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# Solving Quadratic Equations by Factoring

by Ron Kurtus (updated 18 January 2022)

One method of* solving a quadratic equation* is by

*it into two linear equations and then solving each of those equations. For example, the first expression in the equation*

**factoring****x**can be factored into

^{2}+ 8x + 15 = 0**(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 **ax ^{2} +
bx + c = 0** and then try to factor

**ax**. Each factor can then be set to

^{2}+ bx + c**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 **ax ^{2} +
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 **ax ^{2} + 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 **x ^{2} + 8x + 15 = 0**.

Seeing that **3 * 5 = 15** and **3 + 5 = 8**, you can factor the expression **x ^{2} + 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 = −3andx = −5

### Example 2

Another example of solving by factoring is the equation:

x^{2}− 16 = 0

You can see that **16 = 4 ^{2}** and:

x^{2}− 4^{2}= 0

Thus:

(x + 4)(x − 4) = 0

x + 4 = 0andx − 4 = 0

Solutions:

x = − 4andx = 4

or

x = ± 4

### Example 3

Consider the equation:

2x.^{2}− 3x − 14 = 0

You can factor the expression **2x ^{2} − 3x − 14** into

**(2x − 7)(x + 2)**.

(2x − 7)(x + 2) = 0

Thus:

2x − 7 = 0andx + 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

x^{2}− 5x + 3 = 0

You really can't factor **x ^{2} − 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

*it into two linear equations and then solving each of those equations. For example, the first expression in the equation*

**factoring****x**can be factored into

^{2}+ 8x + 15 = 0**(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 **ax ^{2} +
bx + c = 0** and then try to factor

**ax**. Each factor can then be set to

^{2}+ bx + c**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

## Resources and references

### Websites

**Factoring Quadratics** = MathIsFun

**Factoring Quadratics** - PurpleMath.com

### Books

(Notice: The *School for Champions* may earn commissions from book purchases)

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## Solving Quadratic Equations by Factoring