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 x^{2} + 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:
 Move all terms to one side of the equation, usually the left, using addition or subtraction.
 Factor the equation completely.
 Set each factor equal to zero, and solve.
 List each solution from Step 3 as a solution to the original equation.
First Example
x^{2} + 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.
x^{2} + 3x – 8x = 8x – 8x – 6
x
^{2} – 5x = 6
Now, we will add 6 to each side.
x^{2} – 5x + 6 = 6 + 6
x
^{2} – 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.
x^{2} + 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 nonzero 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
5x^{3} = 45x
Step 1
Move all terms to the left side of the equation. We do this by subtracting 45x from each side.
5x^{3} – 45x = 45x – 45x
5x
^{3} – 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(x^{2} – 9) = 0
Now, (x^{2} – 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
3x^{4} – 288x^{2} – 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(x^{4} – 96x^{2} – 400) = 0
Next, we factor a trinomial.
3(x^{2} + 4)(x^{2} – 100) = 0
Finally, we factor the binomial (x^{2} – 100) as a difference between two squares.
3(x^{2} + 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. x^{2} + 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. (x^{2} + 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
Solve By Factoring Resources
Equation Calculator – Solve By Factoring Practice Problems / Worksheet

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Now, let's delve into the key concepts discussed in the provided article on solving equations by factoring:

Equation Solving Fundamentals: The article emphasizes the fundamental tools for solving equations, such as addition, subtraction, multiplication, and division. These methods are effective for linear equations but may fall short when dealing with equations containing variables with exponents.

Introduction of Factoring: Factoring is introduced as a crucial technique for solving equations with variables carrying exponents. The example equation x^2 + 3x = 8x – 6 is used to illustrate the Solve by Factoring process, which involves four major steps.

Solve by Factoring Process: The Solve by Factoring process includes the following steps:
 Step 1: Move all terms to one side of the equation using addition or subtraction.
 Step 2: Factor the equation completely.
 Step 3: Set each factor equal to zero and solve.
 Step 4: List each solution as a solution to the original equation.

Application of Factoring: The article provides a rationale for why factoring works. It demonstrates that if the product of two values is zero, then at least one of the values must be zero. This property is exploited when solving equations by factoring.

Examples of Equation Solving by Factoring:

First Example (x^2 + 3x = 8x – 6):
 Demonstrates the four steps of the Solve by Factoring process.
 Factors the trinomial (x – 2)(x – 3) = 0 and finds solutions.

Second Example (5x^3 = 45x):
 Factors the left side completely: 5x(x + 3)(x – 3) = 0.
 Sets each factor equal to zero and solves the subproblems.

Third Example (3x^4 – 288x^2 – 1200 = 0):
 Factors the expression and sets each factor equal to zero.
 Identifies invalid solutions and addresses complex solutions.


Resource Mention: The article provides additional resources, including an equation calculator for solving by factoring and practice problems. It also hints at the next lesson on quadratic equations.
In summary, the article comprehensively covers the process of solving equations by factoring, illustrating the application of algebraic techniques to handle equations with varying complexities.