip every sign in the second group to get 4x^2 - 7x + 1 - x^2 + 3x + 6. Then 4 - 1 = 3, and -7 + 3 = -4, and 1 + 6 = 7, giving 3x^2 - 4x + 7.

37. Check yourself: subtraction

Check

Do it on paper first. Rewrite the subtraction before you combine anything.

\[ \left(4x^{2} - 7x + 1\right) - \left(x^{2} - 3x - 6\right) \]

Check your understanding

Simplify (4x^2 - 7x + 1) - (x^2 - 3x - 6).

  • A. 3x^2 - 4x + 7 (correct)
  • B. 3x^2 - 10x - 5
  • C. 3x^2 - 4x - 5
  • D. 5x^2 - 10x - 5

Answer: A

Why: Flip every sign in the second group to get 4x^2 - 7x + 1 - x^2 + 3x + 6. Then 4 - 1 = 3, and -7 + 3 = -4, and 1 + 6 = 7, giving 3x^2 - 4x + 7.

Why B tempts people
The minus was never distributed at all: the second polynomial was copied down as written and then added, giving -7 - 3 = -10 and 1 - 6 = -5.
Why C tempts people
The sign flip stopped after the middle term. The -6 was copied instead of becoming +6, so the constant came out as 1 - 6 = -5.
Why D tempts people
The leading terms were added instead of subtracted, and the minus was never distributed, so both the 5x^2 and the rest are off.

38. Multiplying Polynomials

Section

Part 2

39. One rule powers every multiplication

Concept

There is only one rule for multiplying polynomials: every term in the first factor multiplies every term in the second.

\[ a\left(b + c\right) = ab + ac \]

FOIL, the vertical layout, and the box method are not different rules. They are three ways of making sure you do not skip a pairing.

When you multiply the variable parts, the exponents add. That comes from the product rule for exponents, not from anything new here.

40. Complete the line: Worked example: a monomial times a trinomial

Fill the middle

Fill in the blanks

From Worked example: a monomial times a trinomial — finish the line. Write what belongs on the right of the equals sign before you look.

3x^12x^{5} \cdot 4x^___ = ___

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Three times four is twelve, and the exponents two and three add to five.

41. Worked example: a monomial times a trinomial

Worked example

\[ 3x^{2}\left(4x^{3} - 2x + 5\right) \]

Multiply the monomial by the first term

Why: Three times four is twelve, and the exponents two and three add to five.

\[ 3x^{2} \cdot 4x^{3} = 12x^{5} \]

Multiply the monomial by the second term, carrying its sign

Why: The middle term is negative two times the variable, so three times negative two is negative six, and the exponents two and one add to three.

\[ 3x^{2} \cdot \left(-2x\right) = -6x^{3} \]

Multiply the monomial by the constant

Why: Three times five is fifteen and the squared part is untouched, because the constant contributes no variable.

\[ 12x^{5} - 6x^{3} + 15x^{2} \]

Verify by substituting the value one

Why: The original becomes three times the quantity four minus two plus five, which is three times seven, or twenty-one. The answer becomes twelve minus six plus fifteen, which is also twenty-one.

42. a monomial times a trinomial — line by line

Picture it

Animation

Shows: Each line of the worked example "a monomial times a trinomial", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes three times the quantity four minus two plus five, which is three times seven, or twenty-one. The answer becomes twelve minus six plus fifteen, which is also twenty-one.

43. Multiplying is tiling a rectangle

Intuition

Figure (svg): A rectangle split into four cells by a vertical and a horizontal line, with side labels x and 3 across the top and x and 2 down the left, and cell areas x squared, 3x, 2x, and 6

Give the rectangle a width split into two pieces and a height split into two pieces. The whole area is the product; the four little areas are the four pairings.

\[ \left(x + 3\right)\left(x + 2\right) = x^{2} + 3x + 2x + 6 \]

Two of the tiles are like terms, so they merge. That merge is where the middle term of a trinomial comes from.

\[ x^{2} + 5x + 6 \]

44. FOIL is just the four tiles, named

Concept

For two binomials there are exactly four pairings, and FOIL is a mnemonic for their order: First, Outer, Inner, Last.

\[ \left(a + b\right)\left(c + d\right) = \underbrace{ac}_{F} + \underbrace{ad}_{O} + \underbrace{bc}_{I} + \underbrace{bd}_{L} \]

FOIL works only when both factors are binomials. For anything bigger, go back to the one real rule and pair everything with everything.

45. Grouping, when there are four terms

Picture it

Animation

Shows: Grouping, when there are four terms — a rendered Manim animation.

Rendered with Manim.

Takeaway: The shared bracket appearing is the signal that it worked.

46. Complete the line: Worked example: FOIL with a negative term

Fill the middle

Fill in the blanks

From Worked example: FOIL with a negative term — finish the line. Write what belongs on the right of the equals sign before you look.

2x \cdot 3x = 6x^{2}

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Two times three is six, and the exponents one and one add to two.

47. Worked example: FOIL with a negative term

Worked example

\[ \left(2x + 5\right)\left(3x - 4\right) \]

First terms

Why: Two times three is six, and the exponents one and one add to two.

\[ 2x \cdot 3x = 6x^{2} \]

Outer terms, keeping the negative

Why: The second binomial's last term is negative four, so two times negative four is negative eight.

\[ 2x \cdot \left(-4\right) = -8x \]

Inner terms

Why: Five times three is fifteen and one factor of the variable survives.

Last terms

Why: Five times negative four is negative twenty, a constant.

\[ 6x^{2} - 8x + 15x - 20 \]

Combine the two middle terms

Why: Negative eight plus fifteen is seven. These are the only like terms in the expansion.

\[ 6x^{2} + 7x - 20 \]

Verify by substituting the value two

Why: The original becomes nine times two, which is eighteen. The answer becomes twenty-four plus fourteen minus twenty, which is also eighteen.

48. FOIL with a negative term — line by line

Picture it

Animation

Shows: Each line of the worked example "FOIL with a negative term", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes nine times two, which is eighteen. The answer becomes twenty-four plus fourteen minus twenty, which is also eighteen.

49. Guess the shape of the answer: Worked example: a binomial times a trinomial

Estimation

Predict first

Two terms times three terms means six pairings. FOIL does not apply, so keep a layout that catches all six.

Commit before you compute: what does Worked example: a binomial times a trinomial come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Verify by substituting the value two

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The original becomes five times the quantity eight minus ten plus four, which is five times two, or ten.

50. Worked example: a binomial times a trinomial

Worked example

Two terms times three terms means six pairings. FOIL does not apply, so keep a layout that catches all six.

\[ \left(x + 3\right)\left(2x^{2} - 5x + 4\right) \]

Distribute the first term of the binomial across all three terms

Why: One term at a time keeps the bookkeeping honest. The variable times each term raises each exponent by one.

\[ 2x^{3} - 5x^{2} + 4x \]

Distribute the second term of the binomial across all three terms

Why: Three times each term of the trinomial, signs included, gives the second row.

\[ 6x^{2} - 15x + 12 \]

Stack the two rows and add down the columns

Why: This is the vertical layout: like degrees line up, so combining becomes column addition.

degree 3degree 2degree 1degree 0
2x^3-5x^24x-
-6x^2-15x12
2x^3x^2-11x12

\[ 2x^{3} + x^{2} - 11x + 12 \]

Verify by substituting the value two

Why: The original becomes five times the quantity eight minus ten plus four, which is five times two, or ten. The answer becomes sixteen plus four minus twenty-two plus twelve, which is also ten.

51. a binomial times a trinomial — line by line

Picture it

Animation

Shows: Each line of the worked example "a binomial times a trinomial", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes five times the quantity eight minus ten plus four, which is five times two, or ten. The answer becomes sixteen plus four minus twenty-two plus twelve, which is also ten.

52. Special product: the square of a binomial

Concept

Squaring a binomial happens constantly, so its expansion is worth knowing on sight rather than re-deriving.

\[ \left(a + b\right)^{2} = a^{2} + 2ab + b^{2} \]

\[ \left(a - b\right)^{2} = a^{2} - 2ab + b^{2} \]

The middle term is the giveaway: it is twice the product of the two pieces, and it carries the sign of the binomial. The last term is a square, so it is always positive.

53. Something is wrong here: the missing middle term

Anomaly

Predict first

A student writes this, and it looks reasonable:

Squaring a binomial by squaring each piece:

It is wrong. Say what breaks — and say it before you turn the page.

Correct: The power of a product rule does let you square each factor of a product, and it is tempting to believe a sum behaves the same way.

A square is a product, so multiply it out:

Why: The power of a product rule does let you square each factor of a product, and it is tempting to believe a sum behaves the same way. It does not.

54. Trap: the missing middle term

Trap

The trap

Squaring a binomial by squaring each piece:

\[ \left(3x + 4\right)^{2} \stackrel{?}{=} 9x^{2} + 16 \]

Distribute the exponent across the sum

Why: The power of a product rule does let you square each factor of a product, and it is tempting to believe a sum behaves the same way. It does not.

Test the value one

Why: The real value is three plus four, squared, which is forty-nine. This version gives nine plus sixteen, which is twenty-five. It is off by twenty-four.

The fix

A square is a product, so multiply it out:

\[ \left(3x + 4\right)^{2} = \left(3x + 4\right)\left(3x + 4\right) \]

Do all four pairings

Why: First gives nine squared-units, outer and inner each give twelve of the first-power term, and last gives sixteen.

\[ 9x^{2} + 12x + 12x + 16 = 9x^{2} + 24x + 16 \]

Check the value one again

Why: Nine plus twenty-four plus sixteen is forty-nine, matching the true value. The twenty-four that went missing was exactly the doubled cross term.

55. Say it in words: Trap: the missing middle term

Translation

\( \left(3x + 4\right)^{2} = \left(3x + 4\right)\left(3x + 4\right) \)

Draw it

Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.

56. Plan first: Worked example: squaring a difference

Step zero

Discussion prompt

Worked example: squaring a difference — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Identify the two pieces

Answer:

  1. Identify the two pieces
  2. Square the first piece
  3. Double the product of the pieces and attach the minus
  4. Square the second piece
  5. Verify by substituting the value one

57. Worked example: squaring a difference

Worked example

\[ \left(2x - 7\right)^{2} \]

Identify the two pieces

Why: The first piece is two times the variable and the second is seven, with a minus between them. Naming them stops sign slips later.

Square the first piece

Why: Two squared is four and the variable squared comes along, giving four squared-units.

Double the product of the pieces and attach the minus

Why: Two times two times seven is twenty-eight, and the binomial had a minus, so the middle term is negative.

Square the second piece

Why: Seven squared is forty-nine, and a square is never negative, so the last term is positive even though the binomial had a minus.

\[ 4x^{2} - 28x + 49 \]

Verify by substituting the value one

Why: The original becomes two minus seven, squared, which is negative five squared, or twenty-five. The answer becomes four minus twenty-eight plus forty-nine, which is also twenty-five.

58. squaring a difference — line by line

Picture it

Animation

Shows: Each line of the worked example "squaring a difference", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes two minus seven, squared, which is negative five squared, or twenty-five. The answer becomes four minus twenty-eight plus forty-nine, which is also twenty-five.

59. Special product: sum times difference

Concept

Multiply a sum by the matching difference and the two middle terms cancel each other out.

\[ \left(a + b\right)\left(a - b\right) = a^{2} - ab + ab - b^{2} = a^{2} - b^{2} \]

\[ \left(5x - 3\right)\left(5x + 3\right) = 25x^{2} - 9 \]

Two terms in, two terms out, and no middle term at all. Read backwards, this is the difference of squares factoring pattern - the single most useful one in the course.

60. Special product: the cube patterns

Concept

A binomial times a particular trinomial collapses to just two terms. These two are worth recognizing because they run backwards into the cube factoring rules.

\[ \left(a + b\right)\left(a^{2} - ab + b^{2}\right) = a^{3} + b^{3} \]

\[ \left(a - b\right)\left(a^{2} + ab + b^{2}\right) = a^{3} - b^{3} \]

Watch what happened: six pairings, and four of them cancelled in pairs. Only the two cubes survive.

61. Pattern: multiplying any two polynomials

Pattern

  1. Check for a special product first - a square of a binomial, or a sum times its matching difference. If it is one, write the answer down.
  2. Otherwise pair everything with everything. Two binomials means four products; two terms times three terms means six.
  1. Carry every sign into the product; a negative term stays negative.
  2. Add the exponents on matching variables; never multiply them.
  1. Combine like terms and write the result in standard form.
  2. Check with a quick number substitution, usually one or two, in both the original and your answer.

62. Where does it stop working: Pattern: multiplying any two polynomials

Edge cases

Discussion prompt

Pattern: multiplying any two polynomials works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".

Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.

Answer:

  1. Check for a special product first - a square of a binomial, or a sum times its matching difference. If it is one, write the answer down.
  2. Otherwise pair everything with everything. Two binomials means four products; two terms times three terms means six.

63. Check by multiplying back

Picture it

Animation

Shows: Check by multiplying back — a rendered Manim animation.

Rendered with Manim.

Takeaway: Ten seconds of expanding catches almost every factoring error.

64. Check yourself: multiply and simplify

Check

Four pairings. Watch the sign on the last term.

\[ \left(3x - 5\right)\left(2x + 7\right) \]

Check your understanding

Multiply and simplify (3x - 5)(2x + 7).

  • A. 6x^2 - 35
  • B. 6x^2 + 11x - 35 (correct)
  • C. 6x^2 - 11x - 35
  • D. 6x^2 + 11x + 35

Answer: B

Why: First gives 6x^2, outer gives 21x, inner gives -10x, and last gives -35. The middle terms combine as 21x - 10x = 11x, so the product is 6x^2 + 11x - 35.

Why A tempts people
Only the first and last pairings were done, as if the exponent distributed across each binomial. The outer and inner products, 21x and -10x, were never formed.
Why C tempts people
The signs on the outer and inner products were swapped, giving -21x + 10x = -11x instead of 21x - 10x = 11x.
Why D tempts people
The last product was taken as positive 35. Negative five times positive seven is negative thirty-five.

65. Check yourself: which expansion is right?

Check

This is a square of a binomial. Do not let the exponent distribute.

\[ \left(4x - 3\right)^{2} \]

Check your understanding

Which expression equals (4x - 3)^2?

  • A. 16x^2 + 9
  • B. 16x^2 - 12x + 9
  • C. 16x^2 - 24x + 9 (correct)
  • D. 16x^2 - 24x - 9

Answer: C

Why: The square of a difference is the first piece squared, minus twice the product, plus the second piece squared: 16x^2 - 2(4x)(3) + 9 = 16x^2 - 24x + 9. Testing x = 1 gives 1 on both sides.

Why A tempts people
The exponent was distributed across the subtraction, squaring each piece and dropping the middle term entirely. Testing x = 1 gives 25 instead of 1.
Why B tempts people
The middle term was taken as one product of the pieces instead of twice that product, giving -12x where -24x belongs.
Why D tempts people
Negative three was squared as negative nine. Squaring a negative number gives a positive result, so the constant is positive nine.

66. Factoring: Multiplication Run Backwards

Section

Part 3

67. What factoring actually asks for

Concept

factoring — Rewriting a sum of terms as a product of simpler polynomials. It is multiplication run in reverse.

\[ x^{2} + 5x + 6 \quad \longrightarrow \quad \left(x + 2\right)\left(x + 3\right) \]

Because it is the reverse of multiplication, you can always check a factorization by multiplying it back out. There is never a reason to be unsure whether you got it right.

68. Factoring is distributing, backwards

Picture it

Animation

Shows: Factoring is distributing, backwards — a rendered Manim animation.

Rendered with Manim.

Takeaway: The same identity, read in the other direction.

69. Sums hide information, products give it away

Intuition

The number thirty-six tells you almost nothing. The same number written as two times two times three times three tells you everything: its divisors, its square root, how it shrinks a fraction.

Polynomials work the same way. Written as a sum, the useful facts are hidden. Written as a product, the values that make it zero are sitting right there in the factors.

That is why factoring is the workhorse of the rest of the course: solving equations, simplifying fractions, and graphing all start with getting a product.

70. Step zero, always: the greatest common factor

Concept

greatest common factor — The largest number and the highest power of each variable that divide every single term. Pull it out in front before you try anything else.

For the variable part, take the lowest exponent that appears in any term. That is the most every term can spare.

\[ 12x^{4} - 18x^{3} + 30x^{2} \quad \Rightarrow \quad \text{GCF} = 6x^{2} \]

Six is the largest number dividing twelve, eighteen, and thirty. The second power is the smallest power of the variable present.

71. Where does each piece belong: Polynomials and Factoring

Sorting

Sort into buckets

These are the pieces of Polynomials and Factoring, out of order. Put each one back under the part of the lesson it belongs to.

The Language of Polynomials
What a polynomial is; Term, coefficient, degree; Reading one polynomial apart
Multiplying Polynomials
One rule powers every multiplication; Worked example: a monomial times a trinomial; Multiplying is tiling a rectangle
Factoring: Multiplication Run Backwards
What factoring actually asks for; Sums hide information, products give it away; Step zero, always: the greatest common factor
s1
The Language of Polynomials is where Polynomials and Factoring puts What a polynomial is, Term, coefficient, degree, Reading one polynomial apart. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.
s2
Multiplying Polynomials is where Polynomials and Factoring puts One rule powers every multiplication, Worked example: a monomial times a trinomial, Multiplying is tiling a rectangle. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.
s3
Factoring: Multiplication Run Backwards is where Polynomials and Factoring puts What factoring actually asks for, Sums hide information, products give it away, Step zero, always: the greatest common factor. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.

72. Predict the next row: Worked example: factoring out the GCF

Pattern

Predict first

The table runs: 12x^4 | 12 | x^4 · -18x^3 | -18 | x^3

In Worked example: factoring out the GCF, given the rows so far: what is the next one — the row where term is 30x^2?

Correct: 30x^2 | 30 | x^2

termnumber partvariable part
12x^412x^4
-18x^3-18x^3
30x^230x^2

Why: The relationship between the columns, not the individual numbers, is what generates the next row. Twelve, eighteen, and thirty share a factor of six, and nothing larger works, since eighteen is not divisible by twelve.

73. Worked example: factoring out the GCF

Worked example

\[ 12x^{4} - 18x^{3} + 30x^{2} \]

Find the largest number that divides all three coefficients

Why: Twelve, eighteen, and thirty share a factor of six, and nothing larger works, since eighteen is not divisible by twelve.

termnumber partvariable part
12x^412x^4
-18x^3-18x^3
30x^230x^2

Take the lowest power of the variable

Why: Every term contains at least the second power, and the third term contains no more than that, so the second power is the most you can factor out.

Divide each term by the greatest common factor to build the second factor

Why: Twelve over six is two with two powers of the variable left; negative eighteen over six is negative three with one power left; thirty over six is five with nothing left.

\[ 6x^{2}\left(2x^{2} - 3x + 5\right) \]

Check by distributing the greatest common factor back in

Why: Six times two is twelve with four powers of the variable, six times negative three is negative eighteen with three powers, and six times five is thirty with two powers. That is the original polynomial exactly.

74. Why factoring matters

Picture it

Animation

Shows: Why factoring matters — a rendered Manim animation.

Rendered with Manim.

Takeaway: The zero-product property is the reason the whole chapter exists.

75. Something is wrong here: calling it prime before pulling the GCF

Anomaly

Predict first

A student writes this, and it looks reasonable:

Test it as a difference of squares immediately

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Three is not a perfect square, so the first term is not a perfect square either, and twenty-seven is not a perfect square.

Three is not a perfect square, so the first term is not a perfect square either, and twenty-seven is not a perfect square. The pattern seems to fail.

Why: Three is not a perfect square, so the first term is not a perfect square either, and twenty-seven is not a perfect square. The pattern seems to fail.

76. Trap: calling it prime before pulling the GCF

Trap

The trap

The problem:

\[ 3x^{2} - 27 \]

Test it as a difference of squares immediately

Why: Three is not a perfect square, so the first term is not a perfect square either, and twenty-seven is not a perfect square. The pattern seems to fail.

\[ 3x^{2} - 27 \quad \text{declared prime (wrong)} \]

Notice the contradiction

Why: Substituting three gives twenty-seven minus twenty-seven, which is zero. A polynomial that is zero at three must have a factor built from three, so it cannot be prime.

The fix

The same problem:

\[ 3x^{2} - 27 \]

Pull out the greatest common factor of three first

Why: Both terms are divisible by three. Removing it is always the first move, and it often uncovers a pattern that was invisible before.

\[ 3\left(x^{2} - 9\right) \]

Now the difference of squares is obvious

Why: Inside the parentheses both pieces are perfect squares, so it splits into a sum times a difference.

\[ 3\left(x - 3\right)\left(x + 3\right) \]

Check by multiplying back out

Why: The two binomials give the squared term minus nine, and three times that is three squared-units minus twenty-seven. That is the original.

77. Decode the notation: Trap: calling it prime before pulling the GCF

Notation

Annotate

From Trap: calling it prime before pulling the GCF — read this one piece at a time. What is each part doing?

On: \( 3\left(x - 3\right)\left(x + 3\right) \)

  • Three is not a perfect square, so the first term is not a perfect square either, and twenty-seven is not a perfect square. The pattern seems to fail.
  • Substituting three gives twenty-seven minus twenty-seven, which is zero. A polynomial that is zero at three must have a factor built from three, so it cannot be prime.
  • Both terms are divisible by three. Removing it is always the first move, and it often uncovers a pattern that was invisible before.

78. Trinomials with a leading coefficient of one

Concept

When the squared term has a coefficient of one, factoring reduces to a number puzzle.

\[ x^{2} + bx + c = \left(x + m\right)\left(x + n\right) \quad \text{where} \quad mn = c \; \text{ and } \; m + n = b \]

Find two numbers that multiply to the constant and add to the middle coefficient. That is the entire method.

The reason is the tile picture: the last tile is the product of the two numbers and the two middle tiles are their sum.

79. Let the constant's sign do the thinking

Intuition

Before hunting for numbers, read the two signs. They tell you what kind of pair you are looking for, which cuts the search in half or better.

sign of constantsign of middlethe two numbers are
positivepositiveboth positive
positivenegativeboth negative
negativeeitherone of each sign

When the constant is negative the numbers have opposite signs, and the one with the larger size takes the sign of the middle coefficient.

80. Fill in: sign of middle for Let the constant's sign do the thinking

Comparison

Comparison matrix

From Let the constant's sign do the thinking: refill the sign of middle column from what you know. The rest of the table is as it appeared.

sign of constantsign of middlethe two numbers are
positivepositiveboth positive
positivenegativeboth negative
negativeeitherone of each sign

81. Worked example: an easy trinomial

Worked example

\[ x^{2} + 7x + 12 \]

Read the signs

Why: The constant is positive and the middle coefficient is positive, so both numbers are positive. No negatives need to be tested at all.

List the positive factor pairs of twelve and check their sums

Why: Only one pair can add to seven, and going through the list in order guarantees you find it.

pairproductsum
1 and 121213
2 and 6128
3 and 4127

Write the two binomials using the winning pair

Why: Three and four multiply to twelve and add to seven, so they are the constants in the two factors.

\[ \left(x + 3\right)\left(x + 4\right) \]

Check by multiplying back out

Why: First gives the squared term, outer and inner give four and three of the first-power term for seven total, and last gives twelve. That rebuilds the original.

82. an easy trinomial — line by line

Picture it

Animation

Shows: Each line of the worked example "an easy trinomial", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: First gives the squared term, outer and inner give four and three of the first-power term for seven total, and last gives twelve. That rebuilds the original.

83. What has to happen first: Worked example: a trinomial with mixed signs

Ranking

Put in order

Put the moves of Worked example: a trinomial with mixed signs into the order they have to happen.

  1. Read the middle coefficient carefully
  2. Read the signs
  3. Search the factor pairs of twenty with opposite signs
  4. Verify by multiplying back out

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. A bare minus in front of the variable means the coefficient is negative one, not zero.

84. Worked example: a trinomial with mixed signs

Worked example

\[ x^{2} - x - 20 \]

Read the middle coefficient carefully

Why: A bare minus in front of the variable means the coefficient is negative one, not zero. Writing it out prevents the most common slip on this problem.

Read the signs

Why: The constant is negative, so the two numbers have opposite signs. Their sizes must differ by one, since the middle coefficient is negative one.

Search the factor pairs of twenty with opposite signs

Why: Only the pair five and four differs by one, and the negative belongs on the five so the sum is negative.

pairproductsum
-1 and 20-2019
-2 and 10-208
-4 and 5-201
-5 and 4-20-1

\[ \left(x - 5\right)\left(x + 4\right) \]

Verify by multiplying back out

Why: First gives the squared term, outer gives four of the first-power term, inner gives negative five of it, and last gives negative twenty. Four minus five is negative one, which matches the original middle term.

85. When the leading coefficient is not one: the ac method

Concept

With a leading coefficient other than one, the two-number puzzle changes in exactly one place: you multiply the outer two coefficients first.

\[ ax^{2} + bx + c: \quad \text{find } m, n \text{ with } mn = ac \text{ and } m + n = b \]

Then split the middle term into those two pieces and factor the resulting four terms by grouping. The grouping is what recovers the binomials.

This method never guesses. It works every time the trinomial factors over the integers, and it tells you honestly when it does not.

86. When the leading coefficient is not one

Picture it

Animation

Shows: When the leading coefficient is not one — a rendered Manim animation.

Rendered with Manim.

Takeaway: Splitting the middle term turns it into grouping.

87. Predict the next row: Worked example: the ac method in full

Pattern

Predict first

The table runs: -14 and 15 | -210 | 1 · -15 and 14 | -210 | -1

In Worked example: the ac method in full, given the rows so far: what is the next one — the row where pair is -10 and 21?

Correct: -10 and 21 | -210 | 11

pairproductsum
-14 and 15-2101
-15 and 14-210-1
-10 and 21-21011

Why: The relationship between the columns, not the individual numbers, is what generates the next row. Six times negative thirty-five is negative two hundred ten.

88. Worked example: the ac method in full

Worked example

\[ 6x^{2} + 11x - 35 \]

Multiply the leading coefficient by the constant

Why: Six times negative thirty-five is negative two hundred ten. That product, not the constant alone, is what the two numbers must multiply to.

Find two numbers with that product that add to eleven

Why: The product is negative, so the signs are opposite, and the sizes must differ by eleven. Twenty-one and ten fit.

pairproductsum
-14 and 15-2101
-15 and 14-210-1
-10 and 21-21011

Split the middle term into those two pieces

Why: Twenty-one minus ten is eleven, so the value of the expression is unchanged; only its shape is.

\[ 6x^{2} + 21x - 10x - 35 \]

Group the first two terms and the last two, and pull the greatest common factor from each pair

Why: The first pair shares three times the variable; the second pair shares negative five. Taking out the negative is what makes the two parentheses match.

\[ 3x\left(2x + 7\right) - 5\left(2x + 7\right) \]

Factor out the shared binomial

Why: Both pieces now contain the same binomial factor, so it comes out front and what is left behind forms the second factor.

\[ \left(2x + 7\right)\left(3x - 5\right) \]

Check by multiplying back out

Why: First gives six squared-units, outer gives negative ten of the first-power term, inner gives twenty-one of it, and last gives negative thirty-five. Twenty-one minus ten is eleven, matching the original.

89. the ac method in full — line by line

Picture it

Animation

Shows: Each line of the worked example "the ac method in full", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: First gives six squared-units, outer gives negative ten of the first-power term, inner gives twenty-one of it, and last gives negative thirty-five. Twenty-one minus ten is eleven, matching the original.

90. State the rule before it runs: Worked example: the ac method again…

Hypothesis

Predict first

Worked example: the ac method again, faster is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.

Correct: Multiply four by negative fifteen

Why: The target product is negative sixty, and the target sum is negative four.

A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.

91. Worked example: the ac method again, faster

Worked example

\[ 4x^{2} - 4x - 15 \]

Multiply four by negative fifteen

Why: The target product is negative sixty, and the target sum is negative four.

Find the pair

Why: Negative ten and six multiply to negative sixty and add to negative four. Opposite signs are expected because the product is negative.

\[ 4x^{2} - 10x + 6x - 15 \]

Group and pull the greatest common factor from each pair

Why: The first pair shares two times the variable; the second pair shares three. Both leave the same binomial behind, which is the sign that the split was done correctly.

\[ 2x\left(2x - 5\right) + 3\left(2x - 5\right) \]

\[ \left(2x - 5\right)\left(2x + 3\right) \]

Verify by multiplying back out

Why: First gives four squared-units, outer gives six of the first-power term, inner gives negative ten of it, and last gives negative fifteen. Six minus ten is negative four, which matches.

92. the ac method again, faster — line by line

Picture it

Animation

Shows: Each line of the worked example "the ac method again, faster", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: First gives four squared-units, outer gives six of the first-power term, inner gives negative ten of it, and last gives negative fifteen. Six minus ten is negative four, which matches.

93. Four terms? Group them in pairs

Concept

When a polynomial has four terms and no common factor across all of them, split it into two pairs and factor each pair on its own.

If the two pairs leave behind the same binomial, that binomial factors out and you are done. If they do not match, try pairing the terms differently before giving up.

\[ ax + ay + bx + by = a\left(x + y\right) + b\left(x + y\right) = \left(x + y\right)\left(a + b\right) \]

This is the same move that finished the ac method. Learning it once pays twice.

94. Plan first: Worked example: grouping, then keep going

Step zero

Discussion prompt

Worked example: grouping, then keep going — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Split into the first two terms and the last two

Answer:

  1. Split into the first two terms and the last two
  2. Pull the greatest common factor from each pair, taking a negative out of the second
  3. Factor out the shared binomial
  4. Keep factoring: the second factor is a difference of squares
  5. Verify by substituting the value one

95. Worked example: grouping, then keep going

Worked example

\[ x^{3} + 5x^{2} - 4x - 20 \]

Split into the first two terms and the last two

Why: There is no factor common to all four terms, so grouping is the right tool. The minus in front of the third term travels with it.

Pull the greatest common factor from each pair, taking a negative out of the second

Why: The first pair shares the squared term. The second pair shares negative four, and taking the negative is what makes both parentheses read the same.

\[ x^{2}\left(x + 5\right) - 4\left(x + 5\right) \]

Factor out the shared binomial

Why: Both terms now contain the same binomial, so it comes out front.

\[ \left(x + 5\right)\left(x^{2} - 4\right) \]

Keep factoring: the second factor is a difference of squares

Why: Factoring is not finished until no factor can be broken down further. Here the squared term minus four splits again.

\[ \left(x + 5\right)\left(x - 2\right)\left(x + 2\right) \]

Verify by substituting the value one

Why: The original becomes one plus five minus four minus twenty, which is negative eighteen. The factored form becomes six times negative one times three, which is also negative eighteen.

96. grouping, then keep going — line by line

Picture it

Animation

Shows: Each line of the worked example "grouping, then keep going", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes one plus five minus four minus twenty, which is negative eighteen. The factored form becomes six times negative one times three, which is also negative eighteen.

97. Check yourself: a leading coefficient other than one

Check

Use the ac method. The target product and the target sum both come from the coefficients.

\[ 6x^{2} - 7x - 20 \]

Check your understanding

Factor 6x^2 - 7x - 20 completely.

  • A. (2x - 5)(3x + 4) (correct)
  • B. (2x + 5)(3x - 4)
  • C. (6x - 5)(x + 4)
  • D. (2x - 4)(3x + 5)

Answer: A

Why: Six times negative twenty is -120, and -15 plus 8 is -7, so split the middle term: 6x^2 - 15x + 8x - 20 = 3x(2x - 5) + 4(2x - 5) = (2x - 5)(3x + 4). Expanding gives 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20.

Why B tempts people
The two signs were swapped. Expanding gives 6x^2 - 8x + 15x - 20 = 6x^2 + 7x - 20, which has the middle term with the wrong sign.
Why C tempts people
The numbers 5 and 4 were attached to the wrong factors. Expanding gives 6x^2 + 24x - 5x - 20 = 6x^2 + 19x - 20.
Why D tempts people
The first binomial still has a common factor of two, which means the greatest common factor was never removed. Expanding gives 6x^2 + 10x - 12x - 20 = 6x^2 - 2x - 20.

98. The Special Forms

Section

Part 4

99. Difference of squares, read backwards

Concept

Two terms, both perfect squares, with a minus between them. That is the whole recognition test.

\[ a^{2} - b^{2} = \left(a - b\right)\left(a + b\right) \]

A term is a perfect square when its coefficient is a perfect square and its exponent is even. Take the square root of each piece and you have the two binomials.

\[ 49x^{2} \text{ is the square of } 7x, \qquad 16x^{4} \text{ is the square of } 4x^{2} \]

100. The difference of squares

Picture it

Animation

Shows: The difference of squares — a rendered Manim animation.

Rendered with Manim.

Takeaway: A sum of squares does not factor over the reals — only the difference.

101. Guess the shape of the answer: Worked example: two differences of squares

Estimation

Predict first

Now a case where the pattern applies twice:

Commit before you compute: what does Worked example: two differences of squares come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Verify by substituting the value one

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The original becomes sixteen minus eighty-one, which is negative sixty-five.

102. Worked example: two differences of squares

Worked example

\[ 49x^{2} - 64 \]

Take the square root of each piece

Why: Forty-nine is seven squared and the exponent two is even, so the first square root is seven times the variable. Sixty-four is eight squared.

\[ \left(7x - 8\right)\left(7x + 8\right) \]

Now a case where the pattern applies twice:

\[ 16x^{4} - 81 \]

Split it once

Why: Sixteen with a fourth power is the square of four times the squared variable, and eighty-one is nine squared.

\[ \left(4x^{2} - 9\right)\left(4x^{2} + 9\right) \]

Split the difference again, but leave the sum alone

Why: The left factor is another difference of squares. The right factor is a sum of squares and does not factor over the real numbers, so it stays.

\[ \left(2x - 3\right)\left(2x + 3\right)\left(4x^{2} + 9\right) \]

Verify by substituting the value one

Why: The original becomes sixteen minus eighty-one, which is negative sixty-five. The factored form becomes negative one times five times thirteen, which is also negative sixty-five.

103. two differences of squares — line by line

Picture it

Animation

Shows: Each line of the worked example "two differences of squares", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes sixteen minus eighty-one, which is negative sixty-five. The factored form becomes negative one times five times thirteen, which is also negative sixty-five.

104. Something is wrong here: trying to factor a sum of squares

Anomaly

Predict first

A student writes this, and it looks reasonable:

Take the square root of each piece and use two plus signs

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Both terms really are perfect squares, so the pattern looks like it should apply with the signs adjusted.

Both terms really are perfect squares, so the pattern looks like it should apply with the signs adjusted.

Why: Both terms really are perfect squares, so the pattern looks like it should apply with the signs adjusted.

105. Trap: trying to factor a sum of squares

Trap

The trap

Treating a plus like a minus:

\[ x^{2} + 16 \stackrel{?}{=} \left(x + 4\right)\left(x + 4\right) \]

Take the square root of each piece and use two plus signs

Why: Both terms really are perfect squares, so the pattern looks like it should apply with the signs adjusted.

Multiply it back out and watch it fail

Why: The product is the squared term plus eight of the first-power term plus sixteen. That extra middle term was not in the original, so the factorization is false.

\[ \left(x + 4\right)\left(x + 4\right) = x^{2} + 8x + 16 \ne x^{2} + 16 \]

The fix

The honest answer:

\[ x^{2} + 16 \quad \text{is prime over the real numbers} \]

Ask what value could make it zero

Why: A squared real number is never negative, so the expression is at least sixteen for every real input. It never reaches zero, so it has no real linear factors.

Compare with the difference, which does factor

Why: Only the minus version cancels its middle terms. The plus version has nothing to cancel, which is exactly why no factorization exists.

\[ x^{2} - 16 = \left(x - 4\right)\left(x + 4\right) \]

Check the difference by expanding

Why: Outer gives four of the first-power term and inner gives negative four of it, so they cancel and only the squared term minus sixteen survives.

106. Which of these survive contact with Polynomials and Factoring?

Two truths and a lie

Sort into buckets

Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.

Holds up
A polynomial is a sum of terms, and every term is a number multiplied by a variable raised to a whole-number power.; The degree of the polynomial is the largest degree among its terms.; Notice two habits worth building: a bare variable has an invisible coefficient of one, and the minus signs belong to the terms that follow them.
Breaks
Change the sign on the first term only and copy the rest; Squaring a binomial by squaring each piece:
sound
These are stated as this lesson states them — each one survives the edge cases Polynomials and Factoring puts it through.
flawed
Each of these is lifted from a trap in this deck: reasonable-sounding, and wrong in a way that only shows up once you rely on it.

107. Perfect square trinomials

Concept

Three terms, with the first and last both perfect squares and positive, and the middle term equal to twice the product of their square roots.

\[ a^{2} + 2ab + b^{2} = \left(a + b\right)^{2} \]

\[ a^{2} - 2ab + b^{2} = \left(a - b\right)^{2} \]

Always test the middle term before claiming the pattern. If it is not exactly twice the product, this is an ordinary trinomial and the ac method is the tool.

108. Complete the line: Worked example: spotting a perfect square trinomial

Fill the middle

Fill in the blanks

From Worked example: spotting a perfect square trinomial — finish the line. Write what belongs on the right of the equals sign before you look.

2 \cdot 3x \cdot 5 = 30x

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Nine with an even exponent is the square of three times the variable, and twenty-five is the square of five.

109. Worked example: spotting a perfect square trinomial

Worked example

\[ 9x^{2} - 30x + 25 \]

Check that the outer terms are perfect squares

Why: Nine with an even exponent is the square of three times the variable, and twenty-five is the square of five. Both are positive, as the pattern requires.

Test whether the middle term is twice the product of those square roots

Why: Two times three times five is thirty, and the middle term is negative thirty. It matches in size, and the minus tells you which version of the pattern to use.

\[ 2 \cdot 3x \cdot 5 = 30x \]

Write the square of the difference

Why: The minus in the middle term puts the minus inside the binomial. The last term stays positive because it is a square.

\[ \left(3x - 5\right)^{2} \]

Verify by expanding

Why: Three times the variable squared is nine squared-units, twice the product is negative thirty of the first-power term, and negative five squared is positive twenty-five. That is the original.

110. Recognising a perfect square

Picture it

Animation

Shows: Recognising a perfect square — a rendered Manim animation.

Rendered with Manim.

Takeaway: Check the middle term against twice the root of the last.

111. Sum and difference of cubes

Concept

Two terms, both perfect cubes. Unlike squares, the sum case does factor here.

\[ a^{3} + b^{3} = \left(a + b\right)\left(a^{2} - ab + b^{2}\right) \]

\[ a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + ab + b^{2}\right) \]

A memory hook for the signs: Same, Opposite, Always Positive. The binomial takes the same sign as the original, the middle term of the trinomial takes the opposite sign, and the last term is always positive.

Note the trinomial has a single product, not twice the product, so it is never a perfect square trinomial and it does not factor further over the integers.

112. Teach it back: Sum and difference of cubes

Explain it

Discussion prompt

Explain Sum and difference of cubes to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Answer:

Two terms, both perfect cubes. Unlike squares, the sum case does factor here.

113. Cubes factor too

Picture it

Animation

Shows: Cubes factor too — a rendered Manim animation.

Rendered with Manim.

Takeaway: Same signs first, opposite sign in the middle of the quadratic.

114. Worked example: a difference of cubes

Worked example

\[ 8x^{3} - 27 \]

Identify the two cube roots

Why: Eight is two cubed and the exponent three is divisible by three, so the first piece is two times the variable. Twenty-seven is three cubed.

\[ a = 2x, \qquad b = 3 \]

Write the binomial with the same sign

Why: The original had a minus, so the binomial gets a minus. That is the S in the memory hook.

Build the trinomial: first piece squared, opposite sign on the product, then second piece squared

Why: Two times the variable squared is four squared-units, the product of the pieces is six times the variable and takes the opposite sign, so it is positive, and three squared is nine.

\[ \left(2x - 3\right)\left(4x^{2} + 6x + 9\right) \]

Verify by substituting the value one

Why: The original becomes eight minus twenty-seven, which is negative nineteen. The factored form becomes negative one times the quantity four plus six plus nine, which is negative one times nineteen, or negative nineteen.

115. a difference of cubes — line by line

Picture it

Animation

Shows: Each line of the worked example "a difference of cubes", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes eight minus twenty-seven, which is negative nineteen. The factored form becomes negative one times the quantity four plus six plus nine, which is negative one times nineteen, or negative nineteen.

116. Worked example: a sum of cubes

Worked example

\[ 27x^{3} + 64 \]

Identify the two cube roots

Why: Twenty-seven is three cubed with an exponent divisible by three, and sixty-four is four cubed.

\[ a = 3x, \qquad b = 4 \]

Binomial takes the same sign, so it is a sum

Why: The original is a sum of cubes, so the short factor is a sum. The signs inside the trinomial are what change.

Build the trinomial with the opposite sign in the middle

Why: Three times the variable squared is nine squared-units, the product is twelve times the variable and takes the opposite sign, so it is negative, and four squared is sixteen.

\[ \left(3x + 4\right)\left(9x^{2} - 12x + 16\right) \]

Verify by substituting the value one

Why: The original becomes twenty-seven plus sixty-four, which is ninety-one. The factored form becomes seven times the quantity nine minus twelve plus sixteen, which is seven times thirteen, or ninety-one.

117. a sum of cubes — line by line

Picture it

Animation

Shows: Each line of the worked example "a sum of cubes", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes twenty-seven plus sixty-four, which is ninety-one. The factored form becomes seven times the quantity nine minus twelve plus sixteen, which is seven times thirteen, or ninety-one.

118. A Strategy You Can Trust

Section

Part 5

119. Some polynomials really are prime

Concept

prime polynomial — A polynomial that cannot be written as a product of lower-degree polynomials with integer coefficients.

Prime is a legitimate final answer, not an admission of defeat. The trick is earning it: you may only say prime after you have pulled the greatest common factor and worked through every pattern that fits the number of terms.

\[ x^{2} + 16, \qquad x^{2} + x + 1, \qquad x^{2} - x - 1 \]

For the middle one, no pair of integers multiplies to one and adds to one, so the search really does come up empty.

120. By analogy: Some polynomials really are prime

Analogy

Discussion prompt

Explain Some polynomials really are prime by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Answer:

For the middle one, no pair of integers multiplies to one and adds to one, so the search really does come up empty.

121. Quadratic form: hide the mess behind one letter

Concept

Some higher-degree polynomials are quadratics in disguise. The tell is that one exponent is exactly double the other.

\[ x^{4} - 13x^{2} + 36 \quad \text{with} \quad u = x^{2} \quad \Rightarrow \quad u^{2} - 13u + 36 \]

Substitute a single letter for the repeated piece, factor the ordinary trinomial, then put the original piece back and keep factoring.

Putting the piece back is the step people forget. An answer left in terms of the substitute letter is not an answer to the question that was asked.

122. Break it if you can: Quadratic form: hide the mess behind one letter

Counterexample

Discussion prompt

Some higher-degree polynomials are quadratics in disguise. The tell is that one exponent is exactly double the other.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Answer:

Substitute a single letter for the repeated piece, factor the ordinary trinomial, then put the original piece back and keep factoring.

123. Plan first: Worked example: factoring in quadratic form

Step zero

Discussion prompt

Worked example: factoring in quadratic form — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Substitute a single letter for the squared variable

Answer:

  1. Substitute a single letter for the squared variable
  2. Factor the trinomial with the two-number method
  3. Substitute the original piece back in
  4. Keep going: both factors are differences of squares
  5. Verify by substituting the value one

124. Worked example: factoring in quadratic form

Worked example

\[ x^{4} - 13x^{2} + 36 \]

Substitute a single letter for the squared variable

Why: The fourth power is the square of the squared variable, so the whole thing becomes an ordinary trinomial in the new letter.

\[ u^{2} - 13u + 36 \quad \text{where} \quad u = x^{2} \]

Factor the trinomial with the two-number method

Why: The constant is positive and the middle coefficient is negative, so both numbers are negative. Negative four and negative nine multiply to thirty-six and add to negative thirteen.

\[ \left(u - 4\right)\left(u - 9\right) \]

Substitute the original piece back in

Why: The new letter stood for the squared variable, so each parenthesis becomes a difference involving the squared variable.

\[ \left(x^{2} - 4\right)\left(x^{2} - 9\right) \]

Keep going: both factors are differences of squares

Why: Factoring is finished only when nothing can be broken down further, and each of these splits into a sum times a difference.

\[ \left(x - 2\right)\left(x + 2\right)\left(x - 3\right)\left(x + 3\right) \]

Verify by substituting the value one

Why: The original becomes one minus thirteen plus thirty-six, which is twenty-four. The factored form becomes negative one times three times negative two times four, which is also twenty-four.

125. Pattern: the factoring decision flowchart

Pattern

Figure (svg): A five-box vertical flowchart: factor out the GCF, count the terms, two terms means squares or cubes, three terms means the trinomial methods, four terms means grouping

  1. Greatest common factor first, every single time. It is free, and it often reveals a pattern that was hidden.
  2. Count the terms in what is left. The count picks your tool.
  1. Two terms: difference of squares, difference of cubes, or sum of cubes. A sum of squares is prime.
  2. Three terms: try the perfect-square pattern first, then the two-number method or the ac method.
  1. Four terms: group in pairs and look for a shared binomial.
  2. Then factor every factor again until nothing moves, and multiply back out to check.

126. Factoring a simple trinomial

Picture it

Animation

Shows: Factoring a simple trinomial — a rendered Manim animation.

Rendered with Manim.

Takeaway: Product and sum — that pair of conditions is the whole search.

127. What has to happen first: Worked example: running the whole flowchart

Ranking

Put in order

Put the moves of Worked example: running the whole flowchart into the order they have to happen.

  1. Step one: pull the greatest common factor
  2. Step two: count the terms inside, which is two
  3. Step three: factor every factor again
  4. Verify by substituting the value one

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Both terms are divisible by two.

128. Worked example: running the whole flowchart

Worked example

\[ 2x^{4} - 32 \]

Step one: pull the greatest common factor

Why: Both terms are divisible by two. Skipping this would make the first term look like it is not a perfect square, and you would wrongly call the whole thing prime.

\[ 2\left(x^{4} - 16\right) \]

Step two: count the terms inside, which is two

Why: Two terms sends you to the squares and cubes branch. The fourth power is a perfect square and sixteen is a perfect square, with a minus between them.

\[ 2\left(x^{2} - 4\right)\left(x^{2} + 4\right) \]

Step three: factor every factor again

Why: The first parenthesis is another difference of squares. The second is a sum of squares, which is prime over the real numbers, so it stops there.

\[ 2\left(x - 2\right)\left(x + 2\right)\left(x^{2} + 4\right) \]

Verify by substituting the value one

Why: The original becomes two minus thirty-two, which is negative thirty. The factored form becomes two times negative one times three times five, which is also negative thirty.

129. running the whole flowchart — line by line

Picture it

Animation

Shows: Each line of the worked example "running the whole flowchart", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: The original becomes two minus thirty-two, which is negative thirty. The factored form becomes two times negative one times three times five, which is also negative thirty.

130. Check yourself: factor completely

Check

Run the flowchart. Greatest common factor first, then count what is left.

\[ 5x^{3} - 45x \]

Check your understanding

Factor 5x^3 - 45x completely.

  • A. 5x(x^2 - 9)
  • B. x(5x^2 - 45)
  • C. 5x(x - 3)(x + 3) (correct)
  • D. 5x(x - 3)^2

Answer: C

Why: The greatest common factor is 5x, leaving x^2 - 9, which is a difference of squares and splits into (x - 3)(x + 3). Multiplying back: 5x(x^2 - 9) = 5x^3 - 45x.

Why A tempts people
Correct so far but stopped one step early. The leftover x^2 - 9 is a difference of squares, so it still factors.
Why B tempts people
Only the variable was pulled out; the common factor of 5 in both coefficients was left behind, so the greatest common factor is incomplete.
Why D tempts people
This treats x^2 - 9 as a perfect square trinomial. Expanding gives 5x(x^2 - 6x + 9) = 5x^3 - 30x^2 + 45x, which is not the original.

131. Rule out three: Check yourself: which one is prime?

Elimination

Eliminate the wrong options

Which polynomial is prime over the integers?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

  • A. x^2 - 9
  • B. x^2 + 9
  • C. x^2 + 6x + 9
  • D. 2x^2 + 8

Survives elimination: B

Why: A sum of squares with no common factor does not factor over the integers: no pair of integers multiplies to 9 and adds to 0. The other three all factor, as (x - 3)(x + 3), (x + 3)^2, and 2(x^2 + 4).

132. Check yourself: which one is prime?

Check

Three of these factor over the integers. Test each one against the flowchart before you answer.

Check your understanding

Which polynomial is prime over the integers?

  • A. x^2 - 9
  • B. x^2 + 9 (correct)
  • C. x^2 + 6x + 9
  • D. 2x^2 + 8

Answer: B

Why: A sum of squares with no common factor does not factor over the integers: no pair of integers multiplies to 9 and adds to 0. The other three all factor, as (x - 3)(x + 3), (x + 3)^2, and 2(x^2 + 4).

Why A tempts people
This is a difference of squares, the most common factoring pattern in the course. It splits into (x - 3)(x + 3).
Why C tempts people
This is a perfect square trinomial: the outer terms are squares and the middle term is twice the product of x and 3, so it is (x + 3)^2.
Why D tempts people
Both terms share a factor of 2, so the greatest common factor comes out first and gives 2(x^2 + 4). Having a factorable greatest common factor means it is not prime.

133. Connect it up: Polynomials and Factoring

Connect it up

Draw it

One page, no notation unless you need it: draw how these connect — The Language of Polynomials · Multiplying Polynomials · Factoring: Multiplication Run Backwards · The Special Forms · A Strategy You Can Trust. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.

134. What you can do now

Recap

You can name every part of a polynomial, combine and multiply them without losing a sign, and factor with a strategy instead of a guess.

patternfactored form
a^2 - b^2(a - b)(a + b)
a^2 + 2ab + b^2(a + b)^2
a^2 - 2ab + b^2(a - b)^2
a^3 + b^3(a + b)(a^2 - ab + b^2)
a^3 - b^3(a - b)(a^2 + ab + b^2)
a^2 + b^2prime over the reals

Next up: rational expressions, where every simplification starts by factoring the top and the bottom completely.

Sources

  1. OpenStax College Algebra 2e
  2. All algebra, factorizations, and numeric results re-derived and verified by hand; every factorization expanded back out. — Verified 2026-07-31.

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