# Solving Quadratic Equations by Factoring

by Ron Kurtus (updated 18 January 2022)

One method of solving a quadratic equation is by factoring it into two linear equations and then solving each of those equations. For example, the first expression in the equation x2 + 8x + 15 = 0 can be factored into (x + 3)(x + 5), and then those two factors can then be readily solved for x.

The method requires that you first put the equation in the form of ax2 + bx + c = 0 and then try to factor ax2 + bx + c. Each factor can then be set to 0 and solved for x.

Some quadratic equations are not readily factored. In such a case, you can try to solve the equation by the completing the square method or by using the quadratic equation formula.

Questions you may have include:

• What is the general method to solve by factoring?
• What are some examples of solving by factoring?
• When is solving by factoring ineffective?

This lesson will answer those questions.

## General solution

The standard form of a quadratic equation of one variable is ax2 + bx + c = 0, where:

• x is the single variable
• a, b and c are constants
• a is not equal to 0

Factoring the quadratic expression ax2 + bx + c consists of breaking the expression into two sub-expressions in the form of (dx + e)(fx + g). The quadratic equation then is:

(dx + e)(fx + g) = 0

Set each expression equal to 0 and solve them for x to get our two solutions:

dx + e = 0

dx = −e

x = −e/d

Likewise

fx + g = 0

x = −g/f

## Factoring examples

Consider the quadratic equation x2 + 8x + 15 = 0.

Seeing that 3 * 5 = 15 and 3 + 5 = 8, you can factor the expression x2 + 8x + 15 into (x + 3)(x + 5). Thus, the equation is:

(x + 3)(x + 5) = 0

Since (x + 3)*0 = 0 and 0*(x + 5) = 0, you can set both expressions equal to zero and solve:

x + 3 = 0

x + 5 = 0

The solutions of the equation are:

x = −3 and x = −5

### Example 2

Another example of solving by factoring is the equation:

x2 − 16 = 0

You can see that 16 = 42 and:

x2 − 42 = 0

Thus:

(x + 4)(x − 4) = 0

x + 4 = 0 and x − 4 = 0

Solutions:

x = − 4 and x = 4

or

x = ± 4

### Example 3

Consider the equation:

2x2 − 3x − 14 = 0.

You can factor the expression 2x2 − 3x − 14 into (2x − 7)(x + 2).

(2x − 7)(x + 2) = 0

Thus:

2x − 7 = 0 and x + 2 = 0

The solutions are:

x = 7/2 = 3½

and

x = −2

## When solving by factoring does not work

There are some quadratic equations where solving by factoring is not effective. Consider the equation

x2 − 5x + 3 = 0

You really can't factor x2 − 5x + 3 with rational numbers.

In such a case, you can try solving by the Completing the Square method or the Quadratic Formula method.

## Summary

One method of solving a quadratic equation is by factoring it into two linear equations and then solving each of those equations. For example, the first expression in the equation x2 + 8x + 15 = 0 can be factored into (x + 3)(x + 5), and then those two factors can then be readily solved for x.

The method requires that you first put the equation in the form of ax2 + bx + c = 0 and then try to factor ax2 + bx + c. Each factor can then be set to 0 and solved for x.

Some quadratic equations are not readily factored. In such a case, you can try to solve the equation by the completing the square method or by using the quadratic equation formula.

Do the best you can

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