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