Factoring Trinomials

A trinomial is a polynomial with three terms. The most common type you will factor is the quadratic trinomial, which has the form $ax^2 + bx + c$. Factoring these trinomials is one of the most important skills in algebra — it lets you solve quadratic equations, simplify rational expressions, and analyze parabolas.

There are two main cases: trinomials where the leading coefficient $a = 1$ (simpler) and trinomials where $a \neq 1$ (requires the AC method or trial and error). This page covers both.

Case 1: Leading Coefficient Is 1 ($x^2 + bx + c$)

When $a = 1$, the trinomial looks like $x^2 + bx + c$. You need to find two numbers $p$ and $q$ such that:

$p + q = b \qquad \text{and} \qquad p \times q = c$

Then the factored form is:

$x^2 + bx + c = (x + p)(x + q)$

Why this works: When you FOIL $(x + p)(x + q)$, you get $x^2 + qx + px + pq = x^2 + (p + q)x + pq$. So the middle coefficient is the sum and the last term is the product.

Example 1: Factor $x^2 + 7x + 12$

Find two numbers that add to 7 and multiply to 12.

The pair is 3 and 4.

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

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

Example 2: Factor $x^2 - 5x + 6$

Find two numbers that add to $-5$ and multiply to 6.

Since the product is positive and the sum is negative, both numbers must be negative.

The pair is $-2$ and $-3$ (because $-2 + (-3) = -5$ and $(-2)(-3) = 6$).

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

Check: $(x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$ . Correct.

Example 3: Factor $x^2 + 2x - 15$

Find two numbers that add to 2 and multiply to $-15$.

Since the product is negative, one number is positive and one is negative.

The pair is 5 and $-3$ (because $5 + (-3) = 2$ and $5 \times (-3) = -15$).

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

Check: $(x + 5)(x - 3) = x^2 - 3x + 5x - 15 = x^2 + 2x - 15$ . Correct.

Sign Patterns — A Quick Guide

The signs of $b$ and $c$ tell you the signs of $p$ and $q$:

Case 2: Leading Coefficient Is Not 1 ($ax^2 + bx + c$, $a \neq 1$)

When $a \neq 1$, the factoring is harder. The most reliable method is the AC method (also called “factoring by grouping for trinomials”).

The AC Method — Step by Step

1. Multiply $a \times c$ to get the “AC product”
2. Find two numbers that add to $b$ and multiply to $ac$
3. Rewrite the middle term $bx$ as two terms using those numbers
4. Factor by grouping (split into two pairs and factor each pair)
5. Factor out the common binomial

Example 4: Factor $2x^2 + 7x + 3$

Step 1 — Compute AC: $a \times c = 2 \times 3 = 6$

Step 2 — Find two numbers that add to 7 and multiply to 6: The pair is 1 and 6.

Step 3 — Rewrite the middle term:

$2x^2 + 1x + 6x + 3$

Step 4 — Group and factor each pair:

$\underbrace{2x^2 + x}_{\text{group 1}} + \underbrace{6x + 3}_{\text{group 2}}$

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

Step 5 — Factor out the common binomial $(2x + 1)$:

$2x^2 + 7x + 3 = (2x + 1)(x + 3)$

Check: $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$ . Correct.

Example 5: Factor $6x^2 - 11x + 4$

Step 1 — Compute AC: $6 \times 4 = 24$

Step 2 — Find two numbers that add to $-11$ and multiply to 24: Since the product is positive and the sum is negative, both numbers are negative. The pair is $-3$ and $-8$ (because $-3 + (-8) = -11$ and $(-3)(-8) = 24$).

Step 3 — Rewrite:

$6x^2 - 3x - 8x + 4$

Step 4 — Group and factor:

$3x(2x - 1) - 4(2x - 1)$

Step 5 — Factor out $(2x - 1)$:

$6x^2 - 11x + 4 = (2x - 1)(3x - 4)$

Check: $(2x - 1)(3x - 4) = 6x^2 - 8x - 3x + 4 = 6x^2 - 11x + 4$ . Correct.

Example 6: Factor $3x^2 + 10x - 8$

Step 1 — Compute AC: $3 \times (-8) = -24$

Step 2 — Find two numbers that add to 10 and multiply to $-24$: The pair is 12 and $-2$ (because $12 + (-2) = 10$ and $12 \times (-2) = -24$).

Step 3 — Rewrite:

$3x^2 + 12x - 2x - 8$

Step 4 — Group and factor:

$3x(x + 4) - 2(x + 4)$

Step 5 — Factor out $(x + 4)$:

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

Check: $(x + 4)(3x - 2) = 3x^2 - 2x + 12x - 8 = 3x^2 + 10x - 8$ . Correct.

Always Check for a GCF First

Before applying the AC method, always factor out the GCF. This makes the numbers smaller and the factoring easier.

Example 7: Factor $4x^2 + 12x + 8$

Step 1 — Factor out the GCF: The GCF of 4, 12, and 8 is 4.

$4x^2 + 12x + 8 = 4(x^2 + 3x + 2)$

Step 2 — Factor the trinomial inside: Find two numbers that add to 3 and multiply to 2. The pair is 1 and 2.

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

Without factoring out the 4 first, you would have to use the AC method with $ac = 32$, which is much harder.

Trial and Error Method

Some students prefer trial and error for $ax^2 + bx + c$. List the factor pairs of $a$ and $c$, then test combinations until the middle term works out.

Example 8: Factor $5x^2 + 13x + 6$

Factor pairs of 5: $(1, 5)$. Factor pairs of 6: $(1, 6)$, $(2, 3)$.

Try $(x + 2)(5x + 3)$: outer + inner $= 3x + 10x = 13x$ . That works.

$5x^2 + 13x + 6 = (x + 2)(5x + 3)$

Trial and error works well when $a$ has few factor pairs. For larger values of $a$, the AC method is usually faster.

Real-World Application: Electrician — Sizing a Junction Box

An electrician needs to find the dimensions of a junction box. The box must have a cross-sectional area of 24 square inches, and the length must be 2 inches more than the width. Setting up the equation with width $x$:

$x(x + 2) = 24$

Expanding:

$x^2 + 2x = 24$

$x^2 + 2x - 24 = 0$

Factor the trinomial: find two numbers that add to 2 and multiply to $-24$. The pair is 6 and $-4$.

$x^2 + 2x - 24 = (x + 6)(x - 4) = 0$

Setting each factor to zero:

$x + 6 = 0 \implies x = -6 \qquad x - 4 = 0 \implies x = 4$

Since width cannot be negative, $x = 4$ inches and the length is $4 + 2 = 6$ inches. The junction box cross-section is 4 inches by 6 inches.

Common Mistakes to Avoid

1. Forgetting to check for a GCF first. Always look for a common factor before attempting the AC method or trial and error. It makes the arithmetic much simpler.
2. Sign errors in the AC method. Be careful with the signs when finding two numbers that add and multiply. Use the sign pattern table above as a guide.
3. Splitting the wrong term. In the AC method, you split the middle term $bx$, not the first or last term.
4. Grouping that does not produce a common binomial. If your two groups do not share a common factor, try swapping the order of the two middle terms.
5. Declaring a trinomial “unfactorable” too quickly. Some trinomials factor over the integers and some do not. Use the discriminant $b^2 - 4ac$: if it is a perfect square, the trinomial factors over the integers.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• For $x^2 + bx + c$ (leading coefficient 1): find two numbers that add to $b$ and multiply to $c$
• For $ax^2 + bx + c$ (leading coefficient not 1): use the AC method — multiply $a \times c$, find two numbers that add to $b$ and multiply to $ac$, then factor by grouping
• Always factor out the GCF first — it makes the remaining trinomial simpler
• The sign pattern of $b$ and $c$ tells you the signs of the factors
• Check your work by multiplying (FOIL) the factors back together
• If the discriminant $b^2 - 4ac$ is not a perfect square, the trinomial does not factor over the integers

Return to Algebra for more topics in this section.