How do you factor a quadratic trinomial?

Find two numbers that multiply to the constant term and add to the middle coefficient (when the leading coefficient is 1). Those numbers become the constants in your two binomials. Always check by expanding.

What if the leading coefficient isn't 1?

Use the AC method: multiply a by c, find factors of that product that sum to b, split the middle term, and factor by grouping.

Cramzoo is free to start — every unit opens with free skills and no account is needed. This particular skill is part of Premium ($6/month or $30/year), which unlocks every skill plus the parent progress dashboard.

How is this different from IXL?

No punishing score. A wrong answer gives you a step-by-step explanation, not a number that drops — and every question is reviewed by a human before you see it.

Algebra 2 practice

Factoring Quadratic Trinomials — Practice

Practice factoring trinomials of the form $ax^2 + bx + c$ — four choices, one correct, and a worked explanation on every question. Free to start, no account, and every problem is reviewed before it reaches you.

Start practicing — free

No login, no card. Open Algebra 2 and pick the tile.

What you'll practice

A quadratic trinomial is a three-term expression like $x^2 + 7x + 12$ . Factoring rewrites it as a product of two binomials — $(x + 3)(x + 4)$ — the step that unlocks solving equations, finding roots, and graphing parabolas. These questions cover leading coefficient $a = 1$ and $a > 1$ , positive and negative middle terms, and the sign traps that catch most students.

Worked example

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

Find two numbers that multiply to $12$ (the constant) and add to $7$ (the middle coefficient).
$3 \times 4 = 12$ and $3 + 4 = 7$ . ✓
So $x^2 + 7x + 12 = (x + 3)(x + 4)$ .
Check by expanding: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ . ✓ — this check is the habit that catches sign errors.

Common mistakes

Sign flip: giving $(x-3)(x-4)$ for $x^2 + 7x + 12$ . Both numbers must be positive here because the middle and last terms are both positive.
Right product, wrong sum: picking $2$ and $6$ (product $12$ , but sum $8 \neq 7$ ).
Forgetting $a > 1$ : with $2x^2 + 7x + 3$ you can't just split the constant — the leading coefficient changes the split (answer: $(2x+1)(x+3)$ ).

Try a few

1. Factor $x^2 + 8x + 15$ .

$(x+3)(x+5)$
$(x-3)(x-5)$
$(x+1)(x+15)$
$(x+2)(x+6)$

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

$(x-2)(x-3)$
$(x+2)(x+3)$
$(x-1)(x-6)$
$(x-2)(x+3)$

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

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

Want more — with instant scoring and a streak?

FAQ

How do you factor a quadratic trinomial?: Find two numbers that multiply to the constant term and add to the middle coefficient (when the leading coefficient is 1). Those numbers become the constants in your two binomials. Always check by expanding.
What if the leading coefficient isn't 1?: Use the AC method: multiply a by c, find factors of that product that sum to b, split the middle term, and factor by grouping.
Is Cramzoo free?: Cramzoo is free to start — every unit opens with free skills and no account is needed. This particular skill is part of Premium ($6/month or $30/year), which unlocks every skill plus the parent progress dashboard.
How is this different from IXL?: No punishing score. A wrong answer gives you a step-by-step explanation, not a number that drops — and every question is reviewed by a human before you see it.

Related Algebra 2 skills

Ready to practice?

Every question reviewed. Mistakes turn into explanations, never a score that punishes you for trying.

Start practicing — free

No login, no card. Open Algebra 2 and pick the tile.