Homework Help: Mathematics: Algebra: Solve By Factoring

Several previous lessons explain the techniques used to factor expressions. This lesson focuses on an imporatant application of those techniques − solving equations.

Why solve by factoring?

The most fundamental tools for solving equations are addition, subtraction, multiplication, and division. These methods work well for equations like x + 2 = 10 - 2x   and   2(x - 4) = 0.

But what about equations where the variable carries an exponent, like x² + 3x = 8x - 6? This is where factoring comes in. We will use this equation in the first example.

The Solve by Factoring process will require four major steps:

1. Move all terms to one side of the equation, usually the left, using addition or subtraction.
2. Factor the equation completely.
3. Set each factor equal to zero, and solve.
4. List each solution from Step 3 as a solution to the original equation.

First Example

x² + 3x = 8x - 6

Step 1

The first step is to move all terms to the left using addition and subtraction. First, we will subtract 8x from each side.

x² + 3x - 8x = 8x - 8x - 6
x² - 5x = -6

Now, we will add 6 to each side.

x² - 5x + 6 = -6 + 6
x² - 5x + 6 = 0.

With all terms on the left side, we proceed to Step 2.

Step 2

We identify the left as a trinomial, and factor it accordingly:

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

We now have two factors, (x - 2) and (x - 3).

Step 3

We now set each factor equal to zero. The result is two subproblems:

x - 2 = 0

and

x - 3 = 0

Solving the first subproblem, x - 2 = 0, gives x = 2. Solving the second subproblem, x - 3 = 0, gives x = 3.

Step 4

The final step is to combine the two previous solutions, x = 2 and x = 3, into one solution for the original problem.

x² + 3x = 8x - 6
x = 2, 3

Solve by Factoring: Why does it work?

Examine the equation below:

ab = 0

If you let a = 3, then logivally b must equal 0. Similarly, if you let b = 10, then a must equal 0.

Now try letting a be some other non-zero number. You should observe that as long as a does not equal 0, b must be equal to zero.

To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.

When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero. Therefore, we were able to create two equations and determine two solutions from this observation.

A Second Example

5x³ = 45x

Step 1

Move all terms to the left side of the equation. We do this by subtracting 45x from each side.

5x³ - 45x = 45x - 45x
5x³ - 45x = 0.

Step 2

The next step is to factor the left side completely. We first note that the two terms on the left have a greatest common factor of 5x.

5x(x² - 9) = 0

Now, (x² - 9) can be factored as a difference between two squares.

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

We are left with three factors: 5x, (x + 3), and (x - 3). As explained in the "Why does it work?" section, at least one of the three factors must be equal to zero.

Step 3

Create three subproblems by setting each factor equal to zero.

1.   5x = 0
2.   x + 3 = 0
3.   x - 3 = 0

Solving the first equation gives x = 0. Solving the second equation gives x = -3. And solving the third equation gives x= 3.

Step 4

The final solution is formed from the solutions to the three subproblems.

x = -3, 0, 3

Third Example

3x⁴ - 288x² - 1200 = 0

Steps 1 and 2

All three terms are already on the left side of the equation, so we may begin factoring. First, we factor out a greatest common factor of 3.

3(x⁴ - 96x² - 400) = 0

Next, we factor a trinomial.

3(x² + 4)(x² - 100) = 0

Finally, we factor the binomial (x² - 100) as a difference between two squares.

3(x² + 4)(x + 10)(x - 10) = 0

Step 3

We proceed by setting each of the four factors equal to zero, resulting in four new equations.

1.   3 = 0
2.   x² + 4 = 0
3.   x + 10 = 0
4.   x - 10 = 0

The first equation is invalid, and does not yield a solution. The second equation cannot be solved using basic methods. (x² + 4 = 0 actually has two imaginary number solutions, but we will save Imaginary Numbers for another lesson!) Equation 3 has a solution of x = -10, and Equation 4 has a solution of x = 10.

Step 4

We now include all the solutions we found in a single solution to the original problem:

x = -10, 10

This may be abbreviated as

x = ±10

Homework Help: Mathematics: Algebra

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