Lesson 6 Factoring Polynomials: Greatest Common Factor
GCF of a polynomial= The monomial with the highest coefficient and degree that divides evenly into ALL of the terms of the polynomial
4x2y2 + 6x3y3
4x2y2 = 2 2 x x y y
6x +12
6x3y3 =2 3 x x x y y y
­if there is no number or variable that will divide evenly into all of the terms, it cannot be factored by a GCF.
­To "factor a polynomial" means to write it in factored form. Sometimes this will require a number of steps.The first one will always be to look for the greatest common factor!
­If you multiply out the factored form you should get back the original polynomial, you can always check your answers this way.
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Factor each of the following by finding the greatest common factor:
1.7a2 ­ 21a
3. 5x3 + 15x2 + 35x
2. 9b3 + 81
4. 20m2n3 ­10m2n2 ­ 15m2n
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­Sometimes a common factor may be a binomial (will see this later too when we get into grouping). Pretend that the binomial is just a single letter and use it as the greatest common factor.
1.2x(x+y) ­3(x+y)
2. 5a(3x + y) + b(3x + y)
3. 7x(z ­ 2) ­ 5y(z ­ 2)
Lesson 7 Factoring Polynomials by Grouping:
Some polynomials do not have a common factor in all of their terms. These polynomials can sometimes be factored by grouping terms that DO have a common factor. Look to factor by grouping when there are 4 terms in a polynomial
1.If their are 4 terms. Try grouping the first two terms and the last two terms.
ac + bc + ad +bd
2c2 ­4c + 7ac ­ 14a
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2. Sometimes you may need to change the signs of so that you end up with the same common factors.
15a2b ­5a­ 6ab + 2
3. Sometimes you may need to rearrange the terms. 9x2 ­12z2 + 6x + 18xz2
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Practice Work
When asked to factor, the first step is to always look for a GCF!
Write a trinomial that has a GFC of 2xy to be passed in
Pg120 #11­29 odd
Look to factor by grouping whenever there are 4 terms in a polynomial
Pg 120 # 30­35
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Lesson 8 Factoring Short Trinomials:
Factoring ax2 + bx + c ; a=1
Short Trinomial­NO Number in front of x2
Do you see a pattern?
=(x+2)(x+3)
= x2 +3x +2x + 6
Factoring
Expanding (FOIL)
= x2 + 5x + 6
Is this always true?
(x + 5)(x + 2)
(x + 5)(x ­ 3)
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When factoring a short trinomial that looks like x2 + 7x +10:
1.Find two numbers that multiply to equal the 3rd term P= 10 2.And also add to equal the second term. S= 7 2x5=10, 2+5=7, (x+2)(x+5)
­Can always check your factors by multiplying them to get the original polynomial
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Factor: x2 + 8x + 12
1. To get the leading term of x2 each first term in the factors must be x.
(x )(x )
2. P= 12 S= 8
List all the possibilities and look at the sums
Product of 12
Sum
1. x2 ­7x + 12
2. x2 +12x + 32
3. x2 ­2x + ­24
4. z2 + 9z + 20
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Removing a Common Factor
3x2 + 3x ­ 18
Trinomials with Two Variables
x2 + 2xy ­15y2
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1. 4x2 + 4x ­ 24
2. x2 + 11xy + 28y2
3. 4z2 ­ 24zy +36y2
When Asked to factor
1.First step is still to look for a GCF
2.Then look at number of terms
­4 terms, factor by grouping
­3 terms­Short trinomial where a=1,(no number in front of x2)
factors of the third term have the sum of the second
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Practice Work
Pg 126 #1­8 odd
Pg 127 #19­26 odd
#27­49 odd
#51­55 odd
Assignment Due before 1st period Monday
Lesson 9 Factoring Long Trinomials:
Factoring ax2 + bx + c when a > 1
Long trinomials have a number in front of x2 : 6x2 +13x ­5
(2x + 3)(x ­ 5)
Factoring
=2x2 ­ 10x + 3x ­ 15
Expanding (FOIL)
=2x2 ­7x ­15
When: P= ­15 and S= ­7
It does not work out to be able to factor... So we must do something different when we have LONG trinomials.
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Factor : 2x2 ­ 7x­ 15
1. Look for a GCF
2. Look at the number of terms ­ 3
3. Is it a short trinomial or a long trinomial?
4. Number in front of x2, so it is a long trinomial
When factoring LONG Trinomials we break up the middle term, a process called Decomposition, and then factor by grouping.
ax2 + bx + c 1. Find two numbers whose:
Product= First coefficient (a) x Third coefficient (c)
Sum= middle coefficient (b)
2. Write the polynomial again, breaking up the middle term by using the two numbers you found as coefficients for x.
3. Finish factoring by grouping in twos.
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Factor:
1.6m2 + 13m ­ 5
2. 3x2 ­5x ­2
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Removing a Common Factor From a Long Trinomial
10z2 ­22z + 4
Long Trinomials with Two Variables
4x2 ­ 5xy ­6y2
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Practice Work
Pg 130 # 13­31 odd
Pg 131# 43­47 odd
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Lesson 10 Factoring Special Quadratics:
Differences of Squares:
a2 ­ b2 = (a + b)(a ­ b)
Look for two terms that are perfect squares with a minus sign in between them.
ex. 4a2 ­9b2
25x4 ­ 16y2
To Factor:
1. Find the square root of each term
2. Write a sum (+) and difference (­) of the square roots (2 factors)
1. 25x4 ­ 16y2
2. y2 ­16
3. 8x2 ­ 18y2
4. 49x2y8 ­ 16 x6y2z10
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Perfect Square Trinomials:
4x2 + 20x + 25 is a long trinomial
4x2 + 20x + 25 it is also a perfect square trinomial
How to tell:
1. The first and last terms are positive
2. The first and last terms are perfect squares
3. 2 times the square root of the first and last term = the middle term
4. To factor put the square of the first and last term in a brackets and put to the power of 2
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1. x2 + 6x + 9
2.4t2 + 4t + 1
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Practice Work
Pg. 133#1­11 odd
#27­43 odd
#56
If the area of a square is x2 + 10x +16
what are the dimensions (length and width)?
x2 + 10x +16
( ? )
L
W
( ? )
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Lesson 12 Factoring Review
When you are trying to factor polynomials, they will not always be grouped according to the different ways they would be factored. You need to be able to recognize which method to use for which type of polynomial.
2 Terms
Difference of Squares
Ex.4a2 ­ 9b2 = (2a + 3b)(2a ­ 3b)
or
it is Not Possible to factor any further
Factoring Flow Chart
3 Terms
1
Short Trinomial
Ex. x2 ­ 2x ­35 P=­35, S= ­2
=(x­7)(x+5)
2
3
GCF
Check the number of terms,
Always look for and decide if you can factor further
a GFC First
Long Trinomial
Ex. 6m2 ­13m ­5 P= ­30 S= ­2
=6m2 ­2m +15m ­5
=2m(3m­1)+5(3m­1)
=(3m­1)(2m+5)
or
it is Not Possible to factor any further
4 Terms
Grouping
6m2 ­ 2m + 15m ­5
=2m(3m­1)+5(3m­1)
=(3m­1)(2m+5)
or
it is Not Possible to factor any further
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Review
Assignments Due Today end of class
pg 140­141#37­83 odd, 90­93 (optional)
Factoring Practice sheet odd questions only
Factoring Test Tuesday April 28, 2009
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