Guides / Factoring

Factoring trinomials when a ≠ 1 (the AC method)

A reliable, no-guessing way to factor ax² + bx + c, with every step shown and the sign slip that trips up most students.

Why it works

When a = 1 you can look for two numbers that multiply to c and add to b. When a is something else, that shortcut fails because the a gets tangled into the middle term. Multiplying a · c accounts for it. Splitting bx into two pieces doesn't change the expression (you're rewriting 11x as 15x − 4x, for example), but it gives you four terms that pair up neatly.

Example 1: 6x² + 11x − 10

Here a = 6, b = 11, c = −10, so ac = −60. You need two numbers that multiply to −60 and add to 11. The product is negative, so one number is positive and one is negative. Listing factor pairs of 60 and checking differences: 15 and −4 work, since 15 · (−4) = −60 and 15 + (−4) = 11.

1. 6x² + 11x − 10Start.
2. 6x² + 15x − 4x − 10Split 11x into 15x − 4x.
3. (6x² + 15x) + (−4x − 10)Group the first two terms and the last two.
4. 3x(2x + 5) − 2(2x + 5)Factor 3x from the first group, and −2 from the second.
5. (3x − 2)(2x + 5)Both groups share (2x + 5). Factor it out.

Check by multiplying back: (3x − 2)(2x + 5) = 6x² + 15x − 4x − 10 = 6x² + 11x − 10. It matches.

Example 2: 4x² − 4x − 15

ac = 4 · (−15) = −60 again, but now the numbers have to add to −4. That pair is −10 and 6.

1. 4x² − 4x − 15Start.
2. 4x² − 10x + 6x − 15Split −4x into −10x + 6x.
3. 2x(2x − 5) + 3(2x − 5)Factor 2x from the first pair, 3 from the second.
4. (2x + 3)(2x − 5)Factor out the shared (2x − 5).

The order of the split doesn't matter. Writing 4x² + 6x − 10x − 15 instead gives 2x(2x + 3) − 5(2x + 3), the same two factors.

Example 3: take out the common factor first

For 12x² + 10x − 8, every coefficient is even. Pull out the 2 before anything else; the numbers get smaller and the answer comes out fully factored.

1. 12x² + 10x − 8Start.
2. 2(6x² + 5x − 4)Factor out the greatest common factor, 2.
3. 2(6x² + 8x − 3x − 4)ac = 6 · (−4) = −24. The numbers are 8 and −3.
4. 2[2x(3x + 4) − 1(3x + 4)]Group and factor each pair. Write the 1 so the match is visible.
5. 2(2x − 1)(3x + 4)

Common mistakes, and the exact line they happen on

1. The sign slip in the second group

This is the one that makes students think the method “doesn't work.”

1. 6x² + 15x − 4x − 10Fine so far.
2. 3x(2x + 5) − 2(2x − 5)This is the mistake. −2(2x − 5) = −4x + 10, not −4x − 10. The sign of the 5 is wrong.

Now the groups don't match, so there's nothing to factor out, and it looks like a dead end. The split was right, and the numbers were right. The problem is one sign inside one set of parentheses. When you factor a negative out of a group, every sign inside flips: −4x − 10 = −2(2x + 5). Multiply your group back out to check it before moving on.

2. Using c instead of ac

1. 6x² + 11x − 10Start.
2. 5 · (−2) = −10This is the mistake. Searching for numbers that multiply to c = −10 (here 5 and −2, which add to 3). That only works when a = 1.

There's no pair that multiplies to −10 and adds to 11, so this search fails, or worse, lands on something like (x + 5)(x − 2), which multiplies out to x² + 3x − 10. The 6 has to be in the product.

3. Stopping before the common factor is out

(4x − 2)(3x + 4) multiplies out to 12x² + 10x − 8, so it's not wrong, but it isn't fully factored: 4x − 2 still has a 2 in it. Most teachers will want 2(2x − 1)(3x + 4).

Practice

Factor completely. If a trinomial can't be factored with integers, say so.

1. 2x² + 7x + 3

   Show answer

   ac = 6; the numbers are 6 and 1. 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

2. 3x² − 10x + 8

   Show answer

   ac = 24; the numbers are −6 and −4. 3x(x − 2) − 4(x − 2) = (3x − 4)(x − 2).

3. 5x² + 13x − 6

   Show answer

   ac = −30; the numbers are 15 and −2. 5x(x + 3) − 2(x + 3) = (5x − 2)(x + 3).

4. 6x² − x − 2

   Show answer

   ac = −12; the numbers are −4 and 3. 2x(3x − 2) + 1(3x − 2) = (2x + 1)(3x − 2).

5. 8x² + 2x − 3

   Show answer

   ac = −24; the numbers are 6 and −4. 2x(4x + 3) − 1(4x + 3) = (2x − 1)(4x + 3).

6. 4x² + 6x + 3

   Show answer

   It doesn't factor over the integers. ac = 12, and no pair of integers that multiplies to 12 adds to 6 (1 + 12, 2 + 6, 3 + 4, and their negatives all miss). Saying so is the correct answer.