# Factoring Fun!

### By: Kathryn Walsh

## How Many Ways to Factor?!?

## Factor by Grouping!

## Factor by Grouping - Steps

2.) Next, group the first and second terms together and the third and fourth terms together.

3.) Factor out a GFC from each group if possible.** Be sure that the binomials in the parentheses are the same among the two groups.** In the second group, you may have a choice of whether or not to factor out a negative. If it is necessary to factor out a negative to get the binomials to correspond, make sure you do so.

4.) Now that you have corresponding binomials, the GCF's of the two binomials are grouped together to form another binomial. Now, you should have two binomials, one consisting of the original and corresponding binomials and one that is constructed of GCF's.

5.) Finally, using your binomials solve for x, as you would in any equation, by setting the binomials equal to zero.

6.) Don't forget to check your work!! Great job :)

## Factor By Grouping Example

## Practice Problems

1.) x^3-5x^2+3x-15

2.) 4x^2-20x-3xy-15y

3.) 3x^3-6x^2+15x-30

4.) x^2+ab-ax-bx

## Factor 2 Cubes!

## Factor 2 Cubes- Steps

**Do not to forget to include the GCF as part as your final answer.**

2.) Rewrite the original problem as a difference of two perfect cubes.

3.) Use the following sayings to help write the answer.

a. "Write What You See"

b. "Square- Multiply- Square"

c. "Same, Different, End on a Positive"

3.) Use these three pieces to write the final answer. Use the equation given and plug in your values.

## Factor 2 Cubes Example

## Practice Problems

Here are some practice problems, give them a try!

2.) 8x^3 +125

3.) 125x^3 + 216y^3

4.) 81x^3 + 192y^3

## The Factor Theorem!

**if and only if**f(a)=0. This means that if you synthetically divide by (x=a) or algebraically solve for a polynomial by (x=a), and get zero for a remainder, then x-a is a factor of the polynomial. Below there are examples of the factor theorem using both synthetic division and polynomials to solve.

## The Factor Theorem - Steps

**If synthetically dividing:**

1.) Using the polynomial given, begin by factoring solving for x.

2.) Use one solution to synthetically divide the polynomial.

3.) Divide the polynomial by one solution of x using synthetic division.

4.) If the remainder of the problem** is zero**, (x-a), a being one solution, is a factor of the polynomial and x=a is equal to zero.

5.) If the remainder of the problem **is not zero**, (x-a), a being your solution, is not a factor of the polynomial and x=a does not equal zero.

**If solving algebraically: **

1.) Using the polynomial given, begin by factoring and solving for x.

2.) Use one solution to solve algebraically.

3.) Plug in your solution, or (a), in so that f(a) becomes f(c) and solve for f(a)

4.) If f(a) **is equal to zero,** (x-a) must be a factor of the given polynomial and x=a must be a solution.

5.) If f(a) **is not equal to zero**, (x-a) is not a factor of the polynomial and x=a is not a solution.

## Factor Theorem Example

## Practice Problems

Here are some practice problems, give them a try!

1.) 5x^4 + 16x^3 - 15x^2 + 8x + 16

2.) x^2 - 3x - 4

3.) 2x^3 - x^2 - 7x +2

4.) 3x^3 + x^2 + x - 5

## Summary of Concepts

## Upcoming Assessments

Test #2 - Thursday December 10th

## For More Factor Fun...

## Citations

- “Factoring By Grouping.”
*Factoring by Grouping*. Web. 1 Dec. 2015. <http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html> - “Factoring a Sum Of Cubes.”
*Factoring a Sum of Cubes*. Web. 1 Dec. 2015. <http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/sum_of_cubes/sum_of_cubes.html> - “The Factor Theorem.”
*The Factor Theorem*. Web. 1 Dec. 2015. <http://www.purplemath.com/modules/factrthm.htm> - Youtube
- Google Images