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Chapter 1: Factoring and Quadratic Equations
Section 1.1: The Greatest Common Factor
#1-16: Factor out the GCF.
1) 2y – 10
Both 2 and 10 are divisible by 2, so 2 is part of the common factor. Both terms don’t have a y,
so there is no letter in the GCF. Just write a 2 to the left of a parenthesis, and divide the
numbers by 2.
This is an okay answer 2(1y – 5) , however it isn’t necessary to leave a 1 in front of the y.
Solution: 2(y – 5)
3) 14x3 – 7x2 + 7x
Each number is divisible by 7, so 7 is part of the common factor. Each term has an x, and the
smallest power of x that occurs is x1, so x1 is also part of the GCF.
The GCF is 7x
Divide each number by 7, and take one x away from each term to get your answer.
Solution: 7x(2x2 – x + 1)
5) b5 – 3b4 + b3
Each term has a b and the smallest power of b that occurs in the problem is b 3, hence b3 is the
GCF. I will write a b3 to the left of a parenthesis, and take away 3 b’s from each term inside the
parenthesis.
Solution: b3(b2 – 3b + 1)
7) 12x4 – 3x3
Each term has an x, and the smallest power of x that occurs is x3. x3 Is part of the GCF, also each
number is divisible by 3, so 3 is also part of the GCF.
Divide the numbers by 3, and take away three x’s to get the answer.
Solution: 3x3(4x – 1)
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Chapter 1: Factoring and Quadratic Equations
9) 4x3y + 12x2y3
4 divides evenly into both numbers and it is part of the GCF.
Each term has an x and the smallest power of x that occurs is x2, so x2 is part of the GCF
Each term has a y and the smallest power of y that occurs is y, so y is part of the GCF.
The GCF is 4x2y
I will divide each number by 4, take 2 x’s and 1 y away from each term.
Solution: 4x2y(x+3y2)
11) 16a4b2 – 18ab3
2 divides evenly into both numbers and it is part of the GCF.
Each term has an (a) and the smallest power of x that occurs is a1, so a is part of the GCF
Each term has a b and the smallest power of b that occurs is b2, so b2 is part of the GCF.
The GCF is 2ab2
I will divide each number by 2, take an a and 2 b’s away from each term
Solution: 2ab2(8a3 – 9b)
13) 12xyz3 – 14x2y3 – 2xz
Each number is divisible by 2, so 2 is part of the GCF.
the only letter in each term is a x. x1 is part of the GCF.
I will take one x away from each term and divide each number by 2.
Solution: 2x(6yz3 – 7xy3 – z)
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Chapter 1: Factoring and Quadratic Equations
15) 16r2st3 – 4r3st2 + 12rst
4 divides evenly into all of the numbers and it is part of the GCF.
Each term has an r and the smallest power of r that occurs is r1, so r is part of the GCF
Each term has a s and the smallest power of s that occurs is s, so s is part of the GCF.
Each term has a t and the smallest power of t that occurs is t, so s is part of the GCF.
The GCF is 4rst
I will divide each number by 4, take an r, s and t away from each term
Solution: 4rst(4rt2 – r2t + 3)
#17-26: Factor out a (-1) from each polynomial.
17) -x + 2
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(x – 2) or -(x – 2) the second answer is considered better
19) -x – 3
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(x + 3) or -(x + 3) the second answer is considered better
21) -5x + 9
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(5x – 9) or -(5x – 9)
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Chapter 1: Factoring and Quadratic Equations
23) -3x + 6y – 7
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(3x – 6y + 7) or –(3x – 6y + 7)
25) -4x + 6z + 11s
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(4x – 6z – 11s) or –(4x – 6z – 11s)
#27-38: Factor each polynomial by factoring out the opposite of the GCF.
27) -4x3 – 12x2
I will factor out the GCF of -4x2
I will divide each number by -4 and change my signs and take 2 x’s away from each term.
Solution: -4x2(x + 3)
29) -12x3 + 4x2 – 8x
I will factor out the GCF of -4x
I will divide each number by -4 and change my signs and take an x away from each term.
Solution: -4x(3x2 – x + 2)
31) -3z + 6z3
I will factor out the GCF of -3z
I will divide each number by -3 and change my signs and take an z away from each term.
Solution: -3z(1 – 2z2)
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Chapter 1: Factoring and Quadratic Equations
33) -14a4b2 – 6a2b
I will factor out the GCF of -2a2b
I will divide each number by -2 and change my signs and take 2 a’s and 1 b away from each
term.
Solution: -2a2b(7a2b + 3)
35) -8xyz3 – 4x2y3 + 2xyz
I will factor out the GCF of -2xy
I will divide each number by -2 and change my signs and take an x and a y away from each term.
Solution: -2xy(4z3 + 2xy2 – z)
37) -16r2st3 – 4r3st2 – 12rst2
I will factor out the GCF of -4rst2
I will divide each number by -4 and change my signs and take 1 r, 1 s and 2 t’s away from each
term.
Solution: -4rst2(4rt + r2 + 3)
#39 – 52: Factor out the GCF
39) x(x-4) + 3(x-4)
There is an (x – 4) on each side of the plus sign. It is the common factor. When I take the (x-4)
away from the problem I am left with x + 3, and that is what I will put inside a parenthesis.
Solution: (x – 4)(x + 3)
41) x2(y – 2) – 3(y – 2)
There is an (y – 2) on each side of the minus sign. It is the common factor. When I take the
(y - 2) away from the problem I am left with x2 - 3, and that is what I will put inside a
parenthesis.
Solution: (y – 2)(x2 – 3)
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Chapter 1: Factoring and Quadratic Equations
43) 3y(z + 1) – 4(z + 1)
There is an (z + 1) on each side of the minus sign. It is the common factor. When I take the
(z + 1 ) away from the problem I am left with 3y - 4, and that is what I will put inside a
parenthesis.
Solution: (z + 1)(3y - 4)
45) x(3x – 4) – 2(3x – 4)
There is an (3x - 4) on each side of the minus sign. It is the common factor. When I take the
(3x - 4) away from the problem I am left with x - 2, and that is what I will put inside a
parenthesis.
Solution: (3x – 4)(x – 2)
47) 3x(2x – 7) + 4(2x – 7)
There is an (2x - 7) on each side of the minus sign. It is the common factor. When I take the
(2x - 7) away from the problem I am left with 3x + 4, and that is what I will put inside a
parenthesis.
Solution: (2x - 7)(3x + 4)
49) 2x2(3x – 5y) – 5(3x – 5y)
There is an (3x – 5y) on each side of the minus sign. It is the common factor. When I take the
(3x-5y) away from the problem I am left with 2x2 - 5, and that is what I will put inside a
parenthesis.
Solution: (3x – 5y)(2x2 – 5)
51) 8y(y – 5) – 9(y – 5)
There is an (y - 5) on each side of the minus sign. It is the common factor. When I take the
(y - 5) away from the problem I am left with 8y - 9, and that is what I will put inside a
parenthesis.
Solution: (y – 5)(8y – 9)
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