Solve by Factoring Lessons
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:
- 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
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 – 5x = -6
Now, we will add 6 to each side.
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:
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:
and
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, 3
Solve by Factoring: Why does it work?
Examine the equation below:
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
Step 1
Move all terms to the left side of the equation. We do this by subtracting 45x from each side.
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.
Now, (x2 – 9) can be factored as a difference between two squares.
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.
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.
Third Example
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.
Next, we factor a trinomial.
Finally, we factor the binomial (x2 – 100) as a difference between two squares.
Step 3
We proceed by setting each of the four factors equal to zero, resulting in four new equations.
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:
This may be abbreviated as
Solve By Factoring Resources
Equation Calculator – Solve By Factoring Practice Problems / Worksheet
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