The four-integer search for factoring ax squared plus bx plus c, the special patterns when a and c are perfect squares, pulling out a common monomial first, solving equations that must be rewritten in standard form, and maximising revenue through the average of the zeros.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Solve ax^2 + bx + c = 0 by Factoring
Objectives
Five outcomes. The last one is the payoff for all the factoring.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-263 — the lesson these objectives are drawn from
Warm-up
Lesson 4.3 factored trinomials whose leading coefficient was 1. Removing that restriction changes the search, not the idea.
Discussion prompt
In Lesson 4.3 you factored x squared minus 9x plus 20 by finding two numbers with product 20 and sum negative 9. Multiply out the general product of the quantity kx plus m and the quantity lx plus n. What do k, l, m and n now have to satisfy?
Hint: Use FOIL and collect the two middle terms.
Answer:
\[ (kx + m)(lx + n) = klx^2 + (kn + lm)x + mn \]
So k times l must equal a, m times n must equal c, and the middle coefficient is kn plus lm — a mixed expression rather than a plain sum. That mixing is the entire difficulty of this lesson: the four unknowns cannot be chosen independently.
Concept
Factoring a quadratic whose leading coefficient is not 1 uses exactly the reasoning of Lesson 4.3, but with four integers to find instead of two. Because the middle coefficient mixes them, the order in which they are placed now matters, so every shortcut that shrinks the candidate list is worth taking first.
factored completely — An expression is factored completely when no factor other than a number can be factored further. That normally means pulling out a common monomial before anything else.
\[ ax^2 + bx + c = (kx + m)(lx + n) = klx^2 + (kn + lm)x + mn \]
Two shortcuts do most of the work: reading the signs of b and c to decide the signs of m and n, and pulling out any common monomial factor so the numbers being factored are as small as possible.
Figure (svg): Two columns contrasting factoring when the leading coefficient is one against when it is not
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-260
Section
Section 1
Concept
To factor a quadratic whose leading coefficient is not 1, list the ways of factoring a into k times l and the ways of factoring c into m times n, then test each arrangement by expanding. Take k and l positive by convention, so only m and n carry signs.
\[ ax^2 + bx + c = (kx + m)(lx + n), \quad kl = a, \; mn = c, \; kn + lm = b \]
Testing means expanding and comparing the middle term, since the outer and inner terms are automatically right once k, l, m and n have the correct products.
Figure (svg): A table of four candidate factorisations of five x squared minus seventeen x plus six, each expanded, with the matching one highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-259 — Factor ax^2 + bx + c where c > 0
Picture it
The sign of c and then the sign of b cut the list before any expanding.
Figure (svg): A two-branch decision showing how the sign of the constant and the middle coefficient narrow the candidate list
For 5x squared minus 17x plus 6 the constant is positive and the middle coefficient negative, so both m and n are negative — which removes half the candidates immediately.
Worked example
Example 1. Signs first, then four rows.
\[ \text{Factor } 5x^2 - 17x + 6. \]
List the factorisations of a and of c
Why: Five is prime, so k and l must be 5 and 1. Six factors as 6 times 1 or 3 times 2.
\[ k, l = 5, 1 \]
Decide the signs of m and n
Why: The constant is positive so m and n agree in sign, and the middle coefficient is negative, so both are negative.
Test the arrangements
Why: Negative 6 with negative 1 gives a middle term of negative 11; negative 1 with negative 6 gives negative 31; negative 3 with negative 2 gives negative 13.
\[ -11, -31, -13 \]
Find the one that matches
Why: Negative 2 with negative 3 gives a middle term of negative 17, which is what is wanted.
\[ (5 x - 2) (x - 3) \]
Figure (svg): The solution to Worked example a positive constant shown as a ladder of expressions, one row per algebraic move
\[ 5x^2 - 17x + 6 = (5x - 2)(x - 3) \]
Verify: expand and compare all three terms
Why: Five x times x is 5x squared; the outer and inner terms are negative 15x and negative 2x, totalling negative 17x; and negative 2 times negative 3 is 6. All three terms match, so the factorisation is exact. Note that rows three and four use the same pair of numbers in different positions and give different answers.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-259
Matching
Check the middle term by expanding, not by adding.
Match the pairs
Why: In each case the leading coefficient is prime, so k and l are forced and only the placement of m and n is in question. Notice how the last two use the same numbers, 1 and 3, with different signs and different leading coefficients, giving quite different middle terms.
Worked example
Example 2. Opposite signs, so both orders and both assignments must be tried.
\[ \text{Factor } 3x^2 + 20x - 7. \]
List the factorisations
Why: Three is prime so k and l are 3 and 1; seven is prime so m and n are 7 and 1 in some order and with opposite signs.
\[ k, l = 3, 1 \]
Note what the negative constant forces
Why: The product mn is negative, so one of m and n is positive and the other negative — four arrangements in all.
Test each arrangement
Why: Seven with negative 1 gives a middle term of 4; negative 1 with 7 gives 20; negative 7 with 1 gives negative 4; 1 with negative 7 gives negative 20.
\[ 4, 20, -4, -20 \]
Select the match
Why: The second arrangement gives the middle term 20, so m is negative 1 and n is 7.
\[ (3 x - 1) (x + 7) \]
Figure (svg): The solution to Worked example a negative constant shown as a ladder of expressions, one row per algebraic move
\[ 3x^2 + 20x - 7 = (3x - 1)(x + 7) \]
Verify: expand the middle term only
Why: The outer term is 3x times 7, or 21x, and the inner term is negative 1 times x, or negative x; together that is 20x, as required. The four candidate middle terms came in plus-and-minus pairs, which is a useful check that no arrangement was missed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-259
Trap
\[ 5x^2 - 17x + 6 = (5x + m)(x + n) \]
Look for two numbers with product 6 and sum negative 17
Why: The rule from Lesson 4.3 is carried across unchanged.
\[ \text{no such pair exists, so the expression cannot be factored} \quad \text{(wrong)} \]
It does factor. The pair negative 2 and negative 3 has product 6 and sum negative 5, not negative 17 — but the sum is not what the middle term equals here.
\[ (5x + m)(x + n) = 5x^2 + (5n + m)x + mn \]
Compare the middle coefficient with kn plus lm
Why: With k equal to 5 and l equal to 1, the middle coefficient is 5n plus m, not m plus n.
\[ m = -2, \; n = -3 \;\Longrightarrow\; 5(-3) + (-2) = -17 \quad \checkmark \]
The leading coefficient multiplies one of the two constants when the middle term is formed. That is why swapping m and n changes the answer, and why the Lesson 4.3 shortcut stops applying the moment a is not 1.
Sorting
Read the sign of c first, then the sign of b.
Sort into buckets
Sort each trinomial by the signs of m and n.
The same two-glance rule as Lesson 4.3. It does even more work here, because the candidate list is longer to begin with.
Fill the middle
Checking a candidate by expanding.
Fill in the blanks
(5x - 2)(x - 3) = 5x^2 - 15x - 2x + 6 = 5x^2 - 17x + 6
Why: The outer term is 5x times negative 3, or negative 15x, and the inner term is negative 2 times x, or negative 2x; together they give negative 17x. This is the kn plus lm of the general pattern, with the 5 multiplying the negative 3 rather than the two constants simply adding.
Prediction
Commit before reasoning.
Predict first
In Lesson 4.3, (x - 2)(x - 3) and (x - 3)(x - 2) were the same. Why are (5x - 2)(x - 3) and (5x - 3)(x - 2) different?
Correct: The 5 attaches to only one of the constants, so swapping changes the middle term.
\[ (5x-2)(x-3) = 5x^2 - 17x + 6 \qquad (5x-3)(x-2) = 5x^2 - 13x + 6 \]
Why: Multiplication is still commutative — the order of the two brackets is irrelevant. What changes is which constant gets multiplied by the 5. In the first, 5 multiplies negative 3 to give negative 15x; in the second, 5 multiplies negative 2 to give negative 10x. The middle terms come out as negative 17x and negative 13x, so the two products are genuinely different expressions.
Section
Section 2
Concept
If both the leading coefficient and the constant are perfect squares, test the difference-of-squares and perfect-square-trinomial patterns before starting any search. The only change from Lesson 4.3 is that the thing being squared is now a multiple of x rather than x itself.
\[ 9x^2 - 64 = (3x)^2 - 8^2 = (3x + 8)(3x - 8) \]
For a perfect square trinomial the middle term must still be exactly twice the product of the two square roots — so for 4y squared plus 20y plus 25 the test is that 20y equals twice 2y times 5, which it does.
Figure (svg): The special patterns applied when the leading coefficient is itself a perfect square
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260 — Factor with special patterns
Picture it
Example 3, all three parts.
Figure (svg): The special patterns applied when the leading coefficient is itself a perfect square
In each case the square root of the leading coefficient becomes the coefficient of x inside the bracket. Recognising that saves the entire four-integer search.
Worked example
Example 3. Identify the two squares, then check the middle term.
\[ \text{Factor } 9x^2 - 64, \; 4y^2 + 20y + 25, \; 36w^2 - 12w + 1. \]
First: two terms, both squares, subtracted
Why: Nine x squared is the square of 3x and 64 is 8 squared, so the difference of squares pattern applies.
\[ (3 x) ^{2} - 8 ^{2} = (3 x + 8) (3 x - 8) \]
Second: check twice the product
Why: Four y squared is the square of 2y and 25 is 5 squared; twice 2y times 5 is 20y, matching the middle term.
\[ (2 y) ^{2} + 2(2 y) (5) + 5 ^{2} = (2 y + 5) ^{2} \]
Third: the same test with a minus
Why: Thirty-six w squared is the square of 6w and 1 is 1 squared; twice 6w times 1 is 12w, and the sign is negative.
\[ (6 w) ^{2} - 2(6 w) (1) + 1 ^{2} = (6 w - 1) ^{2} \]
State all three
Why: Each was written down from the shape; no candidate arrangements were tested.
Figure (svg): The solution to Worked example factor with special patterns shown as a ladder of expressions, one row per algebraic move
\[ (3x+8)(3x-8), \quad (2y+5)^2, \quad (6w-1)^2 \]
Verify: expand the second one
Why: The quantity 2y plus 5, squared, is 4y squared plus 10y plus 10y plus 25, which is 4y squared plus 20y plus 25. The middle term appears twice, which is exactly why the test asks for twice the product of the roots rather than the product itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260
Sorting
Check both outer terms for being perfect squares first.
Sort into buckets
Sort each expression by the pattern it fits.
Two seconds of looking decides between writing the answer down and testing four arrangements.
Worked example
Guided Practice 7 to 12. One of them needs a monomial pulled out first.
\[ \text{Factor } 16x^2 - 1, \; 9y^2 + 12y + 4, \; 4r^2 - 28r + 49, \; 25s^2 - 80s + 64, \; 49z^2 + 42z + 9, \; 36n^2 - 9. \]
The two differences of squares
Why: Sixteen x squared minus 1 is the square of 4x minus 1 squared. Thirty-six n squared minus 9 has a common factor of 9, giving 9 times the quantity 4n squared minus 1.
\[ (4 x + 1) (4 x - 1); 9(2 n + 1) (2 n - 1) \]
The first two perfect squares
Why: Nine y squared plus 12y plus 4: twice 3y times 2 is 12y, so it is the quantity 3y plus 2, squared. Four r squared minus 28r plus 49: twice 2r times 7 is 28r.
\[ (3 y + 2) ^{2}; (2 r - 7) ^{2} \]
The last two perfect squares
Why: Twenty-five s squared minus 80s plus 64: twice 5s times 8 is 80s. Forty-nine z squared plus 42z plus 9: twice 7z times 3 is 42z.
\[ (5 s - 8) ^{2}; (7 z + 3) ^{2} \]
Note the exception
Why: Only the last of the six needed a common factor removed first, and that is a hint of the next idea.
\[ 9(2 n + 1) (2 n - 1) \]
Figure (svg): The solution to Worked example six more patterns shown as a ladder of expressions, one row per algebraic move
\[ (4x+1)(4x-1); \; (3y+2)^2; \; (2r-7)^2; \; (5s-8)^2; \; (7z+3)^2; \; 9(2n+1)(2n-1) \]
Verify: test one middle term that nearly matches
Why: For 25s squared minus 80s plus 64 the roots are 5s and 8, and twice their product is 80s — exactly the middle term. Had it been 79s or 81s the expression would not be a perfect square at all, so the test has to be an exact match rather than an approximate one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260
Error analysis
A student applies the difference of squares pattern to nine x squared minus 64.
Annotate
On: \( 9x^2 - 64 = (9x + 8)(9x - 8) \)
Take the square root of the whole term, coefficient included. Expanding the answer catches this instantly, which is why the check is never optional.
Fill the middle
Testing the perfect square shape.
Fill in the blanks
49z^2 + 42z + 9 = (7z)^2 + 2(7z)(3) + ___^2 = (7z + ___)^2
Why: Nine is 3 squared, and twice 7z times 3 is 42z, which matches the middle term exactly. Both square roots had to be taken correctly for this to work: 7z from 49z squared and 3 from 9. Getting either root wrong makes the middle-term test fail, which is the built-in safeguard.
Two truths and a lie
All three are about the patterns when a is not 1.
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. Being a square at each end is necessary but not sufficient: the middle term also has to be exactly twice the product of the roots. Four y squared plus 21y plus 25 has square outer terms but the middle term is wrong by 1, and it does not factor at all. The middle-term test is the part that does the real work.
Prediction
Commit before reasoning.
Predict first
Lesson 4.3's pattern was a squared minus b squared. What plays the role of a in 9x squared minus 64?
Correct: The expression 3x.
\[ a^2 - b^2 \text{ with } a = 3x, \; b = 8 \;\Longrightarrow\; (3x+8)(3x-8) \]
Why: The pattern is stated in terms of two quantities whose squares appear, and 9x squared is the square of the whole expression 3x, not of 3 or of x alone. This is the single most useful habit in the lesson: the letters in a pattern stand for whole expressions, not just for single symbols, which is why the same three patterns keep working as the expressions get more complicated.
Section
Section 3
Concept
Before attempting any factorisation, check whether every term shares a common monomial factor. Pulling it out leaves smaller coefficients to factor, which shortens the search dramatically, and it is required anyway for the expression to be factored completely.
\[ 6q^2 - 14q + 8 = 2(3q^2 - 7q + 4) = 2(3q - 4)(q - 1) \]
The common factor may include a variable: negative 5z squared plus 20z becomes negative 5z times the quantity z minus 4, and the whole of the remaining bracket is then linear.
Figure (svg): The same expression factored with and without pulling out the common monomial first
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260 — Factor out monomials first
Picture it
Example 4b: 6q squared minus 14q plus 8, with and without the common factor removed.
Figure (svg): The same expression factored with and without pulling out the common monomial first
The right-hand route tests four arrangements; the left-hand route tests more than twelve. Both reach the same answer, but only one of them reliably.
Worked example
Example 4, parts a and b.
\[ \text{Factor } 5x^2 - 45 \text{ and } 6q^2 - 14q + 8. \]
Find the common factor in the first
Why: Both terms are divisible by 5, leaving x squared minus 9.
\[ 5(x ^{2} - 9) \]
Factor what remains
Why: X squared minus 9 is a difference of two squares.
\[ 5(x + 3) (x - 3) \]
Find the common factor in the second
Why: Six, 14 and 8 are all even, so 2 comes out, leaving 3q squared minus 7q plus 4.
\[ 2(3 q ^{2} - 7 q + 4) \]
Factor the remaining trinomial
Why: Three is prime and 4 factors as 4 times 1 or 2 times 2; the constant is positive and the middle coefficient negative, so both are negative, and negative 4 with negative 1 works.
\[ 2(3 q - 4) (q - 1) \]
Figure (svg): The solution to Worked example pull the monomial out first shown as a ladder of expressions, one row per algebraic move
\[ 5(x+3)(x-3) \quad \text{and} \quad 2(3q-4)(q-1) \]
Verify: expand the second one fully
Why: The quantity 3q minus 4, times the quantity q minus 1, is 3q squared minus 3q minus 4q plus 4, or 3q squared minus 7q plus 4; multiplying by 2 gives 6q squared minus 14q plus 8. The original is recovered, and note that without pulling out the 2 first, k and l could have been 6 and 1, 3 and 2, or 2 and 3 — three times as many rows to test.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260
Ranking
The order that minimises work.
Put in order
Why: The monomial comes out first because it shrinks the numbers every later step depends on. The pattern check comes before the search because it can end the problem outright. Signs are checked before expanding because eliminating a candidate on sight is free, whereas expanding one is not.
Worked example
Example 4 parts c and d, plus Guided Practice 17.
\[ \text{Factor } -5z^2 + 20z, \; 12p^2 - 21p + 3, \; -16n^2 + 12n. \]
First: a common factor of negative 5z
Why: Both terms contain z, and taking the negative out leaves a positive leading coefficient.
\[ -5 z(z - 4) \]
Second: a common factor of 3
Why: Twelve, 21 and 3 are all divisible by 3, leaving 4p squared minus 7p plus 1.
\[ 3(4 p ^{2} - 7 p + 1) \]
Check whether the second bracket factors further
Why: For 4p squared minus 7p plus 1 the only candidates are 4 and 1 or 2 and 2 for k and l, with m and n both negative 1; the middle terms come out as negative 5p and negative 4p, neither of which is negative 7p.
\[ 3(4 p ^{2} - 7 p + 1)\text{ is complete} \]
Third: a common factor of negative 4n
Why: Both terms contain n, and taking the negative out leaves 4n minus 3.
\[ -4 n(4 n - 3) \]
Figure (svg): The solution to Worked example monomials with variables shown as a ladder of expressions, one row per algebraic move
\[ -5z(z-4), \quad 3(4p^2 - 7p + 1), \quad -4n(4n-3) \]
Verify: expand the first and third
Why: Negative 5z times the quantity z minus 4 is negative 5z squared plus 20z, and negative 4n times the quantity 4n minus 3 is negative 16n squared plus 12n. Both match. The middle expression shows something important: pulling out a monomial does not guarantee that what remains will factor, and stopping there is the correct and complete answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 260-260
Error analysis
A student pulls a common factor out of twelve x squared minus 28x minus 24.
Annotate
On: \( 12x^2 - 28x - 24 = 4(3x^2 - 28x - 24) \)
The textbook flags this exact slip in an Avoid Errors note. Multiplying the bracket back by the monomial catches it in one line, every time.
Fill the middle
Guided Practice 14.
Fill in the blanks
8t^2 + 38t - 10 = 2(4t^2 + 19t - 5) = 2(4t - 1)(t + 5)
Why: With 4t and negative 1 in place, the outer term is 4t times n and the inner term is negative t; for the total to be 19t you need 4n minus 1 to equal 19, so n is 5. Checking the constant: negative 1 times 5 is negative 5, as required. Pulling the 2 out first is what made 5 the only constant to consider rather than 10.
Comparison
Fill the blanks. Same expression, two amounts of work.
Comparison matrix
| Question | Straight into the search | Monomial out first |
|---|---|---|
| The expression | 6q^2 - 14q + 8 | 2(3q^2 - 7q + 4) |
| Factorisations of a | 6 and 1, or 3 and 2 | 3 and 1 only |
| Factorisations of c | 8 and 1, or 4 and 2 | 4 and 1, or 2 and 2 |
| Rows to test | twelve or more | four |
The candidate count is roughly the product of the two lists, so shrinking both numbers shrinks the work far faster than it looks.
Prediction
Commit before reasoning.
Predict first
Factoring 6q squared minus 14q plus 8 gives 2 times two brackets. Does the 2 contribute a zero?
Correct: No — a non-zero constant factor can never be zero.
\[ 2(3q-4)(q-1) = 0 \;\Longrightarrow\; q = \tfrac{4}{3} \text{ or } q = 1 \]
Why: The zero product property asks which factors can equal zero, and the constant 2 never can, so it contributes nothing to the solution set. That is exactly why Example 5b is allowed to divide both sides by 5: dividing out a non-zero constant changes the expression but not its zeros. It does change the graph, though — the parabola is stretched vertically, so the leading coefficient still matters for everything except the roots.
Section
Section 4
Concept
The zero product property still requires one side to be zero, so any equation with terms on both sides must be rewritten first. Once in standard form, divide out any common numerical factor, factor the rest, and set each bracket to zero.
\[ 5p^2 - 16p + 15 = 4p - 5 \;\Longrightarrow\; 5p^2 - 20p + 20 = 0 \;\Longrightarrow\; p^2 - 4p + 4 = 0 \]
Dividing out a common factor is optional but almost always worth it, because it shrinks both coefficients the factoring search depends on and it cannot change the solutions.
Figure (svg): A five by four quilt surrounded by a border of uniform width x, with the border area labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 261-261 — Solve quadratic equations
Picture it
Example 6: 10 square feet of fabric added around a 5 by 4 quilt.
Figure (svg): A five by four quilt surrounded by a border of uniform width x, with the border area labelled
The border adds twice its width to each dimension, giving 5 plus 2x by 4 plus 2x. Subtracting the quilt's own area leaves the border area, and the equation follows.
Worked example
Example 5. The second one needs rearranging before anything else.
\[ \text{Solve } 3x^2 + 10x - 8 = 0 \text{ and } 5p^2 - 16p + 15 = 4p - 5. \]
The first is already in standard form
Why: Factor it: three is prime, negative 8 gives opposite signs, and the pair negative 2 with 4 gives a middle term of 10x.
\[ (3 x - 2) (x + 4) = 0 \]
Split and solve
Why: Three x minus 2 equal to zero gives two thirds; x plus 4 equal to zero gives negative 4.
\[ x = \frac{2}{3}\text{ or } x = -4 \]
Rewrite the second in standard form
Why: Subtract 4p and add 5 to both sides, giving 5p squared minus 20p plus 20 equals zero.
\[ 5 p ^{2} - 20 p + 20 = 0 \]
Divide by the common factor and factor
Why: Every term is divisible by 5, leaving p squared minus 4p plus 4, which is a perfect square trinomial.
\[ (p - 2) ^{2} = 0 \]
Solve the repeated factor
Why: If the square of an expression is zero, the expression itself is zero, so p minus 2 is zero.
\[ p = 2 \]
Figure (svg): The solution to Worked example two equations shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{2}{3} \text{ or } -4; \qquad p = 2 \]
Verify: substitute one root from each
Why: For the first, at x equal to two thirds: 3 times four ninths is four thirds, plus twenty thirds is 8, minus 8 is 0. For the second, at p equal to 2 the left side is 20 minus 32 plus 15, or 3, and the right side is 8 minus 5, also 3. Both check, and the second has only one root because its factor is repeated.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 261-261
Ranking
Solving when the leading coefficient is not 1.
Put in order
Why: Standard form is first because nothing else is valid without it. Dividing out comes before factoring because it shrinks the search. The final check matters more here than in Lesson 4.3: with four integers to place, a wrong arrangement can look plausible, and substitution is the only thing that catches it.
Worked example
Example 6. A verbal model turns the fabric into an equation.
\[ \text{A } 5 \text{ ft by } 4 \text{ ft quilt gets a uniform border using } 10 \text{ sq ft of fabric. Find its width.} \]
Write the verbal model
Why: The area of the border equals the area of quilt and border together, minus the area of the quilt.
\[ 10 = (5 + 2 x) (4 + 2 x) - (5) (4) \]
Expand with FOIL
Why: The product is 20 plus 10x plus 8x plus 4x squared; subtracting 20 leaves 18x plus 4x squared.
\[ 10 = 20 + 18 x + 4 x ^{2} - 20 \]
Write in standard form and divide by 2
Why: Moving the 10 across gives 4x squared plus 18x minus 10 equals zero, and every term is even.
\[ 2 x ^{2} + 9 x - 5 = 0 \]
Factor and split
Why: Two is prime and negative 5 gives opposite signs; the pair negative 1 with 5 gives a middle term of 9x.
\[ (2 x - 1) (x + 5) = 0 \]
Solve and reject
Why: The roots are one half and negative 5, and a width cannot be negative.
\[ x = \frac{1}{2} \]
Figure (svg): The solution to Worked example the quilt border shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{1}{2} \text{ ft} = 6 \text{ inches} \]
Verify: compute the border area directly
Why: With x equal to one half the outer rectangle is 6 by 5, an area of 30 square feet, and the quilt itself is 20 square feet, so the border uses 10 square feet — exactly the fabric available. Dividing by 2 before factoring turned a search on 4 and 10 into one on 2 and 5, which is why the right arrangement appeared immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 261-261
Trap
\[ 5p^2 - 16p + 15 = 4p - 5 \]
Factor the left side as it stands
Why: The left side does factor, as the quantity 5p minus 3 times the quantity p minus 5.
\[ (5p - 3)(p - 5) = 4p - 5 \;\Longrightarrow\; 5p - 3 = 0 \text{ or } p - 5 = 0 \quad \text{(wrong)} \]
Neither three fifths nor 5 satisfies the original equation. At p equal to 5 the left side is 60 and the right side is 15.
\[ 5p^2 - 16p + 15 - 4p + 5 = 0 \;\Longrightarrow\; 5p^2 - 20p + 20 = 0 \]
Collect everything on one side first, then factor what results
Why: The factorisation of the rearranged expression has nothing to do with the factorisation of the original left side.
\[ 5(p-2)^2 = 0 \;\Longrightarrow\; p = 2 \]
Factoring is only useful because of the zero product property, and that property says nothing at all about a product equal to 4p minus 5. Standard form is what licenses the split, so it comes first without exception.
Fill the middle
Example 5a, at the split.
Fill in the blanks
(3x - 2)(x + 4) = 0 \;\Longrightarrow\; x = 2/3 \;\text___\; x = -4
Why: The first bracket vanishes when 3x equals 2, so x is two thirds. This is the visible difference from Lesson 4.3: a leading coefficient inside a bracket produces a fractional root, and fractional roots are entirely normal here rather than a sign of a mistake.
Prediction
Commit before reasoning.
Predict first
Example 5b divides 5p squared minus 20p plus 20 equals zero by 5. Why is that safe?
Correct: Dividing both sides by a non-zero number preserves the solutions.
\[ 5p^2 - 20p + 20 = 0 \;\Longleftrightarrow\; p^2 - 4p + 4 = 0 \]
Why: Both sides are divided, and the right side is zero, so zero divided by 5 is still zero. Any p satisfying one equation satisfies the other. What must not be divided out is an expression that could itself be zero: dividing by p, for example, would silently discard the root p equal to zero. Constants are safe; variable factors are not.
Comparison
Fill the blanks. One needs rearranging, the other does not.
Comparison matrix
| Step | 3x^2 + 10x - 8 = 0 | 5p^2 - 16p + 15 = 4p - 5 |
|---|---|---|
| Already standard form? | yes | no, terms on both sides |
| After rearranging | unchanged | 5p^2 - 20p + 20 = 0 |
| Common factor to divide out | none | 5 |
| Number of roots | two: 2/3 and -4 | one: p = 2, repeated |
The second equation looks harder and ends up easier, because the rearrangement produced a perfect square. Doing the setup properly is what revealed that.
Section
Section 5
Concept
To find the maximum or minimum of a quadratic function, factor it into intercept form. The vertex lies on the axis of symmetry, which is halfway between the two zeros, so the extreme value occurs at their average — and evaluating the function there gives the value itself.
\[ y = a(x - p)(x - q) \;\Longrightarrow\; x = \frac{p + q}{2} \]
This is Lesson 4.2's intercept form doing real work. The zeros of a revenue model are usually meaningless as prices, but their average is exactly the price that matters.
Figure (svg): The revenue parabola with its two zeros marked and the maximum at their average
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 262-262 — Factoring and zeros
Picture it
Example 7: a magazine with 28,000 subscribers at 10 dollars, losing 2000 for each dollar of increase.
Figure (svg): The revenue parabola with its two zeros marked and the maximum at their average
The zeros at 14 and negative 10 are the price increases that would drive revenue to nothing. Their average, 2, is where revenue peaks.
Worked example
Example 7, in the book's four steps.
\[ \text{At } 10 \text{ dollars there are } 28{,}000 \text{ subscribers, losing } 2000 \text{ per dollar rise. Maximise revenue.} \]
Define the variables
Why: Let x be the price increase in dollars and R of x the annual revenue in dollars.
\[ x =\text{ price increase} \]
Write the verbal model
Why: Annual revenue equals the number of subscribers times the subscription price.
\[ R(x) = (28, 000 - 2000 x) (10 + x) \]
Factor into intercept form
Why: Two thousand comes out of the first bracket as negative 2000, leaving x minus 14, and the second bracket is x plus 10.
\[ R(x) = -2000(x - 14) (x + 10) \]
Identify the zeros and average them
Why: The zeros are 14 and negative 10, whose average is 2, so the price should rise by 2 dollars, to 12.
\[ x = 2;\text{ price } 12\text{ dollars} \]
Evaluate the maximum
Why: R of 2 is negative 2000 times negative 12 times 12, which is 288,000 dollars.
\[ R(2) = 288, 000 \]
Figure (svg): The solution to Worked example maximise the revenue shown as a ladder of expressions, one row per algebraic move
\[ \text{price } 12, \quad R_{\max} = 288{,}000 \]
Verify: test a price on either side
Why: At a 1 dollar rise, revenue is 26,000 times 11, or 286,000; at a 3 dollar rise it is 22,000 times 13, or 286,000. Both are below 288,000, and the equal values on either side are the symmetry of the parabola showing itself, which confirms that 2 really is the axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 262-262
Ranking
Finding the maximum of a quadratic model.
Put in order
Why: This is the book's four-step method with the conversion made explicit at the end. Defining the variables first is what makes the last step possible: the average of the zeros is a value of x, and only the definition says whether that is a price, an increase, a time or a width.
Worked example
Guided Practice 22. The answer is more interesting than it looks.
\[ \text{Repeat Example 7 if the magazine starts at } 11 \text{ dollars, with } 26{,}000 \text{ subscribers.} \]
Write the new model
Why: At 11 dollars there are 26,000 subscribers, still losing 2000 for each further dollar.
\[ R(x) = (26, 000 - 2000 x) (11 + x) \]
Factor into intercept form
Why: Negative 2000 comes out of the first bracket, leaving x minus 13.
\[ R(x) = -2000(x - 13) (x + 11) \]
Average the zeros
Why: The zeros are 13 and negative 11, whose average is 1, so the price rises by 1 dollar, to 12.
\[ x = 1;\text{ price } 12\text{ dollars} \]
Evaluate
Why: R of 1 is negative 2000 times negative 12 times 12, which is 288,000 dollars again.
\[ R(1) = 288, 000 \]
Explain why the answer repeated
Why: Both models describe the same relationship between price and subscribers, just measured from different starting points, so they have the same maximum.
Figure (svg): The solution to Worked example the same magazine, a different starting price shown as a ladder of expressions, one row per algebraic move
\[ \text{price } 12, \quad R_{\max} = 288{,}000 \]
Verify: compare the two models at the same price
Why: The first model at a 2 dollar rise and the second at a 1 dollar rise both describe a 12 dollar subscription with 24,000 subscribers, and 24,000 times 12 is 288,000. The two functions are the same parabola shifted horizontally by one unit, which is why every feature except the position of x equal to zero is identical.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 262-262
Error analysis
A student finishes Example 7 and reports the wrong number.
Annotate
On: \( \text{the average of the zeros is } 2, \text{ so the magazine should charge } 2 \text{ dollars} \)
Step 1 of the book's method is defining the variables, and it exists so that step 4 can be read correctly. Write down what x means, and reread that line before answering.
Fill the middle
Example 7, step 3.
Fill in the blanks
R(x) = -2000(x - 14)(x + 10) \;\Longrightarrow\; x = \frac2___ = ___
Why: Fourteen plus negative 10 is 4, and half of that is 2. The axis of symmetry is always the average of the two zeros, whatever the leading coefficient is, because the parabola is symmetric about the midpoint of any two points at the same height — and the zeros are two points at the same height, namely zero.
Two truths and a lie
All three are about the revenue model.
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. Averaging applies to the x values, not to the function values. Both zeros give a revenue of zero, and their average is zero — nowhere near the true maximum of 288,000. The average locates the axis; the function must then be evaluated there to get the value.
Prediction
Commit before reasoning.
Predict first
Raising the price loses subscribers. Why does revenue still go up at first?
Correct: Because the price rises faster than subscribers fall, at first.
\[ R(x) = (28{,}000 - 2000x)(10 + x): \; \text{one factor falls, one rises} \]
Why: Revenue is a product of two competing quantities, one rising and one falling, and a product of that kind is exactly what makes a parabola. Near 10 dollars the proportional gain in price beats the proportional loss in subscribers, so revenue climbs; past 12 dollars the losses dominate and it falls. The vertex is where the two effects balance, which is what makes it a maximum rather than an arbitrary point.
Comparison
Fill the blanks. The idea is identical; the bookkeeping is not.
Comparison matrix
| Question | a = 1 (Lesson 4.3) | a is not 1 (Lesson 4.4) |
|---|---|---|
| Unknowns to find | m and n | k, l, m and n |
| The middle coefficient equals | m + n | kn + lm |
| Does the order of m and n matter? | no | yes |
| First move on any expression | check for a common monomial | check for a common monomial |
| Typical roots | integers | often fractions |
The last row is worth remembering when checking your work: a fractional root is normal here and is not evidence of a mistake.
Pattern
One routine for factoring anything quadratic.
Step one is the step most often skipped and the one that saves the most work. Step five's rewriting is the step whose omission produces confidently wrong answers.
OpenStax Algebra and Trigonometry 2e, §1.5 Factoring Polynomials §1.5
Check
Factoring with a leading coefficient. Expand to check the middle term.
Check your understanding
Factor 3x^2 + 5x - 12.
Answer: A
Why: The outer term is 9x and the inner term is -4x, giving 5x, and -4 times 3 is -12. Both match.
Check
Factor completely. Something comes out first.
Check your understanding
Factor 12x^2 - 28x - 24 completely.
Answer: A
Why: Every term is divisible by 4, leaving 3x^2 - 7x - 6, which factors as (3x + 2)(x - 3).
Check
Solving. Standard form first.
Check your understanding
Solve 12x^2 + 7x + 2 = x + 8.
Answer: A
Why: Rearranging gives 12x^2 + 6x - 6 = 0, then 2x^2 + x - 1 = 0 after dividing by 6, which factors as (2x - 1)(x + 1).
Real world
A cinema sells 240 tickets a night at 9 dollars each. Market research suggests that every 1 dollar cut in price brings in 40 more customers.
Discussion prompt
Write nightly revenue as a function of the price cut, factor it, and find the price that maximises revenue. What is the maximum?
Hint: Let x be the price cut in dollars, so the price is 9 minus x and attendance is 240 plus 40x.
Answer:
\[ R(x) = (240 + 40x)(9 - x) = -40(x + 6)(x - 9) \]
\[ \text{zeros } -6 \text{ and } 9 \;\Longrightarrow\; x = \frac{-6 + 9}{2} = 1.5 \]
The price should be cut by 1.50 dollars, to 7.50 dollars, drawing 300 customers for a revenue of 2250 dollars — up from 2160 dollars at the original price.
Two things are worth noticing. The 40 came out of the first bracket as a negative, which flipped its contents into x plus 6 and made the parabola open downward, as a maximum requires. And the optimal cut is not a whole number of dollars, which is normal: the average of two zeros is only an integer when the zeros happen to share a parity.
Commit first
Answer, then rate your confidence honestly.
Predict first
You factor an expression as (12x + 8)(x - 3) and it expands correctly. Is the job finished?
Correct: No — the first bracket still has a common factor of 4.
\[ (12x + 8)(x - 3) = 4(3x + 2)(x - 3) \]
Why: Expanding correctly makes an answer true but not necessarily complete. Factored completely means no factor other than a number can be factored further, and 12x plus 8 is 4 times the quantity 3x plus 2. Writing 4 times the quantity 3x plus 2, times the quantity x minus 3 is the finished form. Pulling the common monomial out at the start, as the procedure says, makes this situation impossible to reach.
Explain it
They can factor x squared plus bx plus c and think this lesson is the same thing.
Discussion prompt
In four sentences or fewer, explain what changes when the leading coefficient is not 1, and what stays the same.
Hint: Point at the middle term.
Answer:
What stays the same is the whole idea: you are looking for two brackets that multiply back to the original, and once you have them the zero product property does the rest. What changes is the middle term. With a leading coefficient of 1 it is just m plus n, so the order does not matter; otherwise it is kn plus lm, so the same two constants placed differently give different answers.
That is why you now have a list of arrangements to test rather than a list of pairs. Pull out any common factor first — it makes both numbers smaller and the list much shorter.
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 arrangements, write the four rows out in a table as the book does rather than doing them in your head. For patterns, take the square root of the whole leading term, coefficient included. For the monomial, make it the first thing you look at on every expression, before you read anything else. For standard form, get into the habit of asking is the right side zero before touching the left. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take 12x squared minus 28x minus 24 and work it across the page. Top: write it out and circle the common monomial factor, then write what remains. Left: write the four-integer table for the remaining trinomial, one row per arrangement, with the expanded middle term beside each and the matching row circled. Right: set the whole thing equal to zero and solve, showing the split and both roots. Bottom: sketch the parabola y equals that expression, marking both x-intercepts, the axis of symmetry at their average, and the vertex you get by substituting. In a margin, write one sentence saying what the common factor of 4 did and did not change about the graph.
If your margin sentence says the 4 changed the roots, check it: dividing by 4 leaves the zeros exactly where they were and only compresses the parabola vertically.
Recap
Five things, and the first one makes the other four easier.
| If you see | Then |
|---|---|
| A common factor in every term | Pull it out first, sign included |
| a and c both perfect squares | Test a pattern before searching |
| A leading coefficient that is not 1 | The middle term is kn + lm, not m + n |
| Terms on both sides of an equation | Collect them before factoring anything |
| A request for a maximum or minimum | Factor, then average the zeros |
| A fractional root | Normal here, not a mistake |
Every method so far has needed the quadratic to factor over the integers, and most do not. Lesson 4.5 starts on the equations that do not need factoring at all, beginning with those that are simply a square equal to a number.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.4 Solve ax^2 + bx + c = 0 by Factoring §4.4, pp. 259-263 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.