Factoring a monic trinomial by finding two numbers with the right product and sum, the difference of two squares and perfect square trinomial patterns, the zero product property that turns factors into roots, a doubling-the-area model, and the zeros of a quadratic function.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Solve x^2 + bx + c = 0 by Factoring
Objectives
Five outcomes. The third one is why the first two matter.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-257 — the lesson these objectives are drawn from
Warm-up
Lesson 4.2 read the x-intercepts straight off intercept form. This lesson earns that form.
Discussion prompt
Lesson 4.2 told you the graph of y equals the quantity x plus 3, times the quantity x minus 4, crosses the horizontal axis at negative 3 and 4. If instead you are handed y equals x squared minus x minus 12, what would you have to do before you could read the crossings the same way?
Hint: Multiply the two brackets out and compare.
Answer:
\[ (x + 3)(x - 4) = x^2 - 4x + 3x - 12 = x^2 - x - 12 \]
The two expressions are the same function written two ways, so the crossings are the same. What is missing is a way to go backwards, from the sum of three terms to the product of two brackets. That reverse move is called factoring, and it is the whole of this lesson.
Concept
A quadratic equation cannot be solved by undoing operations, because x appears twice. But if the quadratic can be written as a product of two brackets equal to zero, then one of those brackets must be zero, and each bracket is a linear equation you already know how to solve.
quadratic equation — An equation that can be written in the standard form ax squared plus bx plus c equals zero, where a is not zero. Its solutions are called the roots of the equation.
\[ x^2 + 3x - 28 = 0 \;\Longrightarrow\; (x-4)(x+7) = 0 \;\Longrightarrow\; x = 4 \text{ or } x = -7 \]
Everything in this lesson serves that one move. Factoring is the price of admission; the zero product property is what you buy with it.
Figure (svg): A product equal to zero branching into two separate linear equations
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-253
Section
Section 1
Concept
To factor x squared plus bx plus c, look for integers m and n such that the expression equals the quantity x plus m, times the quantity x plus n. Multiplying that product out shows what m and n must satisfy: their product is c and their sum is b.
trinomial — The sum of three monomials, such as x squared plus 11x plus 28. A monomial is a number, a variable, or a product of a number and one or more variables; a binomial is the sum of two monomials.
\[ x^2 + bx + c = (x+m)(x+n) = x^2 + (m+n)x + mn \]
The middle line is the whole derivation. Comparing it term by term with the original forces m plus n to equal b and m times n to equal c, so the search has exactly two conditions and both must hold.
Figure (svg): A table of every factor pair of twenty beside its sum, with the pair summing to negative nine highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-252 — Factor trinomials of the form x^2 + bx + c
Picture it
Two glances at the trinomial decide the shape of the answer.
Figure (svg): Three panels showing how the signs of the constant and the middle coefficient decide the signs of m and n
The sign of the constant says whether m and n agree in sign; the sign of the middle coefficient then says which sign. Half the candidate pairs are eliminated before a single sum is computed.
Worked example
Example 1a. Product first, then sum.
\[ \text{Factor } x^2 - 9x + 20. \]
Name what you are looking for
Why: You want x squared minus 9x plus 20 to equal the quantity x plus m, times the quantity x plus n, where mn is 20 and m plus n is negative 9.
\[ m n = 20, m + n = -9 \]
Use the signs to narrow the search
Why: The constant is positive, so m and n have the same sign; the middle coefficient is negative, so both are negative.
List the negative factor pairs of 20 and their sums
Why: Negative 1 and negative 20 sum to negative 21; negative 2 and negative 10 sum to negative 12; negative 4 and negative 5 sum to negative 9.
\[ -4\text{ and } -5 \]
Write the factorisation
Why: Substituting m equal to negative 4 and n equal to negative 5 gives the two brackets.
\[ (x - 4) (x - 5) \]
Figure (svg): The solution to Worked example factor a trinomial shown as a ladder of expressions, one row per algebraic move
\[ x^2 - 9x + 20 = (x - 4)(x - 5) \]
Verify: multiply the brackets back out
Why: Using FOIL: x times x is x squared, x times negative 5 is negative 5x, negative 4 times x is negative 4x, and negative 4 times negative 5 is positive 20. Combining the middle terms gives x squared minus 9x plus 20 — the original expression, so the factorisation is exact.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-252
Matching
For each, ask for the product first, then the sum.
Match the pairs
Why: The first and last share the same constant and differ only in the sign of the middle term, which flips both signs inside the brackets. The middle two have negative constants, so their brackets carry opposite signs, and there the larger factor takes the sign of b.
Worked example
Example 1b. The search is finite, so failing is a legitimate result.
\[ \text{Factor } x^2 + 3x - 12 \text{ if possible.} \]
State the two conditions
Why: You want mn equal to negative 12 and m plus n equal to 3.
\[ m n = -12, m + n = 3 \]
Use the sign of the constant
Why: The constant is negative, so m and n have opposite signs; every pair will therefore have a sum smaller in size than the larger factor.
List all six pairs and their sums
Why: Negative 1 with 12 gives 11; 1 with negative 12 gives negative 11; negative 2 with 6 gives 4; 2 with negative 6 gives negative 4; negative 3 with 4 gives 1; 3 with negative 4 gives negative 1.
Compare against the target sum
Why: None of the six sums is 3, and there are no other integer pairs whose product is negative 12.
\[ \text{no pair sums to } 3 \]
Figure (svg): Two columns separating trinomials that factor over the integers from those that do not
\[ x^2 + 3x - 12 \text{ cannot be factored} \]
Verify: confirm the list was exhaustive
Why: Twelve has exactly three factor pairs of positive integers — 1 and 12, 2 and 6, 3 and 4 — and each can be signed two ways to give a negative product, so six pairs is all of them. Because the list was complete, cannot be factored is a proved conclusion rather than a failure to spot something.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-252
Trap
\[ x^2 - 9x + 20 = (x + m)(x + n) \]
Find a pair whose product is 20 and stop
Why: Four and five multiply to 20, so take m equal to 4 and n equal to 5.
\[ (x + 4)(x + 5) \quad \text{(wrong)} \]
Expanding gives x squared plus 9x plus 20. The constant is right, but the middle term has the wrong sign, because only the product condition was checked.
\[ mn = 20 \quad \text{and} \quad m + n = -9 \]
Test both conditions on every candidate pair
Why: Four and five have the right product but their sum is positive 9, not negative 9, so the pair is rejected.
\[ (-4)(-5) = 20 \quad \text{and} \quad -4 + (-5) = -9 \;\Longrightarrow\; (x - 4)(x - 5) \]
Both conditions are constraints on the same pair. A pair that satisfies one and not the other is not a near miss; it is simply not a solution.
Sorting
Read the sign of the constant, then the sign of the middle coefficient.
Sort into buckets
Sort each trinomial by the signs its factors must have.
Two glances at signs cut the candidate list roughly in half before any arithmetic is done.
Fill the middle
Guided Practice 3.
Fill in the blanks
r^2 + 2r - 63 = (r + 9)(r - 7)
Why: Nine times negative 7 is negative 63, matching the constant, and 9 plus negative 7 is 2, matching the middle coefficient. Both conditions hold, so the factorisation is correct. Notice the positive middle term goes with the larger factor being positive.
Prediction
Commit before reasoning.
Predict first
You have tried every integer pair whose product is c and none sums to b. What have you shown?
Correct: The trinomial does not factor over the integers.
\[ x^2 + 3x - 12 = 0 \;\Longrightarrow\; x = \frac{-3 \pm \sqrt{57}}{2} \]
Why: An integer has finitely many factor pairs, so the search terminates and its failure is a proof, not a stall. But it is a statement about integers only: x squared plus 3x minus 12 has perfectly good real zeros, which Lesson 4.8's quadratic formula will produce as irrational numbers. Not factorable over the integers and has no real zeros are different claims, and confusing them is a common error later in the chapter.
Section
Section 2
Concept
A difference of two squares factors into a sum times a difference. A perfect square trinomial — one whose middle term is twice the product of the two square roots — factors into a single bracket squared. Recognising either shape replaces the search entirely.
\[ a^2 - b^2 = (a+b)(a-b), \quad a^2 \pm 2ab + b^2 = (a \pm b)^2 \]
The middle term is the tell. In a perfect square trinomial the first and last terms are squares and the middle term is exactly twice the product of their roots; if it is anything else, the expression is not a perfect square and the general search applies.
Figure (svg): The three special factoring patterns, each beside a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 253-253 — Special Factoring Patterns
Picture it
Example 2, all three parts.
Figure (svg): The three special factoring patterns, each beside a worked instance
Each pattern is just a particular case of the general method, memorised because it comes up constantly. Nothing would go wrong if you searched for m and n instead — it would only take longer.
Worked example
Example 2. Identify the shape, then write the answer.
\[ \text{Factor } x^2 - 49, \; d^2 + 12d + 36, \; z^2 - 26z + 169. \]
Rewrite the first as a difference of squares
Why: Forty-nine is 7 squared, so the expression is x squared minus 7 squared and the pattern applies with a equal to x and b equal to 7.
\[ x ^{2} - 7 ^{2} = (x + 7) (x - 7) \]
Test the second against the perfect square shape
Why: Thirty-six is 6 squared and the middle term 12d is twice d times 6, so the trinomial is d squared plus twice d times 6, plus 6 squared.
\[ d ^{2} + 2(d) (6) + 6 ^{2} = (d + 6) ^{2} \]
Test the third the same way
Why: One hundred sixty-nine is 13 squared and 26z is twice z times 13, with a minus sign, so the negative pattern applies.
\[ z ^{2} - 2(z) (13) + 13 ^{2} = (z - 13) ^{2} \]
State all three
Why: Each was written down once the shape was recognised; no factor pairs were listed.
Figure (svg): The solution to Worked example factor with special patterns shown as a ladder of expressions, one row per algebraic move
\[ (x+7)(x-7), \quad (d+6)^2, \quad (z-13)^2 \]
Verify: expand one of the squares
Why: Expanding the quantity d plus 6, squared, gives d squared plus 6d plus 6d plus 36, which is d squared plus 12d plus 36 — the original. The middle term appearing twice is exactly why the pattern requires twice the product of the roots.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 253-253
Sorting
Look at the first and last terms first.
Sort into buckets
Sort each expression by the pattern it fits.
Spending two seconds sorting saves the search on roughly half of all textbook trinomials.
Worked example
Guided Practice 4 to 7.
\[ \text{Factor } x^2 - 9, \; q^2 - 100, \; y^2 + 16y + 64, \; w^2 - 18w + 81. \]
First two: differences of squares
Why: Nine is 3 squared and 100 is 10 squared, so each is a sum times a difference.
\[ (x + 3) (x - 3), (q + 10) (q - 10) \]
Third: check twice the product
Why: Sixty-four is 8 squared and 16y is twice y times 8, so the positive pattern applies.
\[ (y + 8) ^{2} \]
Fourth: check twice the product again
Why: Eighty-one is 9 squared and 18w is twice w times 9, with a minus sign.
\[ (w - 9) ^{2} \]
Note what all four share
Why: In each, both the first and last terms are perfect squares. That is the signal to test a pattern before starting a search.
Figure (svg): The solution to Worked example four more of the same shown as a ladder of expressions, one row per algebraic move
\[ (x+3)(x-3), \; (q+10)(q-10), \; (y+8)^2, \; (w-9)^2 \]
Verify: check one against the general method
Why: For y squared plus 16y plus 64 the general search wants mn equal to 64 and m plus n equal to 16; the pair 8 and 8 satisfies both, giving the quantity y plus 8, times itself. The pattern and the search agree, as they must — the pattern is only a shortcut through the same reasoning.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 253-253
Error analysis
A student applies the difference-of-squares pattern to an expression that is a sum.
Annotate
On: \( x^2 + 49 = x^2 + 7^2 = (x + 7)(x - 7) \)
A sum of two squares has no real factorisation. Lesson 4.6 will introduce complex numbers, which do give it one — but nothing in this lesson can.
Fill the middle
Testing the perfect square shape.
Fill in the blanks
w^2 - 18w + 81 = w^2 - 2(w)(9) + ___^2 = (w - ___)^2
Why: Eighty-one is 9 squared, and twice w times 9 is 18w, which matches the middle term exactly. Both checks pass, so the trinomial is a perfect square. If the middle term had been 17w or 19w the expression would not be a perfect square, and neither number is far from the truth — the test has to be exact.
Two truths and a lie
All three are about the special patterns.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The quantity x plus 5, squared, expands to x squared plus 10x plus 25, not x squared plus 25 — the middle term cannot simply be dropped. A sum of two squares with no middle term has no real factorisation at all, which is why B is false while both A and C are correct.
Prediction
Commit before reasoning.
Predict first
If the general factor-pair search finds the same answer, what do the special patterns actually buy you?
Correct: Speed, and recognition of the shape in later work.
\[ x^2 + 6x \;\to\; x^2 + 6x + 9 = (x+3)^2 \qquad \text{Lesson 4.7} \]
Why: The search would find 8 and 8 for y squared plus 16y plus 64 just as the pattern does. What the pattern adds is recognition: completing the square in Lesson 4.7 works by manufacturing a perfect square trinomial deliberately, and you cannot manufacture a shape you do not recognise. The pattern is an investment in the next three lessons, not just a shortcut in this one.
Section
Section 3
Concept
If the product of two expressions is zero, then at least one of them is zero. Applied to a factored quadratic equal to zero, this splits one quadratic equation into two linear equations, each of which can be solved by the methods of Lesson 1.3.
root of an equation — A solution of the equation. For a quadratic written in standard form, the roots are the values of x that make the expression zero.
\[ AB = 0 \;\Longrightarrow\; A = 0 \text{ or } B = 0 \]
The property is about zero specifically. If the product were 12 you could say nothing about the individual factors, which is precisely why the equation must be rewritten in standard form before any factoring is attempted.
Figure (svg): A product equal to zero branching into two separate linear equations
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 253-253 — Zero Product Property
Picture it
Example 3, after the factoring is done.
Figure (svg): A product equal to zero branching into two separate linear equations
The word between the two branches is or, not and. A root only has to satisfy one of the branches, which is why a quadratic can have two different solutions.
Worked example
Example 3. Standard form, factor, split, solve.
\[ \text{Find the roots of } x^2 + 3x - 28 = 0. \]
Confirm the equation is in standard form
Why: All terms are on one side with zero on the other, which the zero product property requires.
\[ x ^{2} + 3 x - 28 = 0 \]
Factor the left side
Why: You need mn equal to negative 28 and m plus n equal to 3; the pair negative 4 and 7 satisfies both.
\[ (x - 4) (x + 7) = 0 \]
Apply the zero product property
Why: The product is zero, so x minus 4 is zero or x plus 7 is zero.
\[ x - 4 = 0\text{ or } x + 7 = 0 \]
Solve each linear equation
Why: The first gives x equal to 4; the second gives x equal to negative 7.
\[ x = 4\text{ or } x = -7 \]
Figure (svg): The solution to Worked example find the roots shown as a ladder of expressions, one row per algebraic move
\[ x = 4 \quad \text{or} \quad x = -7 \]
Verify: substitute both roots into the original equation
Why: At x equal to 4: 16 plus 12 minus 28 is 0. At x equal to negative 7: 49 minus 21 minus 28 is 0. Both check, and note that the signs of the roots are the opposites of the signs inside the brackets — a useful sanity check that catches sign slips instantly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-254
Ranking
Solving a quadratic equation by factoring.
Put in order
Why: Standard form must come first, because factoring a side of an equation that is not equal to zero gains you nothing. The check comes last and is not optional here: factoring can go wrong silently, and substituting catches it in seconds.
Worked example
Guided Practice 8, the same four moves.
\[ \text{Solve } x^2 - x - 42 = 0. \]
State the two conditions
Why: You want mn equal to negative 42 and m plus n equal to negative 1.
\[ m n = -42, m + n = -1 \]
Search the pairs
Why: Forty-two is 6 times 7, and negative 7 with 6 gives a product of negative 42 and a sum of negative 1.
\[ m = -7, n = 6 \]
Factor and split
Why: The equation becomes the quantity x minus 7, times the quantity x plus 6, equals zero, so one bracket must vanish.
\[ (x - 7) (x + 6) = 0 \]
Solve both branches
Why: x minus 7 equal to zero gives 7; x plus 6 equal to zero gives negative 6.
\[ x = 7\text{ or } x = -6 \]
Figure (svg): The solution to Worked example a second equation shown as a ladder of expressions, one row per algebraic move
\[ x = 7 \quad \text{or} \quad x = -6 \]
Verify: check the sum and product of the roots
Why: The roots 7 and negative 6 sum to 1 and multiply to negative 42. For x squared plus bx plus c the roots always sum to negative b and multiply to c, and here negative b is 1 and c is negative 42, so both agree. Substituting confirms it: 49 minus 7 minus 42 is 0.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-254
Trap
\[ (x - 4)(x + 7) = 12 \]
Set each factor equal to 12
Why: The reasoning that worked for zero is copied across to another number.
\[ x - 4 = 12 \text{ or } x + 7 = 12 \;\Longrightarrow\; x = 16 \text{ or } x = 5 \quad \text{(wrong)} \]
Substituting x equal to 5 gives 1 times 12, which is 12 — a coincidence. Substituting x equal to 16 gives 12 times 23, which is 276, not 12.
\[ (x - 4)(x + 7) = 12 \;\Longrightarrow\; x^2 + 3x - 28 = 12 \;\Longrightarrow\; x^2 + 3x - 40 = 0 \]
Move everything to one side first, then factor afresh
Why: Only a product equal to zero licenses the split, so the equation must be rewritten before any factors are used.
\[ (x + 8)(x - 5) = 0 \;\Longrightarrow\; x = -8 \text{ or } x = 5 \]
Twelve has many factor pairs, so nothing follows from a product equalling 12. Zero is the only number whose factorisations all contain a zero, and that uniqueness is the whole content of the property.
Fill the middle
Example 3, at the branching step.
Fill in the blanks
(x - 4)(x + 7) = 0 \;\Longrightarrow\; x = 4 \;\text-7\; x = ___
Why: The second bracket vanishes when x plus 7 is zero, which is x equal to negative 7. The root is the opposite of the number inside the bracket, which is worth internalising: brackets of the form x plus something always give a negative root.
Prediction
Commit before reasoning.
Predict first
Solving by factoring gives two brackets. Must that always give two different roots?
Correct: No — the two brackets can be identical.
\[ x^2 + 12x + 36 = 0 \;\Longrightarrow\; (x+6)^2 = 0 \;\Longrightarrow\; x = -6 \text{ only} \]
Why: A perfect square trinomial such as x squared plus 12x plus 36 factors as the quantity x plus 6, times itself, so both branches give the same root, negative 6. That single value is called a repeated root, and geometrically it is the case from Lesson 4.2 where the vertex sits exactly on the horizontal axis so the parabola touches rather than crosses.
Comparison
Fill the blanks. The difference is why a new method was needed.
Comparison matrix
| Question | Linear equation | Quadratic equation |
|---|---|---|
| How many times does x appear? | once, to the first power | twice, as a square and a first power |
| Can you undo operations? | yes, step by step | no, x is in two places |
| How many solutions? | one, or none, or infinitely many | at most two |
| The method here | isolate x | factor, then split at zero |
The zero product property exists because isolation fails. It converts a problem you cannot do into two copies of a problem you can.
Section
Section 4
Concept
If a described situation changes two quantities by the same unknown amount, multiplying them produces a quadratic. Write the relationship in words, substitute the expressions, expand, put it in standard form, factor, and then reject any root the situation forbids.
\[ 2(600)(400) = (600 + x)(400 + x) \]
The final rejection step is part of the mathematics, not an afterthought. The equation has two roots; the situation has one answer, and deciding which root to keep is a modelling judgement the algebra cannot make for you.
Figure (svg): A rectangular field with an added strip of width x along two sides, doubling its area
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-254 — Use a quadratic equation as a model
Picture it
Example 4: a 600 by 400 metre field enlarged by the same amount x in each direction.
Figure (svg): A rectangular field with an added strip of width x along two sides, doubling its area
The added region is L-shaped, and its area contains an x squared corner. That corner is exactly why the model is quadratic and a linear equation would give the wrong answer.
Worked example
Example 4. A verbal model first, then algebra.
\[ \text{A } 600 \text{ m by } 400 \text{ m field is widened by } x \text{ in both directions to double its area. Find } x. \]
Write the verbal model
Why: New area equals new length times new width, where the new area is twice the old one.
\[ 2(600) (400) = (600 + x) (400 + x) \]
Expand the right side with FOIL
Why: 600 times 400 is 240,000; the cross terms are 600x and 400x, giving 1000x; and the last term is x squared.
\[ 480, 000 = 240, 000 + 1000 x + x ^{2} \]
Write in standard form
Why: Subtracting 480,000 from both sides leaves zero on the left.
\[ 0 = x ^{2} + 1000 x - 240, 000 \]
Factor and split
Why: You need a product of negative 240,000 and a sum of 1000; the pair negative 200 and 1200 works, so the brackets are x minus 200 and x plus 1200.
\[ 0 = (x - 200) (x + 1200) \]
Solve and reject
Why: The roots are 200 and negative 1200, but a length cannot be increased by a negative amount, so negative 1200 is rejected.
\[ x = 200 \]
Figure (svg): The solution to Worked example double the area of a field shown as a ladder of expressions, one row per algebraic move
\[ x = 200 \;\Longrightarrow\; 800 \text{ m by } 600 \text{ m} \]
Verify: compute the new area
Why: Eight hundred times 600 is 480,000 square metres, and the original field was 600 times 400, or 240,000 square metres. The new area is exactly twice the old one, so both the root and the rejection were right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-254
Ranking
Turning a described situation into an answer.
Put in order
Why: The verbal model comes first because it is the only step that uses the situation rather than the algebra; getting it wrong makes everything after it irrelevant. The last step returns to the situation, and it does two jobs — rejecting an impossible root and converting x back into whatever was actually asked for.
Worked example
Guided Practice 9. The structure repeats; only the numbers change.
\[ \text{Repeat Example 4 for a field measuring } 1000 \text{ m by } 300 \text{ m.} \]
Set up the same verbal model
Why: Twice the old area equals the product of the two increased dimensions.
\[ 2(1000) (300) = (1000 + x) (300 + x) \]
Expand and simplify
Why: The left side is 600,000; the right expands to 300,000 plus 1300x plus x squared.
\[ 600, 000 = 300, 000 + 1300 x + x ^{2} \]
Write in standard form and factor
Why: Subtracting gives x squared plus 1300x minus 300,000 equals zero; the pair 1500 and negative 200 has the right product and sum.
\[ (x + 1500) (x - 200) = 0 \]
Solve and reject the negative root
Why: The roots are negative 1500 and 200; only the positive one describes an enlargement.
\[ x = 200 \]
Answer the question that was asked
Why: The question asks for the new dimensions, not for x, so add 200 to each original dimension.
\[ 1200 m\text{ by } 500 m \]
Figure (svg): The solution to Worked example the same field, different dimensions shown as a ladder of expressions, one row per algebraic move
\[ 1200 \text{ m by } 500 \text{ m} \]
Verify: compare the two areas
Why: The new area is 1200 times 500, which is 600,000 square metres, and the original was 1000 times 300, or 300,000. The ratio is exactly 2. Interestingly x came out as 200 in both problems, which is a coincidence of the numbers chosen rather than a general rule.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-254
Error analysis
A student solves Example 4 correctly and then stops one line too early.
Annotate
On: \( x = 200, \text{ so the new dimensions are } 200 \text{ m by } 200 \text{ m} \)
In every modelling problem, write down what the variable means before solving, and reread the question after. A correct root answering the wrong question scores nothing.
Fill the middle
Example 4, at the FOIL step.
Fill in the blanks
(600 + x)(400 + x) = 2401000000 + ___x + x^2
Why: The two cross terms are 600 times x and 400 times x, which combine to 1000x. This is the m plus n of the general pattern showing up in a concrete setting: the middle coefficient is always the sum of the two constants, and the last term is always their product.
Prediction
Commit before reasoning.
Predict first
The equation gives x equal to 200 or x equal to negative 1200. Why is the second discarded?
Correct: It describes an impossible situation, though it solves the equation.
\[ (600 - 1200)(400 - 1200) = (-600)(-800) = 480{,}000 \quad \checkmark \]
Why: Substituting negative 1200 satisfies the equation exactly: 600 minus 1200 is negative 600, 400 minus 1200 is negative 800, and their product is 480,000. The algebra is perfectly happy. What fails is the interpretation — a field cannot have negative side lengths. The equation is a model of the situation, not the situation itself, so it can produce answers the situation cannot use.
Comparison
Fill the blanks. Only one of these is quadratic.
Comparison matrix
| Enlargement | New area | Equation type |
|---|---|---|
| Add x to the length only | (600 + x)(400) | linear |
| Add x to both dimensions | (600 + x)(400 + x) | quadratic |
| Which has an x^2 term? | only the second | from the corner square |
| Doubling the area needs | x = 400 in the first case | x = 200 in the second |
Adding to both dimensions is more efficient because the corner contributes area too. The x squared term in the model is that corner.
Section
Section 5
Concept
Factoring a quadratic function rewrites it in the intercept form of Lesson 4.2. The values that make each bracket vanish are the zeros of the function, the x-intercepts of its graph, and the roots of the corresponding equation — three descriptions of one pair of numbers.
zero of a function — A value of x for which the function's value is zero. If k is a zero of y equals a quadratic, then k is an x-intercept of the graph and a root of the equation formed by setting the quadratic to zero.
\[ y = x^2 - x - 12 = (x+3)(x-4) \;\Longrightarrow\; \text{zeros } -3, \, 4 \]
This closes the loop opened in Lesson 4.2. That lesson could read intercepts off a factored form but could not produce one; factoring produces it, so any factorable quadratic can now be graphed from its intercepts.
Figure (svg): A parabola crossing the horizontal axis at negative three and four, with the factored form beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 254-255 — Zeros of a function
Picture it
Example 5a: y equals x squared minus x minus 12.
Figure (svg): A parabola crossing the horizontal axis at negative three and four, with the factored form beside it
The graph passes through negative 3 and 4 on the horizontal axis, exactly the numbers the factoring produced. The picture is a check on the algebra, and the algebra is a shortcut past the picture.
Worked example
Example 5. Factor, then read.
\[ \text{Find the zeros of } y = x^2 - x - 12 \text{ and } y = x^2 + 12x + 36. \]
Factor the first function
Why: You need a product of negative 12 and a sum of negative 1; the pair 3 and negative 4 works.
\[ y = (x + 3) (x - 4) \]
Read its zeros
Why: Each bracket vanishes at the value that makes it zero, giving negative 3 and 4.
\[ \text{zeros } -3\text{ and } 4 \]
Factor the second function
Why: Thirty-six is 6 squared and 12x is twice x times 6, so this is a perfect square trinomial.
\[ y = (x + 6) (x + 6) \]
Read its zero
Why: Both brackets vanish at the same value, so the function has one zero rather than two.
\[ \text{zero } -6 \]
Figure (svg): The solution to Worked example find the zeros shown as a ladder of expressions, one row per algebraic move
\[ -3 \text{ and } 4; \qquad -6 \text{ (repeated)} \]
Verify: evaluate each function at its zeros
Why: For the first: at x equal to negative 3, 9 plus 3 minus 12 is 0; at x equal to 4, 16 minus 4 minus 12 is 0. For the second: at x equal to negative 6, 36 minus 72 plus 36 is 0. The graphs pass through those points, which is what the CHECK screens in the book display.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 255-255
Matching
Factor first, then ask what makes each bracket vanish.
Match the pairs
Why: Three of these have two distinct zeros and one has a repeated zero, which happens exactly when the trinomial is a perfect square. In every case the zeros are the opposites of the constants inside the brackets, and their product is the constant term of the original function.
Worked example
Guided Practice 10 to 12.
\[ \text{Find the zeros of } y = x^2 + 5x - 14, \; y = x^2 - 7x - 30, \; f(x) = x^2 - 10x + 25. \]
First function
Why: A product of negative 14 and a sum of 5 give the pair 7 and negative 2.
\[ (x + 7) (x - 2);\text{ zeros } -7, 2 \]
Second function
Why: A product of negative 30 and a sum of negative 7 give the pair negative 10 and 3.
\[ (x - 10) (x + 3);\text{ zeros } 10, -3 \]
Third function
Why: Twenty-five is 5 squared and 10x is twice x times 5, so it is a perfect square.
\[ (x - 5) ^{2};\text{ zero } 5 \]
Note the pattern across all three
Why: The zeros are always the opposites of the numbers inside the brackets, and a repeated bracket gives a single zero.
Figure (svg): The solution to Worked example three more functions shown as a ladder of expressions, one row per algebraic move
\[ \{-7, 2\}; \quad \{10, -3\}; \quad \{5\} \]
Verify: use the sum and product relations
Why: For the first, the zeros negative 7 and 2 sum to negative 5 and multiply to negative 14 — matching negative b and c. For the second, 10 and negative 3 sum to 7 and multiply to negative 30. For the third, 5 and 5 sum to 10 and multiply to 25. All three agree, so the factorisations are right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 255-255
Error analysis
A student factors correctly and then misreads the answer.
Annotate
On: \( y = x^2 - x - 12 = (x + 3)(x - 4), \text{ so the zeros are } 3 \text{ and } -4 \)
Same trap as the sign of h in Lesson 4.2, and the same fix: ask what makes the bracket zero rather than reading the sign off the page.
Two truths and a lie
All three are about zeros, roots and intercepts.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A zero of 4 corresponds to the factor x minus 4, not x plus 4, because x minus 4 is what vanishes at x equal to 4. The sign flips between a zero and its factor, and B states the relationship without that flip. Statements A and C are two different ways of saying that zeros, x-intercepts and roots are the same numbers seen from three angles.
Fill the middle
Guided Practice 11.
Fill in the blanks
y = x^2 - 7x - 30 = (x - 10)(x + 3) \;\Longrightarrow\; \text-3 10 \text___ ___
Why: The second bracket vanishes at x equal to negative 3. Checking with the original function: 9 plus 21 minus 30 is zero, so negative 3 really is a zero. The two zeros multiply to negative 30, which is the constant term, and sum to 7, which is the opposite of the middle coefficient.
Prediction
Commit before reasoning.
Predict first
Lesson 4.2 could graph from intercept form but not produce it. What has changed?
Correct: Factoring produces intercept form from standard form.
\[ y = (x+3)(x-4) \;\Longrightarrow\; \text{axis } x = \tfrac{-3+4}{2} = \tfrac{1}{2} \;\Longrightarrow\; \text{vertex } \left(\tfrac{1}{2}, -\tfrac{49}{4}\right) \]
Why: Given y equals x squared minus x minus 12, factoring turns it into the quantity x plus 3 times the quantity x minus 4, which is intercept form with p equal to negative 3 and q equal to 4. From there Lesson 4.2's whole method applies: the intercepts are read off, the axis is their average — one half — and substituting gives the vertex. Factoring is the bridge between the two forms.
Comparison
Fill the blanks. One pair of numbers, four vocabularies.
Comparison matrix
| Language | Object | For y = x^2 - x - 12 |
|---|---|---|
| Algebra of expressions | the factors | (x + 3) and (x - 4) |
| Algebra of equations | the roots | -3 and 4 |
| Functions | the zeros | -3 and 4 |
| Graphs | the x-intercepts | (-3, 0) and (4, 0) |
The words differ because they come from different chapters of mathematics, but a problem phrased in any one of them can be answered by the method of this lesson.
Pattern
One routine, whether you are asked to factor, to solve, or to find zeros.
Step two before step three is the only ordering that saves time. Step five's check is the only thing that catches a wrong factorisation.
OpenStax Algebra and Trigonometry 2e, §1.5 Factoring Polynomials §1.5
Check
Factoring. Product first, then sum.
Check your understanding
Factor x^2 - 3x - 18.
Answer: A
Why: Negative 6 times 3 is negative 18 and negative 6 plus 3 is negative 3, so both conditions hold. Expanding confirms it.
Check
The zero product property. Watch the signs.
Check your understanding
What are the roots of x^2 + 3x - 28 = 0?
Answer: A
Why: The expression factors as (x - 4)(x + 7), so the roots are 4 and -7. Substituting: 16 + 12 - 28 = 0 and 49 - 21 - 28 = 0.
Check
Zeros of a function. Ask what makes each bracket vanish.
Check your understanding
What are the zeros of y = x^2 + 5x - 14?
Answer: A
Why: The function factors as (x + 7)(x - 2), which vanishes at -7 and at 2. Checking: 49 - 35 - 14 = 0.
Real world
A rectangular garden 12 feet by 16 feet is to be surrounded by a uniform gravel path. The gardener has enough gravel for 288 square feet of path.
Discussion prompt
How wide should the path be? Set up the equation, solve it, and say which root you keep and why.
Hint: The outer rectangle is wider than the garden by twice the path width, because the path runs along both sides.
Answer:
\[ (12 + 2w)(16 + 2w) - (12)(16) = 288 \]
\[ 192 + 56w + 4w^2 - 192 = 288 \;\Longrightarrow\; 4w^2 + 56w - 288 = 0 \;\Longrightarrow\; w^2 + 14w - 72 = 0 \]
\[ (w + 18)(w - 4) = 0 \;\Longrightarrow\; w = -18 \text{ or } w = 4 \]
The path is 4 feet wide. The root negative 18 solves the equation but describes a negative width, so it is rejected exactly as in Example 4.
Two details are worth noticing. The path adds twice its width to each dimension because it runs along both sides — a very common setup error. And dividing through by the common factor of 4 before factoring made the search far easier, which is the first move of Lesson 4.4.
Commit first
Answer, then rate your confidence honestly.
Predict first
If (x - 2)(x + 5) = 0, must x be either 2 or -5?
Correct: Yes — one factor has to be zero.
\[ AB = 0 \text{ and } A \neq 0 \;\Longrightarrow\; B = \tfrac{0}{A} = 0 \]
Why: This is the whole content of the zero product property, and it depends on a fact about the real numbers: there are no zero divisors, meaning two non-zero numbers can never multiply to zero. The property fails in some other number systems, but it holds throughout the real and complex numbers, which is why this method is reliable for everything in this course.
Explain it
They can solve 3x plus 5 equals 20 but have never seen an x squared.
Discussion prompt
In four sentences or fewer, explain why you cannot solve x squared plus 3x minus 28 equals zero by undoing operations, and what factoring does instead.
Hint: Point at how many times x appears.
Answer:
In a linear equation x appears once, so you can peel operations off it one at a time until it is alone. Here x appears twice, in the square and in the middle term, so there is nothing to isolate — moving one x changes the other.
Factoring rewrites the left side as two brackets multiplied together. Since their product is zero, one of them must be zero, and each bracket is a linear equation with x appearing exactly once — so you are back to a problem you can already do, twice over.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For signs, read c first for whether they agree and b second for which sign wins. For perfect squares, check that the middle term is exactly twice the product of the two roots. For word problems, always write the verbal model in words before any algebra. For roots, ask what makes each bracket zero rather than reading the number off the page. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the single expression x squared minus x minus 12 and follow it across the page in four directions. Top left: factor it, showing the factor-pair search with every pair of factors of negative 12 and its sum, and circle the pair that works. Top right: set it equal to zero and solve, showing the split into two linear equations. Bottom left: treat it as a function, sketch its parabola, and mark both x-intercepts, the axis of symmetry at their average, and the vertex you get by substituting. Bottom right: write the four vocabulary words — factors, roots, zeros, x-intercepts — and beside each write which of your three panels produced it. Finally, in a margin, write one trinomial that does NOT factor over the integers and show the exhausted list of pairs that proves it.
If your factor-pair list has fewer than six rows, you have missed the sign variations: every pair of positive factors of 12 can be signed two ways to give a negative product.
Recap
Five things, and the third one is the reason for the other four.
| If you see | Then |
|---|---|
| x^2 + bx + c to factor | Find m and n with mn = c and m + n = b |
| Two squares with a minus between | Difference of squares: (a + b)(a - b) |
| A middle term twice the product of the roots | Perfect square: (a +- b)^2 |
| A product equal to zero | Set each factor to zero |
| A product equal to anything else | Move it all to one side first |
| A negative root in a length problem | Reject it and keep the positive one |
Every trinomial here had a leading coefficient of 1, which is what made the search a search over pairs. Lesson 4.4 removes that restriction, and the extra coefficient changes the method more than you would expect.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.3 Solve x^2 + bx + c = 0 by Factoring §4.3, pp. 252-257 — everything on these slides traces back here
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