Factoring Complex Trinomials (ax² + bx + c)

When the coefficient of $x^2$ isn’t $1$, as in $6x^2 - x - 12$, the quick product-and-sum shortcut from factoring trinomials doesn’t work directly. These are sometimes called complex trinomials. You’ll learn two reliable methods: decomposition, which always works if the trinomial factors, and inspection, which is faster once you get the feel for it.

Key ideas

Always look for a common factor first

Before anything else, take out any common factor. It makes the numbers smaller and sometimes turns the problem into an easier one:

$$6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x + 1)(x + 2)$$

If the $x^2$ coefficient is negative, take out a negative factor so that $a$ is positive.

Decomposition (splitting the middle term)

To factor $ax^2 + bx + c$:

1. Multiply $a \times c$.
2. Find two integers $m$ and $n$ with product $mn = ac$ and sum $m + n = b$.
3. Decompose (split) the middle term: rewrite $bx$ as $mx + nx$.
4. Factor the four terms by grouping (see common factoring).
5. Check by expanding.

For example, for $2x^2 + 7x + 3$: $ac = 2 \times 3 = 6$, and $1$ and $6$ have product $6$ and sum $7$.

$$\begin{aligned} 2x^2 + 7x + 3 &= 2x^2 + x + 6x + 3 && \text{split } 7x \text{ into } x + 6x \\ &= x(2x + 1) + 3(2x + 1) && \text{group} \\ &= (2x + 1)(x + 3) \end{aligned}$$

Why does this work? Splitting the middle term doesn’t change the expression ($x + 6x$ is still $7x$), and choosing numbers with product $ac$ is exactly what makes the two groups share a bracket. If the brackets don’t match after grouping, recheck your arithmetic or your signs.

It doesn’t matter which order you write $mx$ and $nx$ in. Writing $2x^2 + 6x + x + 3$ also works: $2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)$.

Inspection (guess and check)

The factors must look like $(px + q)(rx + s)$, where

• $p \times r = a$ (the first terms multiply to $ax^2$),
• $q \times s = c$ (the last terms multiply to $c$),
• the outside and inside products add to the middle term: $ps + qr = b$.

Try combinations of factors of $a$ and $c$ until the middle term comes out right. The first and last terms are easy to get; the middle term is the real test.

Sign hints

The same sign rules from $x^2 + bx + c$ still help:

• If $c \gt 0$, both constants have the same sign as $b$.
• If $c \lt 0$, the constants have opposite signs.

Not every trinomial factors

If no pair of integers has product $ac$ and sum $b$, the trinomial doesn’t factor over the integers. For example, $2x^2 + 3x + 4$: $ac = 8$, and the pairs $1, 8$ and $2, 4$ (and their negatives) have sums $\pm 9$ and $\pm 6$. None is $3$.

Worked examples

Example 1: Decomposition with positive terms

Factor $3x^2 + 8x + 4$.

Solution. There’s no common factor. $ac = 3 \times 4 = 12$. We need product $12$ and sum $8$: the pair $2, 6$ works.

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

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

Example 2: Decomposition with negatives

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

Solution. $ac = 6 \times (-12) = -72$. We need product $-72$ and sum $-1$. The numbers have opposite signs and are close together in size: $-9$ and $8$ work, since $(-9)(8) = -72$ and $-9 + 8 = -1$.

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

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

Example 3: Inspection

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

Solution. The first terms must be $3x$ and $x$. Since $c = 8 \gt 0$ and $b = -10 \lt 0$, both constants are negative. The negative factor pairs of $8$ are $-1, -8$ and $-2, -4$, and each pair can go in either order. Test the middle term of each:

Try	Outside + inside	Middle term
$(3x - 1)(x - 8)$	$-24x - x$	$-25x$
$(3x - 8)(x - 1)$	$-3x - 8x$	$-11x$
$(3x - 2)(x - 4)$	$-12x - 2x$	$-14x$
$(3x - 4)(x - 2)$	$-6x - 4x$	$-10x$ ✓

$$3x^2 - 10x + 8 = (3x - 4)(x - 2)$$

Check: $(3x - 4)(x - 2) = 3x^2 - 6x - 4x + 8 = 3x^2 - 10x + 8$. ✓

Example 4: Common factor first

Factor fully: $12x^2 + 2x - 4$.

Solution. Every term is even, so take out $2$:

$$12x^2 + 2x - 4 = 2(6x^2 + x - 2)$$

Now factor $6x^2 + x - 2$. $ac = 6 \times (-2) = -12$, and we need sum $1$: the pair $4, -3$ works.

$$\begin{aligned} 6x^2 + x - 2 &= 6x^2 + 4x - 3x - 2 \\ &= 2x(3x + 2) - 1(3x + 2) && \text{take out } -1 \text{ so the brackets match} \\ &= (3x + 2)(2x - 1) \end{aligned}$$

So the full answer is

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

Check with $x = 1$: the original is $12 + 2 - 4 = 10$, and $2(5)(1) = 10$. ✓

Common mistakes

Skipping the common factor. $12x^2 + 2x - 4 = (6x + 4)(2x - 1)$ is true, but not fully factored, because $6x + 4 = 2(3x + 2)$. Take out the common factor first and the numbers stay smaller too.

Using c instead of a × c. For $3x^2 + 8x + 4$, you need two numbers with product $ac = 12$, not $4$. The product-and-sum shortcut with just $c$ only works when $a = 1$.

Sign errors in the second group. In $6x^2 + 4x - 3x - 2$, the second group is $-3x - 2 = -1(3x + 2)$. Writing $-1(3x - 2)$ is a common slip. Expand the group to check that you get back $-3x - 2$.

Checking only the first and last terms. With inspection, $(3x - 2)(x - 4)$ gives the right $3x^2$ and the right $+8$, but the wrong middle term ($-14x$). The middle term is the one that decides it.

Dividing everything by a. You can’t just divide $2x^2 + 7x + 3$ by $2$ to make it “nicer”. That changes the expression. Only take out a factor that divides every term, and keep it in the answer.

Practice

1. (Warm-up) For each trinomial, find two integers with product $ac$ and sum $b$.

• (a) $3x^2 + 8x + 4$
• (b) $2x^2 - 3x - 5$

Solution

(a) $ac = 12$, $b = 8$: the integers are $2$ and $6$.

(b) $ac = -10$, $b = -3$: the integers are $-5$ and $2$.

2. (Warm-up) Factor $2x^2 - 3x - 5$ by decomposition, using your answer to question 1(b).

Solution

$$\begin{aligned} 2x^2 - 3x - 5 &= 2x^2 + 2x - 5x - 5 \\ &= 2x(x + 1) - 5(x + 1) \\ &= (x + 1)(2x - 5) \end{aligned}$$

Check: $(x + 1)(2x - 5) = 2x^2 - 5x + 2x - 5 = 2x^2 - 3x - 5$. ✓

3. (Core) Factor $5x^2 + 13x - 6$.

Solution

$ac = -30$, sum $13$: the integers are $15$ and $-2$.

$$\begin{aligned} 5x^2 + 13x - 6 &= 5x^2 + 15x - 2x - 6 \\ &= 5x(x + 3) - 2(x + 3) \\ &= (x + 3)(5x - 2) \end{aligned}$$

4. (Core) Factor.

• (a) $4x^2 - 4x - 15$
• (b) $6a^2 + 11a + 4$

Solution

(a) $ac = -60$, sum $-4$: the integers are $-10$ and $6$.

$$\begin{aligned} 4x^2 - 4x - 15 &= 4x^2 - 10x + 6x - 15 \\ &= 2x(2x - 5) + 3(2x - 5) \\ &= (2x - 5)(2x + 3) \end{aligned}$$

(b) $ac = 24$, sum $11$: the integers are $3$ and $8$.

$$\begin{aligned} 6a^2 + 11a + 4 &= 6a^2 + 3a + 8a + 4 \\ &= 3a(2a + 1) + 4(2a + 1) \\ &= (2a + 1)(3a + 4) \end{aligned}$$

5. (Core) Factor fully: $18x^2 - 33x + 12$.

Solution

Take out the common factor $3$ first: $18x^2 - 33x + 12 = 3(6x^2 - 11x + 4)$.

For $6x^2 - 11x + 4$: $ac = 24$, sum $-11$, so the integers are $-3$ and $-8$.

$$\begin{aligned} 6x^2 - 11x + 4 &= 6x^2 - 3x - 8x + 4 \\ &= 3x(2x - 1) - 4(2x - 1) \\ &= (2x - 1)(3x - 4) \end{aligned}$$

So $18x^2 - 33x + 12 = 3(2x - 1)(3x - 4)$.

6. (Core) Factor $-10x^2 + 11x + 6$.

Solution

Take out $-1$ so the $x^2$ coefficient is positive: $-10x^2 + 11x + 6 = -(10x^2 - 11x - 6)$.

For $10x^2 - 11x - 6$: $ac = -60$, sum $-11$, so the integers are $-15$ and $4$.

$$\begin{aligned} 10x^2 - 11x - 6 &= 10x^2 + 4x - 15x - 6 \\ &= 2x(5x + 2) - 3(5x + 2) \\ &= (5x + 2)(2x - 3) \end{aligned}$$

So $-10x^2 + 11x + 6 = -(5x + 2)(2x - 3)$.

Check with $x = 1$: the original is $-10 + 11 + 6 = 7$, and $-(7)(-1) = 7$. ✓

7. (Core) Factor $3x^2 + 7xy + 2y^2$.

Solution

Treat it like $3x^2 + 7x + 2$, with $y$ travelling along. $ac = 6$, sum $7$: the integers are $1$ and $6$.

$$\begin{aligned} 3x^2 + 7xy + 2y^2 &= 3x^2 + xy + 6xy + 2y^2 \\ &= x(3x + y) + 2y(3x + y) \\ &= (3x + y)(x + 2y) \end{aligned}$$

Check: $(3x + y)(x + 2y) = 3x^2 + 6xy + xy + 2y^2 = 3x^2 + 7xy + 2y^2$. ✓

8. (Challenge) A rectangular poster has an area of $(6x^2 + 17x + 12)$ cm².

• (a) Find expressions for its dimensions.
• (b) Find the dimensions when $x = 2$, and check them against the area.

Solution

(a) $ac = 72$, sum $17$: the integers are $8$ and $9$.

$$\begin{aligned} 6x^2 + 17x + 12 &= 6x^2 + 8x + 9x + 12 \\ &= 2x(3x + 4) + 3(3x + 4) \\ &= (3x + 4)(2x + 3) \end{aligned}$$

The dimensions are $(3x + 4)$ cm by $(2x + 3)$ cm.

(b) When $x = 2$: $3(2) + 4 = 10$ cm and $2(2) + 3 = 7$ cm, so the area is $70$ cm².

Check: $6(2)^2 + 17(2) + 12 = 24 + 34 + 12 = 70$. ✓

9. (Challenge) Find all integers $k$ so that $2x^2 + kx + 3$ can be factored over the integers.

Solution

By decomposition, $k$ must be the sum of two integers whose product is $ac = 2 \times 3 = 6$:

Pair	$1, 6$	$2, 3$	$-1, -6$	$-2, -3$
Sum	$7$	$5$	$-7$	$-5$

So $k$ can be $7$, $5$, $-5$ or $-7$. For example, $2x^2 + 7x + 3 = (2x + 1)(x + 3)$ and $2x^2 + 5x + 3 = (2x + 3)(x + 1)$.