Solving Quadratics by Factoring

Factoring is the fastest way to solve many quadratic equations — and it requires no formula to memorize beyond one simple principle. If you can factor a polynomial (which you learned in the factoring section), you can solve quadratic equations by turning one equation into two simpler ones.

The key idea is the zero product property: if two things multiply to zero, at least one of them must be zero.

The Zero Product Property

The zero product property states:

$\text{If } a \cdot b = 0, \text{ then } a = 0 \text{ or } b = 0 \text{ (or both).}$

Think about why this must be true. If neither $a$ nor $b$ is zero, their product cannot be zero — you would get some nonzero number. The only way to get zero from a multiplication is if at least one factor is zero.

This property is the engine behind solving quadratics by factoring. Once you write a quadratic as a product of two factors equal to zero, you set each factor equal to zero and solve.

The Method: Step by Step

To solve a quadratic equation by factoring:

1. Move all terms to one side so the equation equals zero.
2. Factor the quadratic expression.
3. Set each factor equal to zero (zero product property).
4. Solve each resulting linear equation.
5. Check both solutions in the original equation.

Let’s see this in action.

Solving Quadratics with Leading Coefficient 1

When the $x^2$ coefficient is 1, factoring means finding two numbers that multiply to the constant term and add to the coefficient of $x$.

Example 1: $x^2 + 5x + 6 = 0$

We need two numbers that multiply to $6$ and add to $5$.

Those numbers are $2$ and $3$, since $2 \times 3 = 6$ and $2 + 3 = 5$.

Factor:

$x^2 + 5x + 6 = (x + 2)(x + 3) = 0$

Apply the zero product property:

$x + 2 = 0 \quad \text{or} \quad x + 3 = 0$

Solve each equation:

$x = -2 \quad \text{or} \quad x = -3$

Check: Substitute $x = -2$: $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$ . Correct.

Substitute $x = -3$: $(-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0$ . Correct.

Answer: $x = -2$ or $x = -3$

Example 2: $x^2 - 7x + 12 = 0$

Find two numbers that multiply to $12$ and add to $-7$. Those numbers are $-3$ and $-4$.

Factor:

$(x - 3)(x - 4) = 0$

Set each factor equal to zero:

$x - 3 = 0 \implies x = 3$

$x - 4 = 0 \implies x = 4$

Answer: $x = 3$ or $x = 4$

When the Equation Is Not Already Set to Zero

You must rearrange the equation so one side equals zero before factoring.

Example 3: $x^2 = 3x + 10$

Step 1 — Move all terms to the left:

$x^2 - 3x - 10 = 0$

Step 2 — Factor. Find two numbers that multiply to $-10$ and add to $-3$. Those are $-5$ and $2$.

$(x - 5)(x + 2) = 0$

Step 3 — Set each factor to zero:

$x - 5 = 0 \implies x = 5$

$x + 2 = 0 \implies x = -2$

Check $x = 5$: $5^2 = 25$ and $3(5) + 10 = 25$ . Correct.

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

Answer: $x = 5$ or $x = -2$

Solving with a Leading Coefficient Other Than 1

When the coefficient of $x^2$ is not 1, factoring requires a bit more work — often using the AC method or trial and error.

Example 4: $2x^2 + 7x + 3 = 0$

Using the AC method: $a \cdot c = 2 \times 3 = 6$. Find two numbers that multiply to $6$ and add to $7$. Those are $1$ and $6$.

Rewrite the middle term and factor by grouping:

$2x^2 + x + 6x + 3 = 0$

$x(2x + 1) + 3(2x + 1) = 0$

$(x + 3)(2x + 1) = 0$

Set each factor to zero:

$x + 3 = 0 \implies x = -3$

$2x + 1 = 0 \implies x = -\frac{1}{2}$

Check $x = -3$: $2(-3)^2 + 7(-3) + 3 = 18 - 21 + 3 = 0$ . Correct.

Check $x = -\tfrac{1}{2}$: $2\!\left(-\tfrac{1}{2}\right)^2 + 7\!\left(-\tfrac{1}{2}\right) + 3 = \tfrac{1}{2} - \tfrac{7}{2} + 3 = 0$ . Correct.

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

Special Cases

Difference of Squares

Example 5: $x^2 - 25 = 0$

This is a difference of squares: $x^2 - 5^2 = (x - 5)(x + 5) = 0$.

$x = 5 \quad \text{or} \quad x = -5$

One Solution (Repeated Root)

Example 6: $x^2 - 6x + 9 = 0$

Factor: $(x - 3)(x - 3) = (x - 3)^2 = 0$.

The only solution is $x = 3$. We call this a repeated root or double root — the parabola touches the $x$-axis at exactly one point.

Equations with a Common Factor

Example 7: $3x^2 - 12x = 0$

Factor out the greatest common factor first:

$3x(x - 4) = 0$

Set each factor to zero:

$3x = 0 \implies x = 0$

$x - 4 = 0 \implies x = 4$

Answer: $x = 0$ or $x = 4$

Never divide both sides by $x$ — that would lose the $x = 0$ solution entirely.

Real-World Application: Carpentry — Designing a Deck

A carpenter is building a rectangular deck. The length is 4 feet more than the width. The total area must be 96 square feet to meet the homeowner’s specifications.

Let $w$ = the width in feet. Then the length is $w + 4$.

$w(w + 4) = 96$

$w^2 + 4w = 96$

$w^2 + 4w - 96 = 0$

Find two numbers that multiply to $-96$ and add to $4$. Those are $12$ and $-8$.

$(w + 12)(w - 8) = 0$

$w = -12 \quad \text{or} \quad w = 8$

A width cannot be negative, so $w = 8$ feet. The length is $8 + 4 = 12$ feet.

Answer: The deck should be 8 feet wide by 12 feet long. The carpenter can now calculate lumber quantities: the perimeter is $2(8 + 12) = 40$ feet of border boards, and the area confirms 96 square feet of decking material.

Common Mistakes to Avoid

1. Forgetting to set the equation to zero first. If you factor $x^2 + 5x = 6$ as $x(x + 5) = 6$ and then set $x = 6$ or $x + 5 = 6$, you get wrong answers. The zero product property only works when the product equals zero.

2. Dividing by the variable. Given $x^2 = 5x$, do not divide both sides by $x$. You would lose the solution $x = 0$. Instead, move everything to one side: $x^2 - 5x = 0 \implies x(x - 5) = 0$.

3. Stopping after factoring. Factoring is not the answer — it is a step toward the answer. After factoring, you must set each factor equal to zero and solve.

4. Sign errors when finding factor pairs. If the constant term is positive and the middle coefficient is negative, both factors must be negative. Double-check signs by expanding your factored form.

Practice Problems

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

Key Takeaways

• The zero product property says if $a \cdot b = 0$, then $a = 0$ or $b = 0$ — this is the foundation of solving by factoring
• Always set the equation equal to zero before factoring
• After factoring, set each factor equal to zero and solve the resulting linear equations
• Never divide both sides by the variable — you will lose solutions
• Factoring works best when the equation factors neatly over the integers; when it does not, use the quadratic formula or completing the square instead
• Always check your solutions by substituting back into the original equation

Return to Algebra 1 for more topics in this section.