Solving Quadratic Equations by Factoring

Solving quadratic equations by factoring uses the zero product property: if $ab = 0$ , then $a = 0$ or $b = 0$ . Set the equation equal to 0, factor the quadratic, then set each factor equal to 0 and solve. For example, $x^2 + 5x + 6 = 0$ becomes $(x + 2)(x + 3) = 0$ , so $x = -2$ or $x = -3$ .

Before this lesson: You should know factoring trinomials, factoring completely and solving quadratics by graphing.

The Zero Product Property

If two numbers multiply to 0, at least one of them must be 0. There is no other way to get a product of 0. This is the zero product property.

\text{If } ab = 0, \text{ then } a = 0 \text{ or } b = 0.

So if $(x + 2)(x + 3) = 0$ , then $x + 2 = 0$ or $x + 3 = 0$ . Each small equation gives one solution.

This only works when one side is 0. A product equal to 6 could come from many pairs, like $2 \cdot 3$ or $1 \cdot 6$ .

Four rows of steps. Start with x² + 5x + 6 = 0. Factor to get (x + 2)(x + 3) = 0. Use the zero product property: x + 2 = 0 or x + 3 = 0. Solve each: x = −2 or x = −3. — Factor, then set each factor equal to 0. Each factor gives one solution.

Steps for solving by factoring

Move all terms to one side so the equation equals 0.
Factor the quadratic completely.
Set each factor equal to 0.
Solve each small equation.
Check each answer in the original equation.

Worked Examples

Example 1: Already factored

Solve $(x - 4)(x + 7) = 0$ .

Step 1. The product is 0, so one of the factors must be 0. Set each factor equal to 0.

x - 4 = 0 \text{ or } x + 7 = 0

Step 2. Solve each equation. Add 4 to the first and subtract 7 from the second (see one-step equations).

x = 4 \text{ or } x = -7

Answer: $x = 4$ or $x = -7$

Example 2: Factor a trinomial

Solve $x^2 + 5x + 6 = 0$ .

Step 1. Factor the trinomial (see factoring trinomials). The numbers 2 and 3 multiply to 6 and add to 5.

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

Step 2. Set each factor equal to 0.

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

Step 3. Subtract to solve each one.

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

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

Check: $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$ . And $(-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0$ .

Example 3: A missing constant

Solve $x^2 = 3x$ .

Step 1. Get 0 on one side. Subtract $3x$ from both sides.

x^2 - 3x = 0

Step 2. Factor out the GCF, $x$ (see factoring out the GCF).

x(x - 3) = 0

Step 3. Set each factor equal to 0.

x = 0 \text{ or } x - 3 = 0

Step 4. Add 3 to solve the second one.

x = 0 \text{ or } x = 3

Answer: $x = 0$ or $x = 3$ . Don’t divide both sides by $x$ . That would lose the solution $x = 0$ .

Example 4: Leading coefficient not 1

Solve $2x^2 - x - 6 = 0$ .

Step 1. Use the AC method (see factoring trinomials when a is not 1). $ac = 2(-6) = -12$ , and $-4$ and 3 multiply to $-12$ and add to $-1$ . Split the middle term.

2x^2 - 4x + 3x - 6 = 0

Step 2. Factor each pair (see factoring by grouping).

2x(x - 2) + 3(x - 2) = 0

Step 3. Factor out the common binomial, $x - 2$ .

(2x + 3)(x - 2) = 0

Step 4. Set each factor equal to 0.

2x + 3 = 0 \text{ or } x - 2 = 0

Step 5. Subtract 3 in the first. Add 2 in the second.

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

Step 6. Divide the first by 2.

x = -\frac{3}{2} \text{ or } x = 2

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

Example 5: Word problem

A rectangle’s length is 3 feet more than its width. Its area is 40 square feet. Find its length and width.

Step 1. Let $w$ be the width. The length is $w + 3$ . Area is width times length.

w(w + 3) = 40

Step 2. Distribute the $w$ (see distributive property).

w^2 + 3w = 40

Step 3. Subtract 40 from both sides.

w^2 + 3w - 40 = 0

Step 4. Factor. The numbers 8 and $-5$ multiply to $-40$ and add to 3.

(w + 8)(w - 5) = 0

Step 5. Set each factor equal to 0.

w + 8 = 0 \text{ or } w - 5 = 0

Step 6. Solve each one.

w = -8 \text{ or } w = 5

Step 7. A width can’t be negative, so $w = 5$ . Find the length.

5 + 3 = 8

Answer: The width is 5 feet and the length is 8 feet.

Check: $5 \cdot 8 = 40$ .

Common Mistakes

Mistake	Example of the error	Fix
Not setting the equation to 0	$(x - 2)(x + 1) = 4$ , so $x - 2 = 4$ or $x + 1 = 4$	Expand and subtract 4: $x^2 - x - 6 = 0$ , so $(x - 3)(x + 2) = 0$ and $x = 3$ or $x = -2$ .
Wrong sign on the solution	$x + 7 = 0$ , so $x = 7$	Subtract 7: $x = -7$ .
Dividing by x	$x^2 = 3x$ , so $x = 3$ only	Factor instead: $x(x - 3) = 0$ gives $x = 0$ too.
Ignoring the coefficient	$2x + 3 = 0$ , so $x = -3$	Divide by 2: $x = -\tfrac{3}{2}$ .
Keeping an answer that makes no sense	Width $= -8$ feet	Lengths can’t be negative. Keep only answers that fit the problem.

Special Patterns

Equation	Factor	Solutions
$x^2 - 49 = 0$	$(x + 7)(x - 7) = 0$	$x = \pm 7$
$x^2 - 10x + 25 = 0$	$(x - 5)^2 = 0$	$x = 5$ (one solution)
$3x^2 - 12 = 0$	$3(x + 2)(x - 2) = 0$	$x = \pm 2$

The symbol $\pm$ means plus or minus. The patterns come from factoring the difference of squares and perfect square trinomials. A number factor like the 3 never equals 0, so it gives no solution.

If a quadratic won’t factor, use square roots, completing the square or the quadratic formula. Those come next in this unit.

Practice Problems

Solve $(x + 5)(x - 1) = 0$ .
Solve $x^2 - 7x + 12 = 0$ .
Solve $x^2 + 2x - 15 = 0$ .
Solve $x^2 - 64 = 0$ .
Solve $x^2 + 8x = 0$ .
Solve $x^2 - 12x + 36 = 0$ .
Solve $3x^2 + 7x + 2 = 0$ .
Solve $x^2 = 5x + 14$ .
Solve $2x^2 - 18 = 0$ .
A ball’s height in feet is $h = -16t^2 + 48t$ . Solve $-16t^2 + 48t = 0$ to find when it is on the ground.

Show answers

Set each factor equal to 0: $x + 5 = 0 \text{ or } x - 1 = 0$
Solve each one: $x = -5 \text{ or } x = 1$
Factor: $-3$ and $-4$ multiply to 12 and add to $-7$ : $(x - 3)(x - 4) = 0$
Set each factor equal to 0: $x - 3 = 0 \text{ or } x - 4 = 0$
Solve each one: $x = 3 \text{ or } x = 4$
Factor: 5 and $-3$ multiply to $-15$ and add to 2: $(x + 5)(x - 3) = 0$
Set each factor equal to 0: $x + 5 = 0 \text{ or } x - 3 = 0$
Solve each one: $x = -5 \text{ or } x = 3$
Factor the difference of squares (see factoring difference of squares): $(x + 8)(x - 8) = 0$
Set each factor equal to 0: $x + 8 = 0 \text{ or } x - 8 = 0$
Solve each one: $x = -8 \text{ or } x = 8$
Factor out the GCF, $x$ : $x(x + 8) = 0$
Set each factor equal to 0: $x = 0 \text{ or } x + 8 = 0$
Solve the second one: $x = 0 \text{ or } x = -8$
Factor the perfect square trinomial (see perfect square trinomials): $(x - 6)^2 = 0$
Both factors are the same, so set it equal to 0 once: $x - 6 = 0$
Add 6: $x = 6$
AC method: $ac = 6$ , and 6 and 1 multiply to 6 and add to 7. Split the middle term: $3x^2 + 6x + x + 2 = 0$
Group and factor each pair: $3x(x + 2) + 1(x + 2) = 0$
Factor out the common binomial: $(3x + 1)(x + 2) = 0$
Set each factor equal to 0: $3x + 1 = 0 \text{ or } x + 2 = 0$
Solve each one: $x = -\frac{1}{3} \text{ or } x = -2$
Subtract $5x$ from both sides: $x^2 - 5x = 14$
Subtract 14 from both sides: $x^2 - 5x - 14 = 0$
Factor: $-7$ and 2 multiply to $-14$ and add to $-5$ : $(x - 7)(x + 2) = 0$
Set each factor equal to 0: $x - 7 = 0 \text{ or } x + 2 = 0$
Solve each one: $x = 7 \text{ or } x = -2$
Factor out the GCF, 2: $2(x^2 - 9) = 0$
Factor the difference of squares: $2(x + 3)(x - 3) = 0$
The factor 2 is never 0. Set the other factors equal to 0: $x + 3 = 0 \text{ or } x - 3 = 0$
Solve each one: $x = -3 \text{ or } x = 3$
Factor out the GCF, $-16t$ : $-16t(t - 3) = 0$
Set each factor equal to 0: $-16t = 0 \text{ or } t - 3 = 0$
Solve each one: $t = 0 \text{ or } t = 3$
$t = 0$ is the launch. It lands after 3 seconds.

FAQ

What is the zero product property?

If a product equals 0, then at least one factor equals 0. That is why you set each factor equal to 0 after factoring.

Why does the equation have to equal 0?

The zero product property only works for 0. If $ab = 4$ , you can’t tell what $a$ or $b$ is.

Can you divide both sides by x?

No. Dividing by $x$ loses the solution $x = 0$ . Factor out $x$ instead.

What if the quadratic doesn’t factor?

Then use another method, like square roots, completing the square or the quadratic formula. They work for every quadratic.

Summary

Zero product property: if $ab = 0$ , then $a = 0$ or $b = 0$ .
Set the equation equal to 0 before you factor.
Factor completely, then set each variable factor equal to 0.
Never divide both sides by a variable.
Check answers, and drop any that don’t fit a word problem.

Previous lesson: Solving Quadratics by Graphing

Next lesson: Solving Quadratics with Square Roots

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