# Algebra: Solving Quadratics by Factoring

## Solving Quadratics by Factoring

If you can transform an equation into a factorable quadratic polynomial, it is very simple to solve. Even though this technique will not work for all quadratic equations, when it does, it is by far the quickest and simplest way to get an answer. Therefore, unless a problem specifically tells you to use another technique to solve a quadratic equation, you should try this one first. If you get a prime (unfactorable) polynomial, you can always shift to one of the other techniques you'll learn later in this section.

To solve a quadratic equation by factoring, follow these steps:

##### How'd You Do That?

If you have the equation (*x* - *a*)(*x* - *b*) = 0, Step 3 tells you to change that into the twin equations:

*x* - *a* = 0 or *x* - *b* = 0

Are you wondering why that's allowed? It's thanks to something called the *zero product property*.

Think about it this way: If two things are multiplied togetherin this case the quantities (*x* - *a*) and (*x* - *b*)and the result is 0, then at least one of those two quantities actually has to be equal to 0! There's no way to multiply two or more things to get 0 unless at least one of them equals 0.

**Set the equation equal to 0**. Move*all*of the terms to the left side of the equation by adding or subtracting them, as appropriate, leaving only 0 on the right side of the equation.**Factor the polynomial completely**. Use one of the techniques you learned in Factoring Polynomials to factor; remember to always factor out the greatest common factor first.**Set each of the factors equal to 0**. You're basically creating a bunch of tiny, little equations whose left sides are the factors and whose right sides are each 0. It's good form to separate these little equations with the word "or," because any one of them could be true.**Solve the smaller equations and check your answers**. Each of the solutions to the tiny, little equations is also a solution to the original equation. However, to make sure they actually work, you should plug them back into that original equation to verify that you get true statements.

The hardest part of this technique is actually the factoring itself, and since that's not a new concept, this procedure is very simple and straightforward.

**Example 1**: Solve the equations, and give all possible solutions.

- (a)
*x*^{2}- 6*x*+ 9 = 0 **Solution**: Since this equation is already set equal to 0, start by factoring the left side.- (
*x*- 3)(*x*- 3) = 0 - Now set each factor equal to 0.
*x*- 3 = 0 or*x*- 3 = 0*x*= 3 or*x*= 3- Well, since both factors were the same, both solutions ended up equal, so the equation
*x*^{2}- 6*x*+ 9 = 0 only has one valid solution,*x*= 3. When you get an answer like this, which appears as a possible solution twice, it has a special nameit's called a*double root*.

##### Critical Point

A **double root** is a repeated solution for a polynomial equation; it's the result of a repeated factor in the polynomial.

##### You've Got Problems

Problem 1: Give all the solutions to the equation 4*x*^{3} = 25*x*.

- Check to make sure that 3 is a valid answer by plugging it back into the original equation.
*x*^{2}- 6*x*+ 9 = 0- 3
^{2}- 6(3) + 9 = 0 - 9 - 18 + 9 = 0
- 0 = 0
- There's no doubt that 0 = 0 is a true statement, so you got the answer right.
- (b) 3
*x*^{2}+ 10*x*= -4*x*+ 24 **Solution**: Your first job is to set this equal to 0; to accomplish this, add 4*x*to and subtract 24 from both sides.- 3
*x*^{2}+ 14*x*-24 = 0 - Factor the trinomial using the bomb method discussed in Factoring Polynomials. The two mystery numbers you're looking for are -4 and 18.
- 3
*x*^{2}+ (-4 + 18)*x*- 24 = 0 3*x*^{2}-4*x*+ 18*x*- 24 = 0 *x*(3*x*- 4) + 6(3*x*- 4) = 0- (3
*x*- 4)(*x*+ 6) = 0 - Set each factor equal to 0 and solve.
- 3
*x*- 4 = 0 or*x*+ 6 = 0 *x*=^{4}_{3}or*x*= -6- Both of these answers work when you check them.

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

You can purchase this book at Amazon.com and Barnes & Noble.