1. 6.4 Factoring and Solving Polynomial Equations

2. Factor Polynomial Expressions
In the previous lesson, you factored various polynomial expressions.
Such as:
x3 – 2x2 =
x4 – x3 – 3x2 + 3x = =
Grouping – common factor the first two terms and then the last two terms.
Common Factor
x2(x – 2)
x(x3 – x2 – 3x + 3)
x[x2(x – 1) – 3(x – 1)]
Common Factor
x(x2 – 3)(x – 1)

3. Solving Polynomial Equations
The expressions on the previous slide are now equations:
y = x3 – 2x2 and y = x4 – x3 – 3x2 +3x
To solve these equations, we will be solving for x when y = 0.

4. Solve y = x3 – 2x2 0 = x3 – 2x2 0 = x2(x – 2) x2 = 0 or x – 2 = 0
Let y = 0
y = x3 – 2x2
0 = x3 – 2x2
0 = x2(x – 2)
x2 = 0 or x – 2 = 0
x = 0 x = 2
Therefore, the roots are 0 and 2.
Common factor
Separate the factors and set them equal to zero.
Solve for x

5. Solve Let y = 0 y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x
x = 0 or x – 1 = 0 or x2 – 3 = 0
x = x = x =
Therefore, the roots are 0, 1 and ±1.73
Let y = 0
Common factor
Group
Separate the factors and set them equal to zero.
Solve for x

6. The Quadratic Formula
For equations in quadratic form: ax2 + bx + c = 0, we can use the quadratic formula to solve for the roots of the equation.
This equation is normally used when factoring is not an option.

7. Using the Quadratic Formula
Solve the following cubic equation:
y = x3 + 5x2 – 9x
0 = x(x2 + 5x – 9)
x = 0 x2 + 5x – 9 = 0
We can, however, use the quadratic formula.
Can this equation be factored?
We still need to solve for x here. Can this equation be factored?
YES it can – common factor.
Remember, the root 0 came from an earlier step.
No. There are no two integers that will multiply to -9 and add to 5.
a = 1
b = 5
c = -9
Therefore, the roots are 0, 6.41 and

8. Factoring Sum or Difference of Cubes
If you have a sum or difference of cubes such as a3 + b3 or
a3 – b3, you can factor by using the following patterns.
Note: The first and last term are cubed and these are binomials.

9. Example
Factor x
Note: This is a binomial. Are the first and last terms cubed?
x = (x)3 + (7)3
= ( )( ) x 7 x2 7x 49

10. Example Factor 64a4 – 27a = a(64a3 – 27)
Note: Binomial. Is the first and last terms cubes?
= a( (4a)3 – (3)3)
Note:
= a( )( ) 4a 3
16a2
12a 9

11. Factor by Grouping
Some four term polynomials can be factor by grouping.
Example Factor 3x3 + 7x2 +12x + 28
Step 1 Pair the terms.
Step 2 Factor out common factor from each pair.
Identical factors
Step 3 Factor out common factor from each term.

12. Example Factor 3x3 + 7x2 -12x - 28 Step 1
Note: Subtraction is the same as adding a negative
Step 2
Step 3
Note: This factor can be further factored

13. Solving Polynomial Equations
Solve
Set equation equal to zero.
Factor.
Set each factor equal to zero and solve.