Chapter 5, Section 5



Chapter 5, Section 5

Factoring

The Greatest Common Factor

Factoring by grouping

TEST Three

Difference of Two Squares/Cubes – 5.6

Factoring Trinomials – 5.7

Solving Equations by factoring – 5.9

TEST Four

Now, the reverse process to multiplication: Factoring.

Just like with numbers: multiply/factor

multiply: 2(3)(5)

factor: 12

factor: [pic]

We’ll do this in steps.

First step is always factor out the Greatest Common Factor if there is one, then you apply steps that we’ll show you in the upcoming sections.

Finding the Greatest Common Factor:

Take each term: factor it to primes

identify factors that are common to each

multiply these common factors to form the GCF

factor it out of the polynomial to result in a multiplied form

Example

[pic]

I like to use stacks or tables to organize the prime factors initially so I’ll put in a table:

|[pic] |2 |3 |x |x |x |y | | |

|[pic] | | | | | | | | |

|[pic] | | | | | | | | |

|[pic] | | | | | | | | |

|GCF | | | | | | | | |

So the GCF is _______________________________________________.

Sometimes it is not necessary to go to the trouble to make a chart. Look at

(4xy ( 8x

we can write it out factored and see the GCF.

The GCF is ______________________.

Once the GCF is factored out you will use the techniques from 5.6 and 5.7 to finish the factoring if need be.

Looking at another type of polynomial, let’s explore more factoring.

If you have four terms, you can try factoring by grouping to get it from added form to multiplied form:

Look at ( a + x) ( b + y). When you multiply these using FOIL, you get

ab + ay + xb + xy.

The O term and the I term are not “like” so they don’t combine at all. In order to factor this expression, you use a clever technique called “grouping”.

There’s no GCF for the whole expression but we can group the terms and find a

“semi – GCF”:

( ab + ay ) + (xb + xy) associate pairs of terms that have a factor in common

a ( b + y) + x ( b + y) factor out the common terms for each associated pair

notice that ( b + y) is now a common term and factor it out, getting

(a + x) ( b + y)

Let’s do another following the steps above.

Looking at xy ( xq ( py + pq, notice that there are 4 terms – the O and the I terms don’t combine. There’s no obvious common factor but if we’re careful we can group terms and find a common binomial factor.

(xy ( xq) + ((py + pq) note how careful I am with that “minus sign”…I made it go with the number and put a home made + sign inbetween my grouped terms

x (y ( q) + p ( (y + q) this doesn’t work out. There’s no common binomial factor.

rewrite the problem as

(xy ( xq) + ((py ( ( pq) now you can see that if I factor out a (p from the second group I’ll get a common binomial factor of ( x (q)

x (y ( q) + (p (y ( q)

( y ( q) ( x ( p) done

Let’s try this one: xy + xm ( 3y ( 3m

And this one: PQ ( PT ( 5Q + 5T

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