Factoring Trinomials (x² + bx + c)

A trinomial is a polynomial with three terms, like $x^2 + 7x + 12$. Many of them can be written as a product of two binomials, $(x + 3)(x + 4)$. Factoring trinomials is the key skill for finding the zeros of a parabola and for solving quadratic equations, which is where this unit is heading.

Key ideas

Where the pattern comes from

Expand a product of two binomials and watch what happens to the numbers:

$$(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$$

The middle coefficient $7$ is the sum $3 + 4$, and the last term $12$ is the product $3 \times 4$. In general,

$$(x + m)(x + n) = x^2 + (m + n)x + mn$$

The method

To factor $x^2 + bx + c$ (where the coefficient of $x^2$ is $1$):

1. Find two integers $m$ and $n$ with product $mn = c$ and sum $m + n = b$.
2. Write $x^2 + bx + c = (x + m)(x + n)$.
3. Check by expanding.

It’s easiest to list the factor pairs of $c$ and look for the pair with the right sum. Start with the product, since there are only a few pairs to try.

Sign patterns

The signs of $b$ and $c$ tell you the signs of $m$ and $n$ before you start:

$c$ (product)	$b$ (sum)	Signs of $m$ and $n$	Example
positive	positive	both positive	$x^2 + 7x + 12 = (x + 3)(x + 4)$
positive	negative	both negative	$x^2 - 7x + 12 = (x - 3)(x - 4)$
negative	positive	opposite; the one farther from zero is positive	$x^2 + x - 12 = (x + 4)(x - 3)$
negative	negative	opposite; the one farther from zero is negative	$x^2 - x - 12 = (x - 4)(x + 3)$

Algebra tiles: a rectangle picture

Algebra tiles show factoring as building a rectangle. An $x^2$ tile is a big square, an $x$ tile is a long strip, and a $1$ tile is a small square. To factor $x^2 + 5x + 6$, arrange one $x^2$ tile, five $x$ tiles and six $1$ tiles into a rectangle. Its side lengths are the factors.

The tiles for $x^2 + 5x + 6$ form a rectangle $x + 3$ long and $x + 2$ wide, so $x^2 + 5x + 6 = (x + 2)(x + 3)$.

The six $1$ tiles fill a $2$ by $3$ block (product $6$), and the strips along the sides are $2 + 3 = 5$ $x$ tiles (sum $5$). That’s the product-and-sum rule in picture form.

Common factor first

Always look for a common factor before anything else. Taking it out often leaves a trinomial with an $x^2$ coefficient of $1$:

$$2x^2 - 6x - 20 = 2(x^2 - 3x - 10) = 2(x - 5)(x + 2)$$

Keep the common factor in your final answer. If the $x^2$ term is negative, take out $-1$ first. (For trinomials like $6x^2 - x - 12$, where the $x^2$ coefficient isn’t $1$ and has no common factor, see factoring complex trinomials.)

Some trinomials don’t factor

If no pair of integers has the right product and sum, the trinomial doesn’t factor over the integers. For example, $x^2 + 3x + 5$: the only factor pairs of $5$ are $1, 5$ and $-1, -5$, with sums $6$ and $-6$. Neither is $3$, so it doesn’t factor.

Worked examples

Example 1: Both numbers positive

Factor $x^2 + 7x + 12$.

Solution. We need a product of $12$ and a sum of $7$. Both are positive, so both numbers are positive. List the factor pairs of $12$:

Pair	$1, 12$	$2, 6$	$3, 4$
Sum	$13$	$8$	$7$ ✓

$$x^2 + 7x + 12 = (x + 3)(x + 4)$$

Check: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$. ✓

Example 2: Working with negatives

Factor.

• (a) $x^2 - 9x + 14$
• (b) $x^2 + 2x - 15$
• (c) $x^2 - 4x - 21$

Solution.

(a) Product $14$ (positive), sum $-9$ (negative): both numbers are negative. The pair $-2, -7$ has product $14$ and sum $-9$.

$$x^2 - 9x + 14 = (x - 2)(x - 7)$$

(b) Product $-15$ (negative): one number is positive and one is negative. The sum $2$ is positive, so the number farther from zero is positive. Try $5$ and $-3$: product $-15$, sum $2$. ✓

$$x^2 + 2x - 15 = (x + 5)(x - 3)$$

(c) Product $-21$, sum $-4$: opposite signs, and the number farther from zero is negative. Try $-7$ and $3$: product $-21$, sum $-4$. ✓

$$x^2 - 4x - 21 = (x - 7)(x + 3)$$

Check (c): $(x - 7)(x + 3) = x^2 + 3x - 7x - 21 = x^2 - 4x - 21$. ✓

Example 3: Take out a common factor first

Factor fully.

• (a) $3x^2 - 3x - 36$
• (b) $-x^2 + 2x + 8$

Solution.

(a) Every term is divisible by $3$:

$$\begin{aligned} 3x^2 - 3x - 36 &= 3(x^2 - x - 12) \\ &= 3(x - 4)(x + 3) && -4 \times 3 = -12, \ -4 + 3 = -1 \end{aligned}$$

(b) The $x^2$ term is negative, so take out $-1$. Every sign inside flips:

$$\begin{aligned} -x^2 + 2x + 8 &= -(x^2 - 2x - 8) \\ &= -(x - 4)(x + 2) && -4 \times 2 = -8, \ -4 + 2 = -2 \end{aligned}$$

Check (b) with $x = 1$: the original is $-1 + 2 + 8 = 9$, and $-(1 - 4)(1 + 2) = -(-3)(3) = 9$. ✓

Example 4: Dimensions of a rectangle

A rectangular patio has an area of $(x^2 + 11x + 24)$ m². Find expressions for its length and width, and find the dimensions when $x = 4$.

Solution. Factor the area. We need a product of $24$ and a sum of $11$: the pair $3, 8$ works.

$$x^2 + 11x + 24 = (x + 3)(x + 8)$$

The patio is $(x + 8)$ m long and $(x + 3)$ m wide.

When $x = 4$: the length is $4 + 8 = 12$ m and the width is $4 + 3 = 7$ m.

Check: the area should be $12 \times 7 = 84$ m², and $4^2 + 11(4) + 24 = 16 + 44 + 24 = 84$. ✓

Common mistakes

Mixing up the product and the sum. The numbers multiply to give the last term and add to give the middle coefficient. For $x^2 + 7x + 12$, you need product $12$ and sum $7$, not the other way around.

Getting the signs backwards. For $x^2 - 5x - 14$, the pair is $-7$ and $2$, so the answer is $(x - 7)(x + 2)$. Writing $(x + 7)(x - 2)$ gives a middle term of $+5x$. Check the middle term by expanding.

Forgetting the common factor in the answer. $3x^2 - 3x - 36 = 3(x - 4)(x + 3)$. If you drop the $3$, your answer is only a third of the original expression.

Not looking for a common factor first. If you try to factor $2x^2 + 4x - 30$ directly, the $x^2$ coefficient isn’t $1$ and the product-and-sum method doesn’t apply. Take out the $2$ first: $2(x^2 + 2x - 15) = 2(x + 5)(x - 3)$.

Giving up too early, or never giving up. List all the factor pairs of $c$, including the negative ones. If none has the right sum, the trinomial doesn’t factor. Say so; that’s a correct answer.

Practice

1. (Warm-up) Find two integers with the given product and sum.

• (a) product $18$, sum $9$
• (b) product $-20$, sum $1$
• (c) product $24$, sum $-10$

Solution

(a) $3$ and $6$

(b) $5$ and $-4$

(c) $-4$ and $-6$

2. (Warm-up) Factor.

• (a) $x^2 + 9x + 20$
• (b) $x^2 - 8x + 15$

Solution

(a) Product $20$, sum $9$: $4$ and $5$. So $x^2 + 9x + 20 = (x + 4)(x + 5)$.

(b) Product $15$, sum $-8$: $-3$ and $-5$. So $x^2 - 8x + 15 = (x - 3)(x - 5)$.

3. (Core) Factor.

• (a) $x^2 + 3x - 28$
• (b) $x^2 - 2x - 35$
• (c) $x^2 - 13x + 36$

Solution

(a) Product $-28$, sum $3$: $7$ and $-4$. So $(x + 7)(x - 4)$.

(b) Product $-35$, sum $-2$: $-7$ and $5$. So $(x - 7)(x + 5)$.

(c) Product $36$, sum $-13$: $-4$ and $-9$. So $(x - 4)(x - 9)$.

Check (c): $(x - 4)(x - 9) = x^2 - 9x - 4x + 36 = x^2 - 13x + 36$. ✓

4. (Core) Factor fully.

• (a) $2x^2 + 4x - 30$
• (b) $5a^2 - 40a + 60$

Solution

(a)

$$2x^2 + 4x - 30 = 2(x^2 + 2x - 15) = 2(x + 5)(x - 3)$$

(b)

$$5a^2 - 40a + 60 = 5(a^2 - 8a + 12) = 5(a - 2)(a - 6)$$

5. (Core) Factor $-x^2 - 3x + 10$.

Solution

Take out $-1$ first:

$$-x^2 - 3x + 10 = -(x^2 + 3x - 10) = -(x + 5)(x - 2)$$

Check with $x = 0$: the original is $10$, and $-(5)(-2) = 10$. ✓

6. (Core) Factor $x^2 + 7xy + 10y^2$.

Solution

This works the same way, with $y$ travelling along: we need two numbers with product $10$ and sum $7$, which are $2$ and $5$.

$$x^2 + 7xy + 10y^2 = (x + 2y)(x + 5y)$$

Check: $(x + 2y)(x + 5y) = x^2 + 5xy + 2xy + 10y^2 = x^2 + 7xy + 10y^2$. ✓

7. (Core) Does $x^2 + 4x + 6$ factor over the integers? Explain.

Solution

No. We need product $6$ and sum $4$. The factor pairs of $6$ are $1, 6$ (sum $7$), $2, 3$ (sum $5$), $-1, -6$ (sum $-7$) and $-2, -3$ (sum $-5$). None has a sum of $4$, so it doesn’t factor.

8. (Challenge) Find all integers $k$ so that $x^2 + kx + 12$ can be factored over the integers.

Solution

$k$ must be the sum of a pair of integers whose product is $12$:

Pair	$1, 12$	$2, 6$	$3, 4$	$-1, -12$	$-2, -6$	$-3, -4$
Sum	$13$	$8$	$7$	$-13$	$-8$	$-7$

So $k$ can be $13$, $8$, $7$, $-7$, $-8$ or $-13$.

9. (Challenge) Factor $(x + 1)^2 - 5(x + 1) + 6$. (Hint: let $u = x + 1$.)

Solution

With $u = x + 1$, the expression is $u^2 - 5u + 6$. Product $6$, sum $-5$: $-2$ and $-3$.

$$u^2 - 5u + 6 = (u - 2)(u - 3)$$

Now replace $u$ with $x + 1$:

$$(x + 1 - 2)(x + 1 - 3) = (x - 1)(x - 2)$$

Check by expanding the original: $x^2 + 2x + 1 - 5x - 5 + 6 = x^2 - 3x + 2$, and $(x - 1)(x - 2) = x^2 - 3x + 2$. ✓