Solving Quadratic Equations by Factoring

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Hello again. In the previous lesson, you learned to factor polynomials by taking out a greatest common factor, factoring trinomials, and recognizing a difference of squares. This lesson uses that skill for a new purpose: solving quadratic equations.

By the end, you should be able to take a quadratic equation, put it in the correct form, factor it completely, and use the zero-product property to find every solution. This is one of the most efficient methods when a quadratic factors neatly.

Why a product of zero is special

The zero-product property states:

If A \cdot B = 0, then A = 0 or B = 0.

A product can equal zero only if at least one of its factors is zero. For example,

0 \cdot 8 = 0

but no multiplication of two nonzero real numbers can produce $0$ .

This matters because factoring turns a quadratic expression into a product. Once that product equals zero, each factor gives a possible value of the variable.

For instance, suppose you have

(x - 4) (x + 3) = 0.

The factors are $x - 4$ and $x + 3$ . Apply the zero-product property by setting each factor, not each term, equal to zero:

x - 4 = 0

x = 4

x + 3 = 0

x = - 3.

So the solutions are

x = 4 or x = - 3 .

A useful sign rule follows from this:

Factor	Value that makes it zero
$x - r$	$x = r$
$x + r$	$x = - r$
$a x + b$	$x = - \frac{b}{a}$

The sign inside a factor is the opposite of the root when the coefficient of $x$ is $1$ .

Solving a quadratic by factoring | Quadratic equations | Algebra I | Khan Academy

Watch “Solving a quadratic by factoring” from Khan Academy for a visual introduction to how factoring reveals the solutions of a quadratic equation, including the connection to points where a graph meets the horizontal axis.

Watch the graph idea to see why solving a quadratic means finding where its value is zero. Then watch a basic factorization, focusing on the moment each factor is set equal to zero. Continue with a repeated factor to see why a quadratic can sometimes have only one distinct solution, and finish with sign choices for a quadratic whose factors have opposite signs.

The required setup: one side must be zero

You may use the zero-product property only when all of the following are true:

One side of the equation equals $0$ .
The other side is written as a product of factors.
The expression is factored completely.

The standard form of a quadratic equation is

a x^{2} + b x + c = 0,

where $a \neq = 0$ .

Here is the full process:

Move all terms to one side so the other side is $0$ .
Factor the quadratic expression completely.
Set every nonconstant factor equal to $0$ .
Solve the resulting linear equations.
Check the solutions in the original equation.

Example: a quadratic that is already in standard form

Solve:

6 x^{2} + x - 15 = 0.

From the previous lesson’s AC method, the trinomial factors as

6 x^{2} + x - 15 = (3 x + 5) (2 x - 3) .

So rewrite the equation:

(3 x + 5) (2 x - 3) = 0.

Set each factor equal to zero:

3 x + 5 = 0

3 x = - 5

x = - \frac{5}{3} .

For the other factor:

2 x - 3 = 0

2 x = 3

x = \frac{3}{2} .

Therefore,

x = - \frac{5}{3} or x = \frac{3}{2} .

Notice that the factors need not be of the form $x + r$ . When a factor contains a coefficient, such as $3 x + 5$ , solve it as an ordinary linear equation.

A quick check uses the factored form. If

x = - \frac{5}{3},

then

3 x + 5 = 3 (- \frac{5}{3}) + 5 = 0.

Thus the entire product equals zero. Similarly, when

x = \frac{3}{2},

the factor $2 x - 3$ becomes zero.

Factoring patterns become solution methods

The actual solving step is always the same: factor, set each variable-containing factor equal to zero, and solve. What changes is the factoring pattern you use first.

A greatest common factor can create a solution

Solve:

3 x^{2} - 9 x = 0.

Both terms share a factor of $3 x$ :

3 x (x - 3) = 0.

Now apply the zero-product property:

3 x = 0

x - 3 = 0.

The first equation gives

x = 0.

The second gives

x = 3.

So the solutions are

x = 0 or x = 3 .

Do not ignore the $3 x$ factor. It is exactly what produces the solution $x = 0$ . However, you do not need to set the constant $3$ alone equal to zero, since

3 \neq = 0.

You may simplify the factor equation

3 x = 0

by dividing both sides by $3$ .

Difference of squares

Solve:

4 x^{2} - 81 = 0.

This is a difference of squares:

4 x^{2} = (2 x)^{2}

and

81 = 9^{2} .

Factor:

(2 x - 9) (2 x + 9) = 0.

Set each factor equal to zero:

2 x - 9 = 0

2 x = 9

x = \frac{9}{2} .

For the other factor:

2 x + 9 = 0

2 x = - 9

x = - \frac{9}{2} .

Therefore,

x = \frac{9}{2} or x = - \frac{9}{2} .

How To Solve Quadratic Equations By Factoring - Algebra Introduction

The Organic Chemistry Tutor’s “How To Solve Quadratic Equations By Factoring” gives short, worked examples of two especially useful structures: equations with a greatest common factor and equations that are differences of squares.

Watch GCF factoring, paying attention to why the factor containing x creates the root x=0. Then watch difference of squares and compare the two conjugate factors with the two opposite-sign solutions.

A repeated factor produces one distinct solution

Solve:

x^{2} + 10 x + 25 = 0.

This is a perfect-square trinomial:

x^{2} + 10 x + 25 = (x + 5)^{2} .

Write the repeated factor explicitly:

(x + 5) (x + 5) = 0.

Both factors lead to the same equation:

x + 5 = 0.

Thus,

x = - 5 .

This is called a double root because the same factor occurs twice. It is still a quadratic equation, but it has only one distinct real solution.

When the equation does not initially equal zero

A common mistake is trying to use the zero-product property too early. Consider:

x (x + 5) = 14.

The left side is factored, but it does not equal zero. You cannot set $x = 0$ or $x + 5 = 0$ , because neither conclusion follows from a product equal to $14$ .

Instead, rewrite the equation in standard form. First expand:

x^{2} + 5 x = 14.

Then subtract $14$ from both sides:

x^{2} + 5 x - 14 = 0.

Now factor:

(x + 7) (x - 2) = 0.

Set each factor equal to zero:

x + 7 = 0

x = - 7

x - 2 = 0

x = 2.

The solutions are

x = - 7 or x = 2 .

Both are valid algebraic solutions. Unlike a measurement or geometry problem, there is no real-world condition here that would eliminate the negative value.

Avoiding the most common errors

Use this short diagnostic list whenever you solve by factoring.

Check	What it prevents
Is one side equal to $0$ ?	Using the zero-product property when it does not apply
Did you factor out the GCF first?	Missing a factor and possibly missing a solution such as $x = 0$
Is the quadratic fully factored?	Stopping at an incomplete factorization
Did you set factors, not terms, equal to zero?	Incorrectly treating addition as multiplication
Did you solve both factor equations?	Losing one of the roots
Did you keep signs opposite when solving?	Saying $x = 5$ from $x + 5 = 0$
Did you check in the original equation?	Catching factoring or arithmetic errors

One final limitation is worth remembering: factoring is powerful, but not every quadratic factors neatly using integers. That does not mean such an equation has no solution. Later, you will learn methods that work even when integer factoring does not.

Key takeaways

To solve a quadratic equation by factoring:

Write it with $0$ on one side.
Factor the other side completely.
Apply the zero-product property.
Solve each resulting linear equation.
Check every solution in the original equation.

The central idea is:

A \cdot B = 0

means

A = 0

B = 0.

Factoring turns a quadratic into that useful product form. A GCF, a factorable trinomial, a difference of squares, or a perfect-square trinomial can all reveal the factors needed to solve the equation.

You have now completed the core factoring-and-solving work of this algebra module. Next, the course begins working with functions: recognizing when a relation is a function and evaluating expressions using function notation.