One of the three gaps from the Algebra 2 pre-class diagnostic. Why factored form hands you the zeros, the GCF that has to come first, factoring four terms by grouping, the sum and difference of cubes with their three signs, and quartics that are quadratics in disguise - finishing with the complete-factoring procedure and solving by the zero-product property.
Subject: Algebra 2 · 68 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Algebra 2 · Session 1
Pull out, group, recognise: the four moves that break any polynomial apart
Objectives
This is one of the three gaps the diagnostic found. Everything here is pattern recognition, not new arithmetic.
Five moves. The hard part is choosing which one, so the deck spends most of its time on the choosing.
Section
Section 1
Warm-up
Algebra 1 ended here, so start by pulling it back up.
Discussion prompt
Without writing anything down: what two numbers multiply to give -12 and add to give 1? And which factoring problem were you just solving?
Hint: Think of a trinomial whose middle coefficient is 1 and whose constant is -12.
Answer:
\[ 4 \cdot (-3) = -12 \qquad 4 + (-3) = 1 \]
\[ x^2 + x - 12 = (x + 4)(x - 3) \]
That trinomial move is still the engine. Everything in this deck is about getting a messier polynomial into a shape where you can use it.
Concept
A polynomial written as a product tells you exactly where its graph meets the x-axis, with no solving left to do.
\[ y = (x + 3)(x + 1)(x - 2) \]
zero of a polynomial — An input that makes the whole polynomial equal zero. On a graph it is an x-intercept.
Set each factor to zero on its own and read the three answers straight off.
Picture it
Look at where the curve meets the axis and then at the factors beside it.
Figure (svg): A cubic curve crossing the x-axis at negative three, negative one and two, with each crossing labelled by the factor that produced it
This is the whole reason factoring is worth learning. Standard form hides the zeros; factored form prints them.
Intuition
If two numbers multiply to zero, at least one of them was already zero. Nothing else works.
\[ A \cdot B = 0 \;\Longrightarrow\; A = 0 \text{ or } B = 0 \]
That is the entire trick. It fails for any other target, which is why you cannot set the factors equal to six.
zero-product property — If a product equals zero, at least one of the factors equals zero. It is only true for zero.
Socratic
Answer in a sentence before you move on.
\[ x^2 + x - 12 = 6 \]
Discussion prompt
Why can you not solve this by writing (x + 4)(x - 3) = 6 and then saying x + 4 = 6 or x - 3 = 6?
Hint: What is special about zero that six does not share?
Answer:
Because six has many factor pairs: 6 times 1, 3 times 2, 12 times one half, and infinitely many more. Knowing a product is six tells you nothing about either factor.
Zero is the only number with the property that a product hits it only when a factor is already there. Move everything to one side first, then factor.
\[ x^2 + x - 18 = 0 \]
Matching
Match each polynomial to the largest number of real zeros it could have.
Match the pairs
Why: The degree caps the number of linear factors, and each linear factor can supply at most one crossing. A degree-4 polynomial can have fewer than four real zeros, and often does, because some factors are irreducible over the reals. But it can never have five.
Section
Section 2
Prediction
Commit to an answer, then check it.
\[ 12x^5 - 18x^4 + 30x^3 \]
Predict first
What is the single largest factor common to all three terms?
Correct: 6x cubed.
\[ \gcd(12, 18, 30) = 6 \qquad \min(5, 4, 3) = 3 \]
\[ \text{GCF} = 6x^3 \]
Why: The number part is the greatest common divisor of 12, 18 and 30, which is 6. The variable part is the lowest power of x that appears anywhere, which is x cubed, because you cannot pull out more x than the poorest term has. Put them together and you get 6x cubed.
Concept
greatest common factor (GCF) — The largest expression that divides every term of the polynomial exactly: the greatest common divisor of the coefficients, times the lowest power of each shared variable.
Two separate jobs. Handle the numbers, then handle the letters, then multiply the two answers together.
| part | rule | here |
|---|---|---|
| numbers | greatest common divisor | 6 divides 12, 18 and 30 |
| letters | lowest power present | x cubed, because one term only has three |
Worked example
\[ 12x^5 - 18x^4 + 30x^3 \]
Find the greatest common divisor of 12, 18 and 30
Why: All three are even and all three are multiples of 3, so 6 divides each of them. Nothing larger does, since 12 is not divisible by 9 or 12 beyond 6 in common with 30.
Find the lowest power of x
Why: The three powers are five, four and three. Three is the smallest, so x cubed is the most you can take out of every term.
\[ \text{GCF} = 6x^3 \]
Divide each term by the GCF and write what is left inside a bracket
Why: Twelve over six is two and five minus three is two, giving 2x squared. Eighteen over six is three, giving -3x. Thirty over six is five, giving 5.
\[ 6x^3(2x^2 - 3x + 5) \]
Verify: multiply back out
Why: Six x cubed times 2x squared is 12x to the fifth, times -3x is -18x to the fourth, times 5 is 30x cubed. That is the original polynomial, so the factoring is right.
Picture it
Every factoring problem in this deck is a ladder: one layer of structure removed per rung.
Figure (svg): A three-rung ladder showing a polynomial, its greatest common factor pulled out, and the simpler quadratic left inside
The bracket left at the bottom is a plain quadratic. Notice how much easier it is to look at than the fifth-degree line above it.
Trap
\[ 4x^3 - 36x \]
Pull out an x and stop there
Why: The x is obvious, so it feels like the job is done.
\[ x(4x^2 - 36) \]
Try to factor the bracket as a difference of squares
Why: Four x squared minus thirty-six looks like a difference of squares, so it becomes (2x - 6)(2x + 6).
\[ x(2x - 6)(2x + 6) \]
Call it finished
Why: It is a product of three things, so it looks complete. But each of those brackets still has a common factor of 2 hiding in it.
\[ 4x^3 - 36x \]
Pull out the full GCF, numbers included
Why: Four divides both coefficients and x divides both terms, so the GCF is 4x, not just x.
\[ 4x(x^2 - 9) \]
Now factor the bracket as a difference of squares
Why: X squared minus nine is x squared minus three squared, so it splits into (x - 3)(x + 3).
\[ 4x(x - 3)(x + 3) \]
Verify: compare the two answers
Why: The wrong route gives x(2x - 6)(2x + 6) which equals 4x(x - 3)(x + 3) once you pull a 2 out of each bracket. Same value, but not fully factored, and a grader marks it wrong.
Error analysis
One line of this is wrong. Say which, and what it should be.
Annotate
On: \( 15x^4 + 25x^3 - 10x^2 = 5x^2(3x^2 + 5x - 10) \)
The habit that catches this every time is multiplying back out. It takes ten seconds and it is the only check that never lies.
Fill the middle
The outside and the answer are given. Supply the missing bracket.
Fill in the blanks
8x^4 - 20x^3 + 12x^2 = 4x^2\left(2x^2 - 5x + 3\right)
Why: Divide term by term: 8 over 4 is 2 and four minus two is two, giving 2x squared. Negative 20 over 4 is -5 and three minus two is one, giving -5x. Twelve over 4 is 3 and the x cancels entirely, giving 3. The bracket then factors again into (2x - 3)(x - 1), which is worth noticing.
Explain it to yourself
Explain it out loud as if to someone who has not seen it.
\[ 12x^5 - 18x^4 + 30x^3 = 6x^3(2x^2 - 3x + 5) \]
Discussion prompt
Nothing was added or removed, yet the expression looks completely different. Why is this allowed?
Hint: What property lets you write a sum of products as a product with a sum inside?
Answer:
It is the distributive property run backwards. Multiplying out is distributing forwards; factoring is undoing that.
\[ ab + ac = a(b + c) \]
Because it is an identity, it holds for every value of x. That is why multiplying back out is a valid check, and why the graph does not move.
Section
Section 3
Step zero
No algebra yet. Just describe the plan in words.
\[ x^3 + 3x^2 - 4x - 12 \]
Discussion prompt
There are four terms and no common factor across all of them. In plain English, what would you try, and what would tell you it worked?
Hint: You have two tools: pairs, and the GCF you just practised.
Answer:
Split the four terms into two pairs. Factor the GCF out of each pair separately. If the brackets left behind are identical, that shared bracket is a factor of the whole thing.
The signal that it worked is the repeat. If the two brackets differ, you either paired the wrong terms or made a sign error.
Concept
Four terms with nothing in common overall can still have something in common two at a time.
\[ ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y) \]
You manufacture the common factor by working locally, then harvest it globally.
factoring by grouping — Splitting a polynomial into pairs, factoring each pair, and then factoring out the bracket both pairs now share.
Intuition
After factoring each pair you have exactly two terms. Two terms only combine if they share something.
The bracket is that shared thing. If the brackets are different, the two terms are unlike and the expression simply does not factor that way.
So a mismatch is information, not failure. It tells you to try a different pairing before you give up.
Worked example
\[ x^3 + 3x^2 - 4x - 12 \]
Split into the first two terms and the last two
Why: There is no factor common to all four, so work two at a time and hope a shared bracket appears.
\[ (x^3 + 3x^2) + (-4x - 12) \]
Factor the GCF out of each pair
Why: The first pair shares x squared. The second pair shares -4; pulling out the negative is what makes the brackets match.
\[ x^2(x + 3) - 4(x + 3) \]
Factor out the repeated bracket
Why: Both terms now contain (x + 3), so treat that bracket as a single common factor and pull it out.
\[ (x + 3)(x^2 - 4) \]
Keep going: the second bracket is a difference of squares
Why: X squared minus four is x squared minus two squared, so it splits further. Factoring is not finished until nothing moves.
\[ (x + 3)(x - 2)(x + 2) \]
Verify: test x equals 2 in the original
Why: Eight plus twelve minus eight minus twelve is zero, and x = 2 is exactly what the factor (x - 2) predicts. The factoring checks out.
Picture it
The picture of what just happened.
Figure (svg): A four-term polynomial split into two pairs, each pair factored, revealing the same bracket twice
Circle the repeated bracket when you do this by hand. It is the moment the problem is solved, and it is easy to walk past.
Worked example
\[ 6x^3 + 9x^2 - 2x - 3 \]
Split into pairs
Why: Again there is nothing common to all four terms, so pair them up as they stand.
\[ (6x^3 + 9x^2) + (-2x - 3) \]
Factor the GCF from the first pair
Why: Three and x squared divide both terms, so pull out 3x squared and leave 2x plus 3 inside.
\[ 3x^2(2x + 3) - 2x - 3 \]
Factor -1 out of the second pair
Why: There is no number bigger than one in common, but pulling out -1 turns -2x - 3 into -(2x + 3), which matches the first bracket exactly.
\[ 3x^2(2x + 3) - 1(2x + 3) \]
Factor out the shared bracket
Why: Both terms contain (2x + 3). What is left over is 3x squared minus 1.
\[ (2x + 3)(3x^2 - 1) \]
Verify: multiply back out
Why: (2x + 3)(3x squared - 1) gives 6x cubed minus 2x plus 9x squared minus 3, which rearranges to the original. Note that 3x squared minus 1 does not factor over the integers, so this is complete.
Picture it
Two routes through the same polynomial, both legal.
Figure (svg): Two grouped pairs whose leftover brackets do not match, shown beside a re-ordered version where they do
If your two brackets disagree, re-order the terms and try a different pairing before concluding it does not factor.
Trap
\[ x^3 + 2x^2 - 3x - 6 \]
Factor x squared out of the first pair
Why: That part is standard and gives x squared times the bracket x plus 2.
\[ x^2(x + 2) + 3(x - 2) \]
Pull 3 out of -3x - 6 and keep the plus sign in front
Why: Three does divide both terms, so it feels correct - but 3 times (x - 2) is 3x - 6, not -3x - 6.
Get stuck
Why: The brackets are (x + 2) and (x - 2), which do not match, so the problem looks unfactorable. The polynomial was fine; the sign was not.
\[ x^3 + 2x^2 - 3x - 6 \]
Factor x squared out of the first pair
Why: Identical first move: x squared times the bracket x plus 2.
\[ x^2(x + 2) - 3(x + 2) \]
Pull out -3, not 3
Why: Negative three times x is -3x and negative three times 2 is -6. Taking the sign with the number is what makes the bracket come out as (x + 2).
\[ (x + 2)(x^2 - 3) \]
Verify: multiply back out
Why: (x + 2)(x squared - 3) gives x cubed - 3x + 2x squared - 6, which is the original. When the second pair starts with a minus, pull out a negative.
Sorting
Sort by whether a straightforward pairing produces a repeated bracket.
Sort into buckets
Drag each polynomial into the right column.
A fast pre-check: compare the ratio of the first two coefficients with the ratio of the last two. If they match, grouping will work.
Faded example
The first two lines are done for you. Fill in the last.
\[ 2x^3 - 10x^2 + 3x - 15 = 2x^2(x - 5) + 3(x - 5) \]
Fill in the blanks
= \left(x - 5\right)\left(2x^2 + 3\right)
Why: The repeated bracket is (x - 5), so that comes out front. What sat in front of each copy was 2x squared and then +3, and those two pieces form the second factor. The bracket 2x squared plus 3 is a sum of squares in disguise and does not factor further over the reals, so the answer is complete.
Elimination
Only one of these is a correct complete factoring.
\[ 3x^3 - 12x^2 - 4x + 16 \]
Eliminate the wrong options
Which is the fully factored form?
Survives elimination: A
Why: Group as 3x cubed minus 12x squared, then minus 4x plus 16. The first pair gives 3x squared times (x - 4); the second gives -4 times (x - 4). Pulling out the shared bracket leaves (x - 4)(3x squared - 4). Since 3x squared minus 4 is not a difference of integer squares, that is as far as it goes.
Explain it
Write two or three sentences, in your own words, no symbols.
Discussion prompt
How would you explain factoring by grouping to a student who has only ever factored trinomials?
Hint: Start from what they already do: pulling out a common factor.
Answer:
A model answer: you already know how to pull a common factor out of two terms. Grouping is that same move, done twice, on a polynomial that has four terms instead of two.
The catch is that you have to pull out the right thing each time, so that the leftovers match. When they match, you pull the matching bracket out and you are done.
Section
Section 4
Notation
Every symbol here does a job. Name them before you memorise the line.
Annotate
On: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
The trinomial factor never factors again. If you find yourself trying, you have misremembered it as a perfect square.
Concept
Both terms must be perfect cubes: the coefficient a cube number, and every exponent a multiple of three.
| expression | cube root | why |
|---|---|---|
| 8x cubed | 2x | 8 is 2 cubed and the exponent 3 divides by 3 |
| 27 | 3 | 27 is 3 cubed |
| 64x to the sixth | 4x squared | 64 is 4 cubed and six divided by three is two |
| 125 | 5 | 125 is 5 cubed |
The cube numbers worth knowing on sight are 1, 8, 27, 64, 125, 216 and 1000.
Prediction
Commit before checking.
\[ 16x^3 - 27 \]
Predict first
Is this a difference of cubes?
Correct: No, because 16 is not a perfect cube.
\[ 2^3 = 8, \quad 3^3 = 27, \quad \text{but } 16 \text{ is neither} \]
Why: Twenty-seven is a perfect cube, but sixteen is not - the cube numbers near it are 8 and 27. The cube root of 16 is not a whole number, so the formula does not apply. There is no common factor either, so over the integers this expression does not factor at all. Recognising a dead end quickly is as useful as recognising a pattern.
Worked example
\[ 8x^3 - 27 \]
Identify a and b as the cube roots of each term
Why: The cube root of 8x cubed is 2x, and the cube root of 27 is 3. Write them down before touching the formula.
\[ a = 2x, \quad b = 3 \]
Write the binomial factor with the same sign as the problem
Why: The problem is a difference, so the binomial is a minus b, which is 2x minus 3.
\[ (2x - 3)(\;\square\;) \]
Fill the trinomial: a squared, then plus ab, then plus b squared
Why: A squared is 2x squared which is 4x squared. The product ab is 2x times 3 which is 6x, and its sign is opposite to the binomial, so it is plus. B squared is 9.
\[ 8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9) \]
Verify: multiply back out
Why: 2x times the trinomial gives 8x cubed plus 12x squared plus 18x. Negative 3 times it gives -12x squared minus 18x minus 27. The middle terms cancel in pairs and you are left with 8x cubed minus 27.
Picture it
The formula is not arbitrary - it is a big cube with a small one carved out.
Figure (svg): A large cube of side two x with a smaller cube of side three removed from one corner, labelled as the difference of two cubes
The binomial is the difference of the side lengths. The trinomial is the leftover shell, which is why it never simplifies further.
Worked example
\[ 27x^3 + 64 \]
Identify a and b
Why: The cube root of 27x cubed is 3x, and the cube root of 64 is 4, since four cubed is sixty-four.
\[ a = 3x, \quad b = 4 \]
Write the binomial with the same sign as the problem
Why: The problem is a sum, so the binomial is a plus b, which is 3x plus 4. Unlike a sum of squares, a sum of cubes does factor.
\[ (3x + 4)(\;\square\;) \]
Fill the trinomial with the opposite middle sign
Why: A squared is 9x squared. The product ab is 12x, and since the binomial was a plus, the middle sign is minus. B squared is 16.
\[ 27x^3 + 64 = (3x + 4)(9x^2 - 12x + 16) \]
Verify: multiply back out
Why: 3x times the trinomial gives 27x cubed minus 36x squared plus 48x. Four times it gives 36x squared minus 48x plus 64. Both middle pairs cancel, leaving 27x cubed plus 64.
Picture it
Three signs, and only the first one is copied from the problem.
Figure (svg): A sign chart for the sum and difference of cubes formulas showing which sign goes in each of the three positions
Say it out loud: Same, Opposite, Always Positive. And there is no 2 in the middle term - that belongs to perfect square trinomials, not cubes.
Trap
\[ x^3 + 8 \]
Recognise it as a sum of cubes with a equal to x and b equal to 2
Why: That part is correct - one is x cubed and eight is two cubed.
Write the trinomial as a squared plus 2ab plus b squared
Why: It looks like the familiar perfect square trinomial, so the 2 gets carried in by reflex.
\[ (x + 2)(x^2 + 4x + 4) \]
Notice something is off
Why: The second bracket factors into (x + 2) squared, so the whole thing would be (x + 2) cubed - which multiplies out to x cubed plus 6x squared plus 12x plus 8, not x cubed plus 8.
\[ x^3 + 8 \]
Recognise it as a sum of cubes with a equal to x and b equal to 2
Why: Identical first move.
Write the trinomial as a squared minus ab plus b squared
Why: Opposite middle sign, and the coefficient is 1, not 2. There is no doubling anywhere in the cube formulas.
\[ (x + 2)(x^2 - 2x + 4) \]
Verify: multiply back out
Why: x times the trinomial gives x cubed minus 2x squared plus 4x. Two times it gives 2x squared minus 4x plus 8. Everything cancels except x cubed plus 8.
Comparison
Fill the blank cells. These three account for most of what you will meet.
Comparison matrix
| pattern | factors as | does the trinomial factor again? |
|---|---|---|
| difference of squares | (a - b)(a + b) | no trinomial appears |
| sum of squares | does not factor over the reals | not applicable |
| difference of cubes | (a - b)(a^2 + ab + b^2) | no, never |
| sum of cubes | (a + b)(a^2 - ab + b^2) | no, never |
The row that surprises people is the second one. A sum of squares is stuck; a sum of cubes is not. The parity of the exponent is what makes the difference.
Discrimination
Speed matters more than accuracy here. Sort, do not compute.
Sort into buckets
Which tool does each one need first?
Counterexample
A classmate makes this claim. Find the counterexample.
\[ x^2 + 4 \;\overset{?}{=}\; (x + 2)^2 \]
Discussion prompt
Someone insists that a sum of squares factors just like a sum of cubes does. Find one value of x that proves them wrong, and explain what went wrong in general.
Hint: Pick any value except zero and evaluate both sides.
Answer:
Take x equal to 1. The left side is 1 plus 4, which is 5. The right side is 3 squared, which is 9. One counterexample is enough.
\[ (x + 2)^2 = x^2 + 4x + 4 \;\neq\; x^2 + 4 \]
In general, squaring a binomial always produces a middle term. A sum of squares has no middle term, so no binomial squared can equal it. Over the real numbers, a sum of squares is irreducible - and that is a legitimate final answer.
Reverse engineer
You are given the answer. Reconstruct the problem.
\[ (5x - 2)(25x^2 + 10x + 4) \]
Fill in the blanks
125x^3 - 8
Why: Read the formula backwards. The binomial (5x - 2) says a is 5x and b is 2, and the minus sign says it was a difference. So the original was a cubed minus b cubed, which is (5x) cubed minus 2 cubed, giving 125x cubed minus 8. The trinomial is the confirmation: 25x squared is a squared, 10x is ab, and 4 is b squared.
Section
Section 5
Hypothesis
Look at the exponents before you look at anything else.
\[ x^4 - 13x^2 + 36 \]
Predict first
The exponents are 4, 2 and 0. What does that pattern let you do?
Correct: Treat x squared as a single variable and factor it like a trinomial.
\[ u = x^2 \;\Longrightarrow\; u^2 - 13u + 36 \]
Why: Four is twice two, and two is twice one, so if you call x squared by a new name - say u - the expression becomes u squared minus 13u plus 36, an ordinary trinomial. Taking square roots of individual terms is never legal, and there is no GCF because the constant 36 has no x in it.
Concept
quadratic form — A polynomial whose three exponents follow the pattern 2n, n and 0. Substituting u for the middle power turns it into an ordinary quadratic.
\[ x^4 - 13x^2 + 36 \qquad x^6 + 7x^3 - 8 \qquad x - 5\sqrt{x} + 6 \]
All three are the same shape. Only the name of the inside piece changes.
The test is arithmetic: is the highest exponent exactly double the middle one, with a plain constant at the end?
Intuition
It does not make the problem easier. It makes the problem look like one you have already solved a hundred times.
Your brain has a trained pattern for a trinomial with a single squared term. Substitution is how you hand the messy version to that trained pattern.
The only cost is remembering to put the original variable back at the end, which is where nearly all the lost marks live.
Worked example
\[ x^4 - 13x^2 + 36 \]
Substitute u for x squared
Why: Then x to the fourth is u squared, because x to the fourth equals x squared, squared. The expression becomes an ordinary trinomial.
\[ u^2 - 13u + 36 \]
Find two numbers multiplying to 36 and adding to -13
Why: Both must be negative to give a positive product and a negative sum. Negative four and negative nine work: their product is 36 and their sum is -13.
\[ (u - 4)(u - 9) \]
Substitute x squared back in
Why: This is the step people skip. The answer must be in terms of x, not u.
\[ (x^2 - 4)(x^2 - 9) \]
Factor each bracket again - both are differences of squares
Why: X squared minus four splits into (x - 2)(x + 2), and x squared minus nine splits into (x - 3)(x + 3).
\[ (x - 2)(x + 2)(x - 3)(x + 3) \]
Verify: test x equals 3
Why: Eighty-one minus 13 times 9 plus 36 is 81 minus 117 plus 36, which is zero. The factor (x - 3) predicted exactly that, so the factoring is right.
Picture it
The graph of the quartic you just factored.
Figure (svg): A quartic curve in a W shape crossing the x-axis four times, at negative three, negative two, two and three
Each of the four linear factors owns one crossing. Sketching this is a fast sanity check on any factoring you are unsure about.
Trap
\[ x^4 - 13x^2 + 36 \]
Substitute and factor
Why: Let u be x squared, then factor the trinomial into (u - 4)(u - 9). All correct so far.
\[ (u - 4)(u - 9) \]
Write that down as the answer
Why: It is factored, and it looks finished. But the question was about x, and u does not appear anywhere in the original problem.
Lose the marks
Why: Even a correct u-answer scores nothing, because it answers a question nobody asked. Worse, if the task was to solve, u equals 4 and u equals 9 are not the solutions.
\[ x^4 - 13x^2 + 36 \]
Substitute and factor
Why: Identical: let u be x squared and factor into (u - 4)(u - 9).
Immediately swap x squared back in
Why: Do it in the same breath as the factoring, before writing anything else. The habit is what protects you under time pressure.
\[ (x^2 - 4)(x^2 - 9) \]
Now factor each difference of squares
Why: Both brackets split further, giving four linear factors and four real zeros.
\[ (x - 2)(x + 2)(x - 3)(x + 3) \]
Verify: count the zeros against the degree
Why: Degree four, four linear factors, four zeros at plus and minus two and plus and minus three. If you had stopped at u you would have reported two solutions instead of four.
Pattern
Step through and predict what the substitution should be each time.
Step through it
Before the last frame: if the exponents were 6, 3 and 0, what would you substitute?
The rule is always the same: rename the middle power. What matters is that the top exponent is exactly double it.
Ranking
Put the steps of a complete factoring into the order that never fails.
Put in order
Why: The GCF comes first because it shrinks everything that follows and because a missed GCF makes an otherwise-correct answer incomplete. Counting terms selects the tool. Applying the pattern does the work. Re-checking is what separates factored from fully factored - a difference of squares hiding inside a bracket is the most common miss. Verifying by multiplying out is the only step that can catch a sign error.
Commit first
Answer, then rate your confidence honestly.
\[ x^6 - 9x^3 + 8 \]
Predict first
Factor this completely. How many linear factors with integer coefficients does the final answer contain?
Correct: Two.
\[ (x^3 - 1)(x^3 - 8) \]
\[ (x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4) \]
Why: Let u be x cubed, giving u squared minus 9u plus 8, which factors as (u - 1)(u - 8). Substituting back gives (x cubed - 1)(x cubed - 8). Each of those is a difference of cubes: the first becomes (x - 1)(x squared + x + 1) and the second becomes (x - 2)(x squared + 2x + 4). So the complete factoring has exactly two linear factors, (x - 1) and (x - 2), plus two irreducible trinomials.
Section
Section 6
Pattern
One order, used every time, whatever the polynomial looks like.
Steps one and six are the ones people skip, and between them they account for most lost marks on this topic.
Picture it
Keep this in front of you for the first ten problems, then stop needing it.
Figure (svg): A decision flowchart for factoring any polynomial, starting with the greatest common factor and branching by the number of terms
Notice that the branches all rejoin at the same place: re-check, then verify.
Worked example
\[ 2x^5 - 32x = 0 \]
Pull out the GCF
Why: Two divides both coefficients and x divides both terms, so the GCF is 2x. This drops the degree from five to four immediately.
\[ 2x(x^4 - 16) = 0 \]
Recognise the bracket as a difference of squares
Why: X to the fourth is x squared, squared, and 16 is 4 squared, so it splits into (x squared - 4)(x squared + 4).
\[ 2x(x^2 - 4)(x^2 + 4) = 0 \]
Factor again - the first bracket is another difference of squares
Why: X squared minus four splits into (x - 2)(x + 2). The bracket x squared plus 4 is a sum of squares and stops there.
\[ 2x(x - 2)(x + 2)(x^2 + 4) = 0 \]
Set each factor to zero
Why: The constant 2 can never be zero, so it contributes nothing. The factor x gives zero, and the two linear factors give 2 and -2. The sum of squares has no real solution.
\[ x = 0, \quad x = 2, \quad x = -2 \]
Verify: substitute x equals 2 into the original
Why: Two times thirty-two is sixty-four, minus thirty-two times two is sixty-four, and sixty-four minus sixty-four is zero. The solution checks.
Picture it
A fifth-degree polynomial, but only three places where the graph touches down.
Figure (svg): A number line marking the three real zeros of the quintic at negative two, zero and two, with the non-real factor noted separately
The degree is the maximum number of real zeros, never a promise. Irreducible factors quietly use up degree without producing crossings.
Check
Work it on paper before clicking anything.
Check your understanding
Factor completely: 8x^3 + 125.
Answer: A
Why: Here a is 2x and b is 5, since 8 is 2 cubed and 125 is 5 cubed. The sum of cubes formula gives (a + b)(a squared - ab + b squared), so the binomial is 2x + 5 and the trinomial is 4x squared - 10x + 25. Multiplying back out, the +20x squared and -20x squared cancel, as do +50x and -50x, leaving 8x cubed + 125.
Check
Watch for the step after the step you think is last.
Check your understanding
Factor completely: 5x^4 - 80.
Answer: B
Why: Pull out the GCF of 5 to get 5(x to the fourth - 16). The bracket is a difference of squares, giving 5(x squared - 4)(x squared + 4). The first of those is another difference of squares, so it splits into (x - 2)(x + 2). The sum of squares x squared + 4 is irreducible over the reals, so the complete factoring is 5(x - 2)(x + 2)(x squared + 4).
Invariant
Step through one polynomial being factored and watch what never changes.
Step through it
Two things stay constant down every row. What are they, and why does that matter?
Degree and value are invariant because factoring is a rewriting, not a transformation. That is exactly why substituting a number back in is a valid check.
Real world
One concrete setting, then abstract it yourself.
Discussion prompt
An open box is made by cutting equal squares of side x from the corners of a 12 by 12 sheet and folding up the sides, giving volume V = x(12 - 2x)^2. For which values of x is the volume zero, and which of those make physical sense?
Hint: The equation is already factored. Use it.
Answer:
\[ V = x(12 - 2x)^2 = 0 \]
The factor x gives x equals 0, and the squared factor gives 12 - 2x equals 0, so x equals 6.
Both are mathematically correct and both are physically meaningless as boxes: cutting nothing gives a flat sheet, and cutting six from each corner leaves no base at all. The useful domain is strictly between 0 and 6, and the zeros are what tell you where that window ends.
This is the everyday use of factoring outside a test: the zeros are the boundaries of the sensible range.
Check
Factor first, then read the solutions off.
Check your understanding
Solve x^4 - 5x^2 + 4 = 0.
Answer: B
Why: Substitute u for x squared to get u squared - 5u + 4, which factors as (u - 1)(u - 4). Substituting back gives (x squared - 1)(x squared - 4), and each of those is a difference of squares: (x - 1)(x + 1)(x - 2)(x + 2). Setting each factor to zero gives four solutions: 1, -1, 2 and -2.
Edge cases
The pattern holds; test where it stops being useful.
\[ x^4 + 4 \]
Discussion prompt
This has the quadratic-form shape and it is a sum. Can you factor it over the real numbers? What does that tell you about which patterns are safe to apply blindly?
Hint: Substitute u for x squared and see what you get.
Answer:
Substituting gives u squared plus 4, a sum of squares, which is irreducible over the reals. So no ordinary factoring works and the answer is that it does not factor - over the reals, with the tools in this deck.
The lesson is that recognising the shape only tells you which tool to try. It does not promise the tool will succeed. Being able to say does not factor with confidence is part of the skill, not an admission of defeat.
For the curious: this one does factor over the reals in a way no Algebra 2 course expects, as (x squared - 2x + 2)(x squared + 2x + 2). Multiply it out if you want to see it.
Exit ticket
The last commitment of the deck. Be honest - this is what the session will open with.
Predict first
Which of these is shakiest for you right now?
Correct: Whichever you picked is the first thing to drill.
Why: All four are process errors rather than understanding errors, which is good news: they respond very quickly to deliberate practice. The cube signs are pure memorisation and take about ten minutes. The other three are habits, and the habit that fixes all of them is the same one - multiply back out before you write the final answer.
Connect it up
One page, drawn by you, that you can rebuild from memory in two minutes.
Draw it
Draw a box for each of the five tools: GCF, difference of squares, sum and difference of cubes, trinomial, and grouping. Under each, write the shape that triggers it - how many terms, what the exponents look like. Then draw an arrow from GCF to every other box, since it always comes first. Finally, write one polynomial of your own that needs at least three of the boxes, and factor it.
The arrows out of GCF are the point of the whole page. Almost every incomplete answer on this topic starts by skipping that box.
Recap
Five tools, one order, and a verification step that catches nearly everything.
| if you remember one thing | it should be |
|---|---|
| about starting | the GCF comes out first, every single time |
| about grouping | the two brackets must match, and a leading minus means pull out a negative |
| about cubes | same, opposite, always positive - and no 2 in the middle |
| about substitution | swap the original variable back in the same breath you factor |
| about finishing | multiply back out; it is ten seconds and it never lies |
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