What if the quadratic doesn't factor nicely?

If the quadratic doesn't factor over the integers, use the Quadratic Formula instead. Not all quadratics factor neatly, and that's perfectly fine—the Quadratic Formula works for any quadratic equation.

Can a quadratic equation have only one solution?

Yes, if the quadratic has a repeated root, like (x - 3)^{2} = 0, which gives only x = 3. A quadratic can also have no real solutions if its graph does not cross the x-axis.

How do I check if my factorization is correct?

Multiply the two binomials back together using FOIL or distribution. If you get the original quadratic, your factorization is correct.

Do I always have to use factoring to solve a quadratic?

No, factoring is just one method. If a quadratic doesn't factor nicely, you can use the Quadratic Formula or complete the square. Different situations call for different approaches.

Solving Quadratics by Factoring

Solve quadratic equations by factoring the polynomial into binomials, then apply the Zero Product Property to find the solutions.

(x - r_1)(x - r_2) = 0 \implies x = r_1, r_2

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Type the problem. The solver will use Solving Quadratics by Factoring where Solving Quadratics by Factoring is the right tool, and tell you when it is not.

What each symbol means

What Solving Quadratics by Factoring takes

Symbol	Meaning
$x$	The variable we are solving for; the value(s) that satisfy the quadratic equation.
$r_1$	The first root (solution) of the equation, representing one value of $x$ that makes the quadratic zero.
$r_2$	The second root (solution) of the equation, representing the other value of $x$ that makes the quadratic zero.

When to use it

Use this method when a quadratic equation can be factored into binomials with integer coefficients.

Level

Usually taught in: Algebra I

Worked examples

1. Solve a quadratic with two negative integer roots

Problem

Solve

x^2 + 7x + 12 = 0

for

x

$(x + 3)(x + 4) = 0$
To factor $x^2 + 7x + 12$ , we look for two numbers that multiply to 12 and add to 7—these are 3 and 4.
$x + 3 = 0 \text{ or } x + 4 = 0$
We apply the Zero Product Property: if $(x + 3)(x + 4) = 0$ , then either $x + 3 = 0$ or $x + 4 = 0$ .
$x = -3 \text{ or } x = -4$
Solving $x + 3 = 0$ gives $x = -3$ , and solving $x + 4 = 0$ gives $x = -4$ .

Answer: $x = -3 \text{ or } x = -4$

This quadratic factors into two binomials with integer coefficients, making it straightforward to apply the Zero Product Property. Once we identify the two factors, we solve each linear equation separately to find both solutions.

2. Solve a quadratic with a fractional root using the AC method

Problem

Solve

2x^2 - 5x - 3 = 0

for

x

$2x(x - 3) + 1(x - 3) = 0$
We use the AC method: find two numbers that multiply to $2 \cdot (-3) = -6$ and add to $-5$ , which are $-6$ and $1$ . We rewrite $-5x$ as $-6x + x$ and factor by grouping.
$(2x + 1)(x - 3) = 0$
The common binomial factor $(x - 3)$ appears in both groups, so we factor it out.
$x = -\frac{1}{2} \text{ or } x = 3$
Setting $2x + 1 = 0$ gives $2x = -1$ , so $x = -\frac{1}{2}$ ; setting $x - 3 = 0$ gives $x = 3$ .

Answer: $x = -\frac{1}{2} \text{ or } x = 3$

This problem requires the AC method and produces a fractional solution. The technique handles coefficients other than 1 and negative constants by strategically splitting the middle term, then using the Zero Product Property on both resulting linear equations.

3. Break-even pricing at a school bake sale

Problem

At a school bake sale, a student sells homemade brownies. If they charge

x

dollars per brownie, their profit is

-2x^2 + 14x - 20

dollars. At what price per brownie will their profit be zero?

$-2x^{2} + 14x - 20 = 0$
To find when profit is zero, we set the profit expression equal to zero.
$-2(x^{2} - 7x + 10) = 0$
We factor out the greatest common factor $-2$ from all three terms, leaving $x^{2} - 7x + 10 = 0$ .
$(x - 5)(x - 2) = 0$
The quadratic $x^{2} - 7x + 10$ factors by finding two numbers that multiply to 10 and add to $-7$ , namely $-5$ and $-2$ .
$x = 5 \text{ or } x = 2$
Setting $x - 5 = 0$ gives $x = 5$ ; setting $x - 2 = 0$ gives $x = 2$ . Both prices result in zero profit.

Answer: $x = 2 \text{ or } x = 5$

Real-world situations often produce quadratics with two solutions. Here, the business breaks even at two different price points—a practical application showing why factoring is useful beyond abstract algebra.

Common mistakes

Where Solving Quadratics by Factoring usually goes wrong

The mistake	Why it is wrong	The fix
Writing $(x - 3)(x + 2) = 0 \implies x = 3$ only, without finding the second solution.	The Zero Product Property requires setting EACH factor equal to zero; if a product equals zero, any of its factors could be the cause.	After factoring, set each factor to zero separately: $x - 3 = 0$ OR $x + 2 = 0$ , giving both $x = 3$ and $x = -2$ .
Expanding the factored form back to standard form, then solving: expanding $(x - 2)(x - 5) = 0$ into $x^{2} - 7x + 10 = 0$ , then solving that.	Once you have successfully factored, expanding it wastes effort and introduces opportunities for arithmetic mistakes.	Never expand a factored quadratic. Apply the Zero Product Property directly to the factored form and solve each linear equation.
For $x^{2} - 3x + 2 = 0$ , guessing factors like $(x + 2)(x + 1) = 0$ because the constant is positive, ignoring that the middle coefficient is negative.	When the constant is positive, both binomial factors must have the same sign (both $+$ or both $-$ ), determined by the middle term's sign.	For $x^{2} - 3x + 2$ , find two numbers that multiply to $+2$ and add to $-3$ . Those are $-2$ and $-1$ , so the factors are $(x - 2)(x - 1)$ .

Tips and when to use something else

If the quadratic doesn't factor neatly over the integers, do not force it—use the Quadratic Formula instead of spending excessive time guessing factors.
Always check your factorization by expanding the binomials back into standard form to confirm you get the original quadratic.
Before you start factoring, make sure the equation is set equal to zero; you cannot factor a quadratic set equal to a nonzero value.
When the leading coefficient is not 1, use the AC method or factor out a common factor first, then factor the simplified quadratic.

Related formulas

Zero Product PropertyA product equals zero if and only if at least one factor equals zero; use this property to solve equations by factoring.Extraneous SolutionsA solution to the transformed equation that doesn't satisfy the original; found when squaring both sides or applying other non-reversible operations.Distributive PropertyThe Distributive Property lets you multiply a number by a sum or difference by distributing the multiplication to each term separately.Greatest Common FactorThe Greatest Common Factor finds the largest whole number that divides evenly into two or more numbers, helping you simplify fractions.

Frequently asked questions

What if the quadratic doesn't factor nicely?: If the quadratic doesn't factor over the integers, use the Quadratic Formula instead. Not all quadratics factor neatly, and that's perfectly fine—the Quadratic Formula works for any quadratic equation.
Can a quadratic equation have only one solution?: Yes, if the quadratic has a repeated root, like $(x - 3)^{2} = 0$ , which gives only $x = 3$ . A quadratic can also have no real solutions if its graph does not cross the $x$ -axis.
How do I check if my factorization is correct?: Multiply the two binomials back together using FOIL or distribution. If you get the original quadratic, your factorization is correct.
Do I always have to use factoring to solve a quadratic?: No, factoring is just one method. If a quadratic doesn't factor nicely, you can use the Quadratic Formula or complete the square. Different situations call for different approaches.

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Reviewed 2026-09-18