Unit 6 · Lesson 3 of 6

Factoring trinomials ax² + bx + c

About 25 minutes · 5 steps · one idea: the ac method

You'll be able to: factor a trinomial like \(6x^2 + x - 2\), where the number in front of \(x^2\) is not 1, and check your answer by multiplying.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first, then check.

Factor \(x^2 - 2x - 15\).

Find two numbers that multiply to \(-15\) and add to \(-2\): \(-5\) and \(3\).

Answer: \((x-5)(x+3)\).

Multiply out \((2x+3)(x-4)\). What is the coefficient of \(x\) in the result?

FOIL: \(2x \cdot x = 2x^2\), \(2x \cdot (-4) = -8x\), \(3 \cdot x = 3x\), \(3 \cdot (-4) = -12\).

The middle terms combine: \(-8x + 3x = -5x\). So the result is \(2x^2 - 5x - 12\), and the coefficient of \(x\) is \(-5\).

Step 2 of 5The idea

In the last lesson the trinomial started with plain \(x^2\). Now the \(x^2\) has a number in front, like \(2x^2 + 7x + 3\). We call the three numbers \(a\), \(b\), and \(c\): here \(a = 2\), \(b = 7\), \(c = 3\).

Guessing gets slow when \(a\) is not 1. The ac method turns the problem into one you already know: factoring by grouping.

The ac method for \(ax^2 + bx + c\)

1. Take out the GCF first, if there is one. If \(a\) is negative, take out a negative too.

2. Multiply \(a \cdot c\).

3. Find two numbers that multiply to \(ac\) and add to \(b\).

4. Use them to split the middle term \(bx\) into two terms.

5. Factor the four terms by grouping.

6. Check: multiply your factors back out. You should get the original.

Splitting the middle term doesn't change the value. \(7x\) and \(x + 6x\) are the same thing. It just gives you four terms, and four terms can be grouped.

Why this shows up on the placement test: factoring trinomials is the key step for solving quadratic equations and simplifying rational expressions, two big topics on the test.

Step 3 of 5Worked examples

Example 1: Factor \(2x^2 + 7x + 3\)

  1. No GCF (2, 7, and 3 share no factor). Here \(a = 2\), \(b = 7\), \(c = 3\).
  2. \(a \cdot c = 2 \cdot 3 = 6\).
  3. Two numbers that multiply to 6 and add to 7: \(1\) and \(6\). (\(1 \cdot 6 = 6\) and \(1 + 6 = 7\).)
  4. Split the middle: \(2x^2 + 1x + 6x + 3\).
  5. Group: \(x(2x + 1) + 3(2x + 1)\). Both parentheses match, so it worked.
  6. Take out the shared \((2x+1)\): \((2x + 1)(x + 3)\).
  7. Check: \((2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3\). Correct.

Example 2: Factor \(12x^2 - 2x - 4\)

  1. GCF first: every term is even, so \(12x^2 - 2x - 4 = 2(6x^2 - x - 2)\).
  2. Work on \(6x^2 - x - 2\): \(a \cdot c = 6 \cdot (-2) = -12\), and \(b = -1\).
  3. Multiply to \(-12\), add to \(-1\): \(3\) and \(-4\). (\(3 \cdot (-4) = -12\) and \(3 + (-4) = -1\).)
  4. Split: \(6x^2 + 3x - 4x - 2\).
  5. Group: \(3x(2x + 1) - 2(2x + 1)\). Take out \(-2\) from the second pair so the parentheses match.
  6. So \(6x^2 - x - 2 = (2x + 1)(3x - 2)\). Don't forget the 2 from step 1: the answer is \(2(2x + 1)(3x - 2)\).
  7. Check: \((2x+1)(3x-2) = 6x^2 - 4x + 3x - 2 = 6x^2 - x - 2\), and \(2(6x^2 - x - 2) = 12x^2 - 2x - 4\). Correct.

Step 4 of 5Practice

Type your answer and press Check. Keep scratch paper next to you for the grouping steps. If you miss twice, open the worked solution. Then mark how it went.

For \(4x^2 + 8x + 3\), find the two numbers that multiply to \(ac\) and add to \(b\). (Just the two numbers, no factoring yet.)

\(a \cdot c = 4 \cdot 3 = 12\) and \(b = 8\).

Pairs that multiply to 12: \(1, 12\) (sum 13), \(2, 6\) (sum 8), \(3, 4\) (sum 7). Answer: \(2\) and \(6\).

Factor \(3x^2 + 7x + 2\).

\(ac = 6\). Multiply to 6, add to 7: \(1\) and \(6\).

Split: \(3x^2 + x + 6x + 2 = x(3x + 1) + 2(3x + 1)\).

Answer: \((3x + 1)(x + 2)\). Check: \(3x^2 + 6x + x + 2 = 3x^2 + 7x + 2\).

Factor \(2x^2 - 11x + 5\).

\(ac = 10\). Multiply to 10, add to \(-11\): \(-1\) and \(-10\).

Split: \(2x^2 - x - 10x + 5 = x(2x - 1) - 5(2x - 1)\). (Take out \(-5\) so the parentheses match.)

Answer: \((2x - 1)(x - 5)\). Check: \(2x^2 - 10x - x + 5 = 2x^2 - 11x + 5\).

Factor \(5x^2 + 13x - 6\).

\(ac = 5 \cdot (-6) = -30\). Multiply to \(-30\), add to 13: \(15\) and \(-2\).

Split: \(5x^2 + 15x - 2x - 6 = 5x(x + 3) - 2(x + 3)\).

Answer: \((x + 3)(5x - 2)\). Check: \(5x^2 - 2x + 15x - 6 = 5x^2 + 13x - 6\).

Factor \(6x^2 + x - 12\).

\(ac = 6 \cdot (-12) = -72\). Multiply to \(-72\), add to 1: \(9\) and \(-8\).

Split: \(6x^2 + 9x - 8x - 12 = 3x(2x + 3) - 4(2x + 3)\).

Answer: \((2x + 3)(3x - 4)\). Check: \(6x^2 - 8x + 9x - 12 = 6x^2 + x - 12\).

Factor completely: \(8x^2 - 20x - 12\).

GCF first: \(8x^2 - 20x - 12 = 4(2x^2 - 5x - 3)\).

Inside: \(ac = -6\). Multiply to \(-6\), add to \(-5\): \(-6\) and \(1\). Split: \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (x - 3)(2x + 1)\).

Answer: \(4(x - 3)(2x + 1)\). "Completely" means the GCF 4 sits in front and neither parenthesis has a common factor left. An answer like \((2x - 6)(4x + 2)\) is equal, but a 2 is still hiding inside each parenthesis.

Which is the completely factored form of \(18x^2 + 3x - 6\)?

  1. \((6x - 3)(3x + 2)\)
  2. \(3(6x^2 + x - 2)\)
  3. \((2x - 1)(9x + 6)\)
  4. \(3(2x - 1)(3x + 2)\)

All four are equal to \(18x^2 + 3x - 6\). Only one is finished.

(a) \(6x - 3\) still has a common factor 3. Not complete.

(b) The trinomial \(6x^2 + x - 2\) still factors. Not complete.

(c) \(9x + 6\) still has a common factor 3. Not complete.

(d) GCF 3 out front, then \(6x^2 + x - 2 = (2x - 1)(3x + 2)\) (\(ac = -12\), numbers \(4\) and \(-3\)). This one.

Factor \(-2x^2 + x + 6\). (Start by taking out \(-1\).)

Take out \(-1\): \(-2x^2 + x + 6 = -(2x^2 - x - 6)\). (Every sign inside flips.)

Inside: \(ac = -12\). Multiply to \(-12\), add to \(-1\): \(-4\) and \(3\). Split: \(2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2) = (x - 2)(2x + 3)\).

Answer: \(-(x - 2)(2x + 3)\). Check: \((x-2)(2x+3) = 2x^2 - x - 6\), and the minus sign gives back \(-2x^2 + x + 6\).

Step 5 of 5Wrap-up

You can now factor any trinomial \(ax^2 + bx + c\): take out the GCF, find two numbers that multiply to \(ac\) and add to \(b\), split the middle, group, and check by multiplying.

Read or watch next: OpenStax Intermediate Algebra 2e, 6.2 Factor Trinomials (free textbook; the ac method is in the second half of the section, with more practice at the end).

Also good: Khan Academy: Factoring quadratics by grouping (short videos). Quick rules: Factoring cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why is it allowed to split 7x into x + 6x?" It knows this course and can explain things another way.