Math Core

When the leading coefficient isn't $1$ , as in $2x^2 + 7x + 3$ , the simple product-and-sum puzzle no longer lines up with the binomials directly. This lesson gives you a dependable method, the ac method, that turns any factorable $ax^2 + bx + c$ into a grouping problem you already know how to solve.

Why the old shortcut fails

Try multiplying two binomials whose $x$ -terms have coefficients:

(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3.

The middle term comes from $6x + x$ , but $6$ and $1$ are not simply the constants $1$ and $3$ . The leading coefficient $2$ gets mixed into one of the products. So instead of guessing binomials, you'll split the middle term into two pieces and factor by grouping.

The ac method

To factor $ax^2 + bx + c$ (after taking out any GCF):

Multiply $a \cdot c$ .
Find two integers $m$ and $n$ with $m \cdot n = ac$ and $m + n = b$ .
Rewrite the middle term: $ax^2 + mx + nx + c$ .
Factor by grouping.

Why does this work? If $ax^2 + bx + c$ factors as $(px + q)(rx + s)$ , then expanding gives $prx^2 + psx + qrx + qs$ . The two middle pieces, $ps$ and $qr$ , add to $b$ , and their product is $ps \cdot qr = pr \cdot qs = a \cdot c$ . So the ac method finds exactly the two middle pieces that the factored form would produce.

Worked example: All positive

Factor $2x^2 + 7x + 3$ .

$ac = 2 \cdot 3 = 6$ .
Two numbers with product $6$ and sum $7$ : $1$ and $6$ .
Split the middle term: $2x^2 + x + 6x + 3$ .
Group:

\begin{aligned} 2x^2 + x + 6x + 3 &= x(2x + 1) + 3(2x + 1) \\ &= (2x + 1)(x + 3) \end{aligned}

It doesn't matter which order you write the split terms. $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3)$ gives the same answer.

Worked example: Negative middle term

Factor $3x^2 - 10x + 8$ .

$ac = 24$ and $b = -10$ . A positive product with a negative sum means both numbers are negative. The pair $-4$ and $-6$ works: $(-4)(-6) = 24$ and $-4 + (-6) = -10$ .

\begin{aligned} 3x^2 - 10x + 8 &= 3x^2 - 4x - 6x + 8 \\ &= x(3x - 4) - 2(3x - 4) \\ &= (3x - 4)(x - 2) \end{aligned}

In the second group, factor out $-2$ (not $2$ ) so the binomial matches $3x - 4$ .

Worked example: Negative constant

Factor $6x^2 + x - 12$ .

$ac = 6 \cdot (-12) = -72$ and $b = 1$ . The numbers have opposite signs and differ by $1$ : $9$ and $-8$ .

\begin{aligned} 6x^2 + x - 12 &= 6x^2 + 9x - 8x - 12 \\ &= 3x(2x + 3) - 4(2x + 3) \\ &= (2x + 3)(3x - 4) \end{aligned}

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

Common mistake

When the second pair begins with a minus sign, the most common mistake is factoring out a positive number and getting binomials that don't match, like $x(3x - 4) + 2(-3x + 4)$ . If the binomials aren't identical, factor a negative out of the second pair. If they still don't match, recheck your split.

GCF first, and negative leading coefficients

Always look for a GCF before you start. Taking it out makes the numbers smaller, so there are fewer factor pairs to check. And if the leading coefficient is negative, factor out $-1$ (or a negative GCF) so the trinomial you work with starts with a positive term.

Worked example: GCF, then ac method

Factor $12x^2 - 2x - 4$ completely.

The GCF is $2$ : $12x^2 - 2x - 4 = 2(6x^2 - x - 2)$ .

Now factor $6x^2 - x - 2$ . Here $ac = -12$ and $b = -1$ , so use $-4$ and $3$ :

\begin{aligned} 6x^2 - x - 2 &= 6x^2 - 4x + 3x - 2 \\ &= 2x(3x - 2) + 1(3x - 2) \\ &= (3x - 2)(2x + 1) \end{aligned}

So $12x^2 - 2x - 4 = 2(3x - 2)(2x + 1)$ .

Guess and check

With practice, you can often factor by trial: the first terms of the binomials must multiply to $ax^2$ and the last terms to $c$ , so try combinations and check the middle term. For $5x^2 + 11x + 2$ , the only options are $(5x + 1)(x + 2)$ and $(5x + 2)(x + 1)$ . The first gives a middle term of $10x + x = 11x$ . ✓ Trial is quick when $a$ and $c$ have few factors; the ac method is more systematic when they have many.

Tip

If the trinomial has no GCF, neither binomial factor can have one either. So when factoring $2x^2 + 7x + 3$ , you can rule out any binomial like $(2x + 2)$ right away.

Practice

Practice 1

Factor $3x^2 + 7x + 2$ .

Practice 2

To factor $2x^2 + 11x + 12$ with the ac method, you need two numbers whose product is $ac$ and whose sum is $11$ . What is the larger of the two numbers?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Factor $5x^2 - 17x + 6$ .

Practice 4

Factor $4x^2 + 4x - 15$ .

Practice 5

The trinomial $6x^2 - 7x - 5$ factors as $(2x + 1)(3x + k)$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Factor $6x^2 - 21x + 18$ completely.

Practice 7

Factor $-2x^2 + 5x + 3$ completely.

Practice 8

A rectangle has area $3x^2 + 14x + 8$ square units, and its width is $x + 4$ units. What is its length when $x = 5$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.