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# Fractions, fractions, fractions!!!

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## Rita Stormm

on 18 November 2013

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#### Transcript of Fractions, fractions, fractions!!!

Fractions are a Hoot!
Definition: The GCF is when a number can be factored into two or more numbers in a term or equation.
Difference Of Square Binomials: An expression where two perfect square terms are being subtracted. It follows a pattern of, a^2-b^2=(a+b)(a-b)
Definition: Where the first and last terms are perfect squares and the middle term is twice the product of the square roots of those terms, you have a perfect square trinomial that can be factored using the following pattern:
a^2 + 2ab + b^2 = (a + b)^2 OR a^2 - 2ab + b^2 = (a - b)^2
Greatest Common Factors : (GCF)
For

Example
: Factor Completely 12vw^4+8v^3w^6+4v^7w^8 .

Special Products: (Difference of Squares Binomials)
Special Products: Perfect Square Trinomials
Factoring Trinomials
Factoring by Grouping: Four-term polynomials
Sum & Difference of Cubes.
Summary.

First, make a list of all the factors for each term.
List:
Factors of 12 Factors of 28 Factors of 4
1*12 1*28 1*4
2*6 2*14 2*2
3*4 4*7

As we can see, each of the terms have one factor in common which is 4. Therefore the GCF is: 4.

Now lets set up our equation
(4)(3)vw^4+(4)(7)v^3w^6+(4)(1)v^7w^8

4(3vw^4+ 7v^3w^6+ v^7w^8)

Continuing GCF
Next we have to find the GCF amongst the variables.
4(3vw^4+7v^3w^6+(4)(1)v^7w^8)
4[(3)(v)(www)+(7)(v v v)(wwwwww)+(vvvvvvv)(wwwwwwww)
4vw^4(3+7v^2w^2+v^6w^4)
Finally we check the factors by distributing
4vw^4(3+7v^2w^2+v^6w^4)
12vw^4+28v^3w^6+4v^7w^8

An Ex: (R+2)(R-2). First you distribute, (R+2)(R-2) = R^2 - 2r + 2r -4. Now we simplify which will give us,
R^2 - 4.
Another ex: 3d^4- 363d^2. First Find the GCF for each term, GCF is 3 so, 3(d^4 - 121d^2). Now we factor out the rest of the equation. 3[dddd-121 (dd)] = 3(dd)[(dd)-121) = 3d^2(d^2-121).
The final answer is: 3d^2(d+11)(d-11). Don't forget to check your answer by first distributing, combine like terms,and then distribute the GCF monomial.
Ex. d^10 + 18d^5 + 81
First we must find the GCF, since there is no number for the first term there is an understood 1 there. Next we need to find the square roots of the first & last terms:
d^10 + 18d^5 +81 = (d^5+9)^2
*
Don't forget to follow the pattern: a^2 + 2ab +b^2=(a+b)^2
Now we check our answer by distributing,
(d^5+9)(d^5-9)= d^10 +9d^5 +9d^5 +81
Now finally, combine like terms:

d^10 +18d^5 +81
Lets factor this example: X^2 + 3x -40
First step is to find the GCF, but in this example there is no number or variable (that is larger ten 1) that all terms have in common. So there is not GCF.
Second step is to determine the binomial factors. Since, X^2 is the first term we'll start with it.
(X )(X )
Now we need to find the factors of -40 and find a pair that adds to the middle term 3.
Now we see that -5 and 8 meet the requirements. So we will place them in our binomial.
Our factored trinomial is(x-5)(x+8)
*Don't forget to check by distributing and then combining like terms!
!!!Reminder!!!

A polynomial that can't be factored is called a prime.
Prime: a number or expression that can be divided by only one and itself.

Factoring by Grouping: Is a method in which pairs of terms are grouped together to factor the entire polynomial.
Lets look at this equation: x^2 +xy - 2x - 2y

Lets start by looking at the first two terms,
x^2 +xy. If this was the only expression how can it be factored? It can be factored by x(x+y).
Now lets look at the last two terms, -2x-2y. How do we factor it? -2(x+y). Our last two answers have something in common, (x+y).
These two answers have a common factor, so we can factor them out which will give us our binomial, (x-2)(x+y).

Continuing Factoring by Grouping Trinomials
1. For factoring Trinomials we first need to find the GCF of all the terms
2. Split the middle term
3. Factor by grouping
Ex: Factor completely, 6a^2 + 16a + 8.
We should always look for GCF before factoring any trinomials. We can see that 6, 16, and 8 are all divisible by 2.
6a^2 + 16a + 8
(2)(3)a^2 + (2)(8)a + (2)(4)
2(3a^2 + 8a + 4)
The GCF is 2, so we can leave it where its at for the rest of the problem. Now lets continue factoring, 2(3a^2 + 8a + 4) = 3*4 = 12.
Now lets find the factors of 12 that total the middle coefficient of 8.
The Pair, 2 and 6 both add up to the middle
coefficient. So lets rewrite the middle term.
2(3a^2 + 8a + 4) = 2(3a^2 + 2a + 6a + 4)
For our last step we need to separate this polynomial into two groups and figure out the factored solution for this equation,

Definitions to remember!
Geometrically: a perfect cube or just cube, is a three-dimensional object whose length, width, and height are all the same measurement.
Algebraically: perfect cube is the product of three identical numbers or variables multiplied.
Cube Root: A cube root is the number of variable which results from separating an expression into three identical groups which are multiplied together. Ex: cube root of 8 is 2.
Sums of cubes
a^3+b^3=(a+b)(a^2-ab+b^2)
Difference of Cubes
a^3-b^3=(a-b)(a^2+ab+b^2)

When dealing with Prime Trinomials
Ex: Factor completely: 2a^2- a + 3.
1. First we must find the GCF. It looks like there isn't a GCF that is greater then 1 so we will go to step 2 which is to split the middle term. Since there isn't a middle term we will ave to create one, 2a^2 - a - 3 2a^2 - 1a + 3
2*3 ?*?=-1
? * ?= 6
Now we list the factors of 6:
Since there is no pair
that will satisfy both
sides of the chart, this
trinomial cannot be
factored. It is a Prime.

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