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 – 7x + 1)
5) b5 – 3b4 + b3
Each term has a b and the smallest power of b that occurs in the problem is b3, 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: b2 – 3b + 1
7) 12x4 – 3x3
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.
3
Chapter 1: Factoring and Quadratic Equations
Solution: 4x2y(x+3y2)
11) 16a4b2 – 18ab3
13) 12xyz3 – 14x2y3 – 2xz
15) 16r2st3 – 4r3st2 + 12rst
#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
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)
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
4
Chapter 1: Factoring and Quadratic Equations
#27-38: Factor each polynomial by factoring out the opposite of the GCF.
27) -4x3 – 12x2
29) -12x3 + 4x2 – 8x
I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the
parenthesis.
Solution: -1(12x3 – 4x2 + 8x) or -(12x3 – 4x2 + 8x)
31) -3z + 6z3
33) -14a4b2 – 6a2b
35) -8xyz3 – 4x2y3 + 2xyz
37) -16r2st3 – 4r3st2 – 12rst2
Section 1.1: The Greatest Common Factor
#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)
5
Chapter 1: Factoring and Quadratic Equations
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)
43) 3y(z + 1) – 4(z + 1)
45) x(3x – 4) – 2(3x – 4)
47) 3x(2x – 7) + 4(2x – 7)
49) 2x2(3x – 5y) – 5x(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 - 5x, and that is what I will put inside a
parenthesis.
Solution: (3x – 5y)(2x2 – 5x)
51) 8y(y – 5) – 9(y – 5)
6
Chapter 1: Factoring and Quadratic Equations
Section 1.2: Factoring by Grouping
#1 – 36: Factor by Grouping, state if a polynomial is prime
1) x2 + 5x + 2x + 10
Think of this as (x2 + 5x) + (2x + 10)
Factor out an x from the first parenthesis and a 2 from the second parenthesis.
= x(x+5) + 2(x+5)
Now factor out a (x+5)
Solution: (x+5)(x+2)
3) x2 – 5x – 2x + 10
This is a bit trickier. Think of this as
(x2 – 5x) + (-2x + 10) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an x from the first parenthesis and -2 from the second
=x(x-5) + (-2)(x – 5) (notice what I did with the signs in the second parenthesis when I factored
out a negative)
This can be written better.
= x(x-5) – 2(x-5)
Now factor out a (x-5)
Solution: (x – 5)(x – 2)
5) x2 + 9x + 4x + 36
7) x2 – 9x – 4x + 36
Think of this as
7
Chapter 1: Factoring and Quadratic Equations
(x2 – 9x) + (-4x + 36) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an x from the first parenthesis and -4 from the second
=x(x-9) + (-4)(x-9) (notice what I did with the signs in the second parenthesis when I factored
out a negative)
This can be written better.
= x(x-9) – 4(x-9)
Now factor out a (x-9)
Solution: (x-9)(x-4)
9) y2 + 10y + 3y + 30
Think of this as (y2 + 10y) + (3y + 30)
Factor out a y from the first parenthesis and a 3 from the second parenthesis.
= y(y+10) + 3(y + 10)
Now factor out a (y+10)
Solution: (y+10)9y+3)
11) y2 – 10y – 3y + 30
13) 5x2 + 5x + 6x + 6
Think of this as (5x2 + 5x) + (6x + 6)
Factor out an 5x from the first parenthesis and a 6 from the second parenthesis.
= 5x(x+1) + 6(x+1)
Now factor out a (x+1)
Solution: (x+1)(5x+6)
15) 5x2 – 5x – 6x+ 6
17) 4y2 + 3y + 4y + 3
8
Chapter 1: Factoring and Quadratic Equations
19) 4y2 – 3y – 4y + 3
Think of this as
(4y2 – 3y) + (-4y + 3) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an x from the first parenthesis and -1 from the second
=y(4y – 3) + (-1)(4y – 3) (notice what I did with the signs in the second parenthesis when I
factored out a negative)
This can be written better.
= y(4y – 3) – 1(4y – 3)
Now factor out a (4y – 3)
Solution: (4y – 3)(y – 1)
21) 2z2 – 2z + 7z – 7
23) 2z2 + 7z – 2z – 7
Think of this as
(2z2 + 7z) + (-2z – 7) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an z from the first parenthesis and -1 from the second
= z(2z+7) + (-1)(2z + 7) (notice what I did with the signs in the second parenthesis when I
factored out a negative)
This can be written better.
= z(2z+7) – 1(2z +7)
Now factor out a (2z + 7)
Solution: (2z+7)(z-1)
9
Chapter 1: Factoring and Quadratic Equations
25) x3 + 2x2 + 6x + 12
27) x3 + 6x + 3x2 + 18x
Think of this as (x3 + 6x) + (3x2 + 18x)
Factor out an x2 from the first parenthesis and a 3x from the second parenthesis.
= x2(x+6) + 3x(x+6)
Now factor out a (x+6)
Okay Solution: (x+6)(x2+3x)
this can be written better, notice the second parenthesis has a common factor of an x. I should
factor that out and write it first.
Best Solution: x(x+6)(x+3)
29) y3 + 4y2 + y + 4
31) z3 + 5z2 – z – 5
Think of this as
(z3 + 5z2) + (-z – 5) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an z2 from the first parenthesis and -1 from the second
= z2(z+5) + (-1)(z+5) (notice what I did with the signs in the second parenthesis when I factored
out a negative)
This can be written better.
= z2(z+5) – 1(z+5)
Now factor out a (z+5)
Solution: (z+5)(z2 – 1) if you factored before, you might know how to simplify this more. If
you wrote (z+5)(z+1)(z-1) it is also correct, but too advanced for now.
10
Chapter 1: Factoring and Quadratic Equations
33) 5x3 + 4x2 + 10x + 8
35) 7t3 + 5t2 + 21t + 15
11
Chapter 1: Factoring and Quadratic Equations
Section 1.3: Factoring Trinomials of the Form x2 + bx + c
#1 – 28: Rewrite as a polynomial with 4 terms (if possible) then factor by grouping and check
your answer, state if a polynomial is prime.
1) x2 + 5x + 6
I need to break the 5x up into two terms that will make factoring by grouping work. I do this by
first listing the factors of the 6.
Factors of last number
1*6
2*3
Choice to break up middle
term
1x + 6x
2x + 3x
What middle column
simplifies to
7x
5x
I will rewrite the problem using 2x + 3x, of course I could write 3x + 2x and still get the correct
answer.
= x2 + 2x + 3x + 6
Now think of this as (x2 + 2x) + (3x + 6) and factor by grouping.
= x(x+2) + 3(x+2)
Solution: (x+2)(x+3)
3) x2 – 5x + 6
Factors of last number
1*6
2*3
-1 * -6
-2 * - 3
Choice to break up middle
term
1x + 6x
2x + 3x
-1x + -6x
-2x + - 3x
What middle column
simplifies to
7x
5x
-7x
-5x
I need to rewrite the -5x as -2x + (-3x)
= x2 – 2x + (-3x) + 6
=(x2 – 2x) + (-3x + 6)
= x(x-2) + (-3)(x-2)
12
Chapter 1: Factoring and Quadratic Equations
= x(x-2) – 3(x-2)
Solution: (x-2)(x-3)
5) y2 + 5y + 4
7) y2 – 5y + 4
9) z2 + 13z + 36
11) z2 – 13z + 36
13) x2 + 5x – 6
Factors of last number
1 * -6
-1*6
2*-3
-2*3
Choice to break up middle
term
1x + -6x
-1x + 6x
2x + -3x
-2x + 3x
What middle column
simplifies to
-5x
5x
-1x
1x
I will pick the -1x + 6x
= x2 + (-1x) + 6x – 6
= (x2 – 1x) + (6x – 6)
= x(x-1) + 6(x-1)
Solution: (x-1)(x+6)
13
Chapter 1: Factoring and Quadratic Equations
15) x2 - 5x – 6
Factors of last number
1 * -6
-1*6
2*-3
-2*3
Choice to break up middle
term
1x + -6x
-1x + 6x
2x + -3x
-2x + 3x
What middle column
simplifies to
-5x
5x
-1x
1x
Choice to break up middle
term
1x + -12x
-1x + 12x
2x + -6x
-2x + 6x
3x + -4x
-3x + 4x
What middle column
simplifies to
-11x
11x
-4x
4x
-1x
1x
I will pick the 1x + -6x
= x2 + 1x + (-6x) – 6
= (x2 + 1x) + (-6x – 6)
= x(x+1) +( -6)(x+1)
=x(x+1) – 6(x+1)
Solution: (x+1)(x-6)
17) x2 + 4x – 12
19) x2 – 4x –12
Factors of last number
1 * -12
-1*12
2*-6
-2*6
3*-4
-3*4
I need to pick the 2x + - 6x
= x2 + 2x + (-6x) – 12
14
Chapter 1: Factoring and Quadratic Equations
= (x2 + 2x) + (-6x – 12)
= x(x+2) + -6(x+2)
= x(x+2) – 6(x+2)
Solution: (x+2)(x-6)
21) x2 + 7x + 2
Factors of last number
1*2
Choice to break up middle
term
1x + 2x
What middle column
simplifies to
3x
There is no way to get a 7x. This tells me that this can’t factor.
Solution: Prime
23) x2 – 7x + 2
Factors of last number
-1 *- 2
Choice to break up middle
term
-1x + -2x
What middle column
simplifies to
-3x
There is no way to get a (-7x). This tells me that this can’t factor.
Solution: Prime
25) y2 + 3y – 11
Factors of last number
1 * -11
-1* 11
Choice to break up middle
term
1y + (-11y)
-1y + 11y
What middle column
simplifies to
-10y
10y
There is no way to get a 3y. This tells me that this can’t factor.
Solution: Prime
15
Chapter 1: Factoring and Quadratic Equations
27) y2 – 3y – 11
Factors of last number
1 * -11
-1* 11
Choice to break up middle
term
1y + (-11y)
-1y + 11y
What middle column
simplifies to
-10y
10y
There is no way to get a (-3y_. This tells me that this can’t factor.
Solution: Prime
#29-58: Factor each trinomial, state if a polynomial is prime.
29) x2 + 11x + 18
Factors of last number
1 * 18
2*9
3*6
What the add to
1 + 18 = 19
2 + 9 = 11
3+6=9
The lasts must be +2 and + 9
Solution: (x+2)(x+9)
31) c2 + 12c + 20
Factors of last number
1 * 20
2 * 10
4*5
What the add to
1 + 20 = 21
2 + 10 = 12
4+5=9
Solution: (x+2)(x+10)
33) r2 + 6r + 8
16
Chapter 1: Factoring and Quadratic Equations
35) y2 – 10y +16
Factors of last number
-1 * - 16
-2 * - 8
-4 * - 4
What the add to
-1 + -16 = -17
-2 + -8 = -10
-4 + -4 = -8
Solution: (x-2)(x-8)
37) x2 – 9x + 20
Factors of last number
-1 * - 20
-2 * - 10
-4 * - 5
What the add to
-1 + -20 = -21
-2 + -10 = -12
-4 + -5 = -9
Solution: (x – 2)(x – 10)
39) x2 – 2x + 3
I will consider the negative factors, as 1*3 has no chance of giving a negative. Also I don’t
consider 1*-3 nor do I consider -1*3 as they multiply to -3 and do not multiply to positive 3.
Factors of last number
-1 * - 3
What the add to
-1 + -3 = -4
No factors give me the -2 that I need, so this does not factor.
Solution: Prime
41) b2 + 4b – 5
17
Chapter 1: Factoring and Quadratic Equations
43) z2 + 5z – 6
Factors of last number
1*-6
-1 * 6
2* - 3
-2*3
What the add to
1 + -6 = -5
-1 + 6 = 5
2 + - 3 = -1
-2 + 3 = 1
Solution: (z – 1)(z+6)
45) x2 + 4x – 12
Factors of last number
1 * - 12
-1 * 12
2* -6
-2*6
3*-4
-3*4
What the add to
1 + -12 = -11
-1 + 12 = 11
2 + -6 = -4
-2 + 6 = 4
3 + -4 = -1
-3 + 4 = 1
Solution: (x-2)(x+6)
47) x2 – 2x – 15
Factors of last number
1 * -15
-1 * 15
3* - 5
-3*5
What the add to
1 + -15 = -14
-1 + 15 = 14
3 + -5 = -2
-3 + 5 = -2
Solution: (x - 3)(x + 5)
49) a2 – 9a – 22
51) x2 – 6x – 16
18
Chapter 1: Factoring and Quadratic Equations
53) x2 + 2x + 8
55) y2 – 2y + 5
I will consider the negative factors, as 1*3 has no chance of giving a negative. Also I don’t
consider 1*-3 nor do I consider -1*3 as they multiply to -3 and do not multiply to positive 3.
Factors of last number
-1 * - 2
What the add to
-1 + -2= -3
No factors give me the -2 that I need, so this does not factor.
Solution: Prime
57) x2 – 5x – 9
Factors of last number
What the add to
1 * -9
1 + -9 = -8
-1 * 9
-1 + 9 = 8
3* - 3
3 + -3 = 0
No factors give me the -5 that I need, so this does not factor.
Solution: Prime
#59-74: Factor each trinomial. Make sure to factor out a negative or the GCF where applicable.
59) -x2 – 7x –10
First factor out a GCF of -1
= -1(x2 + 7x + 10)
Factors of last number
What the add to
1 * 10
1 + 10 = 11
2*5
2+5=7
Solution: =-1(x+2)(x+5) which also may be written -(x+2)(x+5)
19
Chapter 1: Factoring and Quadratic Equations
61) -w2 + 18w – 77
First factor out a GCF of -1
= -1(w2 + 18w + 77)
Factors of last number
What the add to
1 * 77
1 + 77 = 78
11*7
11+7 = 18
Solution: =-1(w+11)(w+7) which also may be written -(w+11)(w+7)
63) 3x2 +12x – 36
First factor out the GCF of 3.
= 3(x2 + 4x – 12)
Factors of last number
1 * - 12
-1 * 12
2* -6
-2*6
3*-4
-3*4
What the add to
1 + -12 = -11
-1 + 12 = 11
2 + -6 = -4
-2 + 6 = 4
3 + -4 = -1
-3 + 4 = 1
Solution: 3(x-2)(x+6)
65) 6z2 – 30z + 24
67) x3 + 6x2 – 7x
First factor out the GCF of x.
20
Chapter 1: Factoring and Quadratic Equations
= x(x2 + 6x - 7)
Factors of last number
1 * -7
-1 * 7
Solution: x(x-1)(x+7)
What the add to
1 + -7= -6
-1 + 7 = 6
69) -2x3 – 10x2 + 12x
71) 20y + 18 + 2y2
First rewrite in descending power order.
= 2y2 + 20y + 18
Next, factor out a GCF of 2.
= 2(y2 + 10y + 9)
Factors of last number
1*9
3*3
Solution: 2(y+1)(y+9)
What the add to
1 + 9 = 10
3+3=6
73) 30z + 3z2 + 45
21
Chapter 1: Factoring and Quadratic Equations
Section 1.4: Factoring Trinomials in the Form ax2 + bx + c where a ≠ 1
#1 – 18: Rewrite as a polynomial with 4 terms (if possible) then factor by grouping and check
your answer, state if a polynomial is prime.
1) 5x2 + 11x + 6
First multiply the 5*6 = 30
Factors of 30
1 * 30
2 * 15
3 * 10
5*6
Choice to break up middle
term
1x + 30x
2x + 15x
3x + 10x
5x + 6x
What middle column
simplifies to
31x
17x
13x
11x
Choice to break up middle
term
-1x + -30x
-2x + -15x
-3x + -10x
-5x + -6x
What middle column
simplifies to
-31x
-17x
-13x
-11x
= 5x2 + 5x + 6x + 6
= (5x2 + 5x) + (6x + 6)
= 5x(x+1) + 6(x+1)
Solution: (x+1)(5x+6)
3) 5x2 – 11x + 6
First multiply the 5*6 = 30
Factors of 30
-1 * -30
-2 * -15
-3 * -10
-5 * -6
= 5x2 - 5x - 6x + 6
= (5x2 - 5x) + (-6x + 6)
= 5x(x-1) + -6(x-1)
Solution: (x-1)(5x-6)
22
Chapter 1: Factoring and Quadratic Equations
5) 4x2 + 7x + 3
First multiply the 5*6 = 30
Factors of 30
1 * 30
2 * 15
3 * 10
5*6
Choice to break up middle
term
1x + 30x
2x + 15x
3x + 10x
5x + 6x
What middle column
simplifies to
31x
17x
13x
11x
Choice to break up middle
term
-1x + - 12x
-2x + - 6x
-3x + - 4x
What middle column
simplifies to
-13x
-8x
-7x
= 5x2 + 5x + 6x + 6
= (5x2 + 5x) + (6x + 6)
= 5(x+1) + 6(x+1)
Solution: (x+1)(5x+6)
7) 4x2 – 7x + 3
First multiply the 4*3 = 12
Factors of 30
-1*-12
-2*-6
-3*-4
= 4x2 + -3x + -4x + 3
= x(4x - 3) + (-1)(4x – 3)
Solution: (4x – 3)(x – 1)
23
Chapter 1: Factoring and Quadratic Equations
9) 2z2 + 5z – 7
First multiply the 2*-7= -14
Factors of 30
-1*14
1*-14
-2*7
2*-7
Choice to break up middle
term
-1z + 14z
1z + -14z
-2z + 7z
2z + -7z
What middle column
simplifies to
13z
-13z
5z
-5z
Choice to break up middle
term
-1z + 14z
1z + -14z
-2z + 7z
2z + -7z
What middle column
simplifies to
13z
-13z
5z
-5z
= 2z2 + -(2z) + 7z – 7
= 2z(z-1) + 7(z-1)
Solution: (z-1)(2z+7)
11) 2z2 – 5z – 7
First multiply the 2*-7= -14
Factors of 30
-1*14
1*-14
-2*7
2*-7
= 2z2 + 2z + - 7z -7
= 2z(z+1) + -7(z+1)
Solution: (z+1)(2z – 7)
13) 6x2 + 23x + 7
15) 3x2 + 10x + 7
17) 5x2 + 13x + 6
24
Chapter 1: Factoring and Quadratic Equations
#19-44: Factor, using bottoms up or the guess and check method, state if a polynomial is prime
(notice #19 – 30 are the same as #1-12, and you should get the same answer regardless of the
technique you use to factor.)
19) 5x2 + 11x + 6
First multiply the 5*6 = 30
Rewrite as x2 + 11x + 30 and factor this
Factors of 30
1 * 30
2 * 15
3 * 10
5*6
Sum of factors
1+30 = 31
2 + 15 = 17
3 + 10 = 13
5 + 6 = 11
(x + 5)(x+6)
Finish divide by 5 / reduce / bottoms up
5
6
(𝑥 + 5) (𝑥 + 5)
Solution: (x+1)(5x+6)
21) 5x2 – 11x + 6
First multiply the 5*6 = 30
Rewrite as x2 - 11x + 30 and factor this
Factors of 30
-1 * -30
-2 * -15
-3 * -10
-5 * -6
Sum of factors
-1+-30 = -31
-2 + -15 = -17
-3 + -10 = -13
-5 + -6 = -11
(x - 5)(x-6)
Finish divide by 5 / reduce / bottoms up
5
6
(𝑥 − 5) (𝑥 − 5)
25
Chapter 1: Factoring and Quadratic Equations
Solution: (x-1)(5x-6)
23) 4x2 + 7x + 3
First multiply 4*3 = 12 and rewrite the problem
x2 + 7x + 12
Factor this
(x+3)(x+4)
Now divide by 4 / reduce and bottoms up
3
4
(𝑥 + 4) (𝑥 + 4)
Solution: (4x+ 3)(x+1)
25) 4x2 – 7x + 3
First multiply 4*3 = 12 and rewrite the problem
x2 - 7x + 12
Factor this
(x-3)(x-4)
Now divide by 4 / reduce and bottoms up
3
4
(𝑥 − 4) (𝑥 − 4)
Solution: (4x- 3)(x-1)
26
Chapter 1: Factoring and Quadratic Equations
27) 2z2 + 5z – 7
First multiply the 2*-7 = -14 and rewrite the problem
z2 + 5z – 14
factor this (z+7)(z-2)
7
2
now divide by 2 / reduce and bottoms up (𝑧 + 2) (𝑧 − 2)
Solution: (2z + 7)(z – 1)
29) 2z2 – 5z – 7
First multiply the 2*-7 = -14 and rewrite the problem
z2 + 5z – 14
factor this (z-7)(z+2)
7
2
now divide by 2 / reduce and bottoms up (𝑧 − 2) (𝑧 + 2)
Solution: (2z - 7)(z + 1)
31) 3x2 – 11x + 10
33) 2b2 – 15b + 7
First multiply 2 * 7 = 14 and rewrite the problem
b2 – 15b + 14 factor this
(b – 14)(b – 1)
Now divide by 2 / reduce and bottoms up
= (𝑏 −
14
2
1
) (𝑏 − 2)
Solution: (b – 7)(2b – 1)
27
Chapter 1: Factoring and Quadratic Equations
35) 6y2 – 7y – 5
37) 8a2 + a – 7
First multiply 8* -7 = -56 and rewrite the problem
a2 + a – 56 and factor this
(a+8)(a-7)
Now divide by 8 / reduce and bottoms up
8
7
(𝑎 + 8) (𝑎 − 8)
Solution: (a + 1)(8a – 7)
39) 2x2 – 5x – 7
41) 3x2 + 5x + 6
43) 2x2 + 5x – 8
#45-64: Factor out the GCF and then factor by bottoms up or the guess and check method.
45) 4m2 + 34m – 18
First factor out the GCF of 2
= 2(2m2 + 17m – 9)
Now bottoms up factor the inside of the parenthesis
2(m2 + 17m – 18)
28
Chapter 1: Factoring and Quadratic Equations
2(m + 18)(m – 1)
2 (𝑚 +
18
2
1
) (𝑚 − 2)
Solution: 2(m+9)(2m-1)
47) 4z3 – 13z2 + 3z
First factor out the GCF of z
= z(4z2 – 13z + 3)
Now bottoms up factor the inside of the parenthesis
z(z2 – 13z + 12)
z(z – 1)(z – 12)
1
𝑧 (𝑧 − 4) (𝑧 −
12
4
)
Solution: z(4z – 1)(z – 3)
49) 20x3 – 18x2 + 4x
51) -16x2 + 44x – 10
First factor out the GCF of -4
= -2(8x2 – 22x +5)
Now bottoms up factor what’s left in the parenthesis.
-2(x2 – 22x + 40)
-2(x – 20)(x – 2)
−2 (𝑥 −
20
8
2
) (𝑥 − 8)
29
Chapter 1: Factoring and Quadratic Equations
5
1
−2 (𝑥 − 2) (𝑥 − 4)
Solution: -2(2x – 5)(4x – 1)
53) 18x2 – 21x – 15
First factor out the GCF of 3x
= 3x(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
3x(x2 – 7x – 30)
3x(x+ 3)(x-10)
3
3𝑥 (𝑥 + 6) (𝑥 −
1
10
6
)
5
3𝑥 (𝑥 + 2) (𝑥 − 3)
Solution: 3x(2x+1)(3x-5)
55) 18x3 – 21x2 – 15x
First factor out the GCF of 3
= 3(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
3(x2 – 7x – 30)
3(x+ 3)(x-10)
3
3 (𝑥 + 6) (𝑥 −
1
10
6
)
5
3 (𝑥 + 2) (𝑥 − 3)
Solution: 3(2x+1)(3x-5)
30
Chapter 1: Factoring and Quadratic Equations
57) -18x2 + 21x + 15
First factor out the GCF of -3
= -3(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
-3(x2 – 7x – 30)
-3(x+ 3)(x-10)
3
−3 (𝑥 + 6) (𝑥 −
1
10
6
)
5
−3 (𝑥 + 2) (𝑥 − 3)
Solution: -3(2x+1)(3x-5)
59) 6x2 + 5x + 6
61) 3x3 + 5x2 + 6x
63) 4b3 + 6b2 -6b
31
Chapter 1: Factoring and Quadratic Equations
Section 1.5: Factoring Sums and Differences of Squares
#1- 42: Completely factor the binomials, remember to factor out the GCF first when applicable
(if a problem is prime say so).
1) x2 – 9
3) x2 + 9
5) y2 – 36
7) y2 + 36
9) 25a2 – 81
11) 25a2 + 81
13) 49x2 – 36
15) 49x2 + 36
17) x3 – 64x
19) x3 + 64x
21) 3x2 – 27
32
Chapter 1: Factoring and Quadratic Equations
23) 3x2 + 27
25) 9 – 25x2
27) 81 – 16x2
29) x4 – 9
31) 16x4 – 25
33) 98y2 – 2x4
35) x4 – 16
37) 2x4 – 512
39) y4 – 2401
41) x4 + 4
Section 1.6: Factoring Sums and Differences of Cubes
#1-42: Completely factor the binomials, remember to factor out the GCF first when applicable
(if a problem is prime say so).
33
Chapter 1: Factoring and Quadratic Equations
1) x3 + 8
3) x3 – 8
5) b3 + 27
7) b3 – 27
9) x3 + 64
11) x3 + 64
13) 8x3 – 27
15) 8x3 + 27
17) 27x3 – 125
19) 64x3 – y3
21) x6 – y3
23) 27x6 – 1
34
Chapter 1: Factoring and Quadratic Equations
25) 125x9 – y6
27) 16x3 – 54
29) 3x3 + 24
31) x4 – 8x
33) 6x4 – 48x
35) 8x5 + 125x2
37) 27 – x3
39) 27 + 64x3
41) 8 + y6
35
Chapter 1: Factoring and Quadratic Equations
Section 1.7: A Review of all the Factoring Strategies – Mixed Up
#1-44: Factor completely, state if a polynomial is prime.
1) a2 + 16
3) 81y2 – 4
5) b3 + 64
7) 64x3 – 1
9) 2x2 – 3x – 9
11) -4x2 + 6x + 18
13) 3x2 – 13x + 10
15) -w2 + 8w – 15
17) x2 – 2x + 15
19) 5x2 + 10x + 6x + 12
21) x2 + 5x + 9
36
Chapter 1: Factoring and Quadratic Equations
23) 6x4 – 6x
25) 2x2 – 8
27) 3x2 + 12
29) 3x2 – 5x – 6x + 10
31) x2 + x + 24x + 24
33) 6x2 + 13x + 6
35) -x2 + 5x + 6
37) -3x2 – 12x +36
39) z2 – 5z + 4
41) x2 – 14x – 15
43) a2 – a – 2
Section 1.8 Solving Quadratic Equations by Factoring
#1 - 21: Solve each equation.
1) (x – 3)(x + 2)=0
3) (x – 1)(x – 7) = 0
37
Chapter 1: Factoring and Quadratic Equations
5) (3x + 12)(2x – 8) = 0
7) (2x – 9)(3x + 10)=0
9) (5x – 7)(3x – 11) = 0
11) 7x(x – 1)(x + 2) = 0
13) 5x(2x + 10)(4x + 20) = 0
15) x(x – 3)(x + 5) = 0
17) 7(x – 1)(x – 2) = 0
19) 2(5x – 30)(3x + 18) = 0
21) 6(2x – 9)(5x – 1) = 0
#22 - 42: Solve each equation.
23) x2 – 14x + 45 = 0
25) x2 – 5x – 6 = 0
27) b2 – 25 = 0
29) 49y2 – 16 = 0
38
Chapter 1: Factoring and Quadratic Equations
31) 5x2 + 9x + 4 =0
33) 3x2 – 5x – 2 = 0
35) 2x2 + 5x + 3 = 0
37) x2 – x – 6 = 0
39) 25y2 – 81 = 0
41) 3y2 + 5y – 2 = 0
39
Chapter 1: Factoring and Quadratic Equations
Section 1.8 Solving Quadratic Equations by Factoring
#43 - 60: Solve each equation. (Remember to factor out the GCF first)
43) 3x3 + 5x2 + 2x = 0
45) 4x2 + 2x – 6 = 0
47) 5x2 – 20 = 0
49) 3x3 – 15x2 + 18x = 0
51) 3x2 – 15x – 18 = 0
53) 10x2 + 18x + 8 = 0
55) 10x3 + 18x2 + 8x = 0
57) -2x2 +10x + 12 = 0
59) x3 – 49x = 0
#61 - 78: Solve each equation.
61) r(r + 1) = 12
63) x(x – 4) = -3
65) (x – 2)(x – 3)= 6
40
Chapter 1: Factoring and Quadratic Equations
67) (x + 1)(x – 4) = -18
69) 3x(x + 1) = -6
71) (2x – 3)(x + 2) = 4
73) x(x – 3) +2 = 30
75) 2x(x – 3) = 5x(x– 4) +8
77) x(x – 4) + 1 = x + 7
41
Chapter 1: Factoring and Quadratic Equations
Section 1.9: Applications that involve factoring
1) A number is 20 less than its square. Find all such numbers.
Let x = a number
then x2 = its square
I will replace “a number” with x
“is” with an equal sign
and “20 less than its square” with x2 - 20
Now I can create an equation.
𝑥 = 𝑥 2 − 20
−𝑥
−𝑥
________________
0 = x2 – x – 20
0 = (x + 4)(x – 5)
x+4=0
x–5=0
x = -4
x=5
Solution: The numbers are -4 and 5.
3) The square of a number is 6 more than the number. Find all such numbers.
A number = x
square of a number = x2
“Is” will turn into an equal sign
6 more than a number = x + 6
𝑥2 = 𝑥 + 6
−𝑥 − 6 − 𝑥 − 6
________________
x2 – x – 6 = 0
(x+2)(x – 3) = 0
x+2=0
x–3=0
x=-2
x=3
Solution: The numbers are -2, 3
42
Chapter 1: Factoring and Quadratic Equations
5) The product of two consecutive numbers is 72. Find all such numbers.
I will call the first of the two numbers x,
First number x
Since they are consecutive numbers the second number will be called x + 1
Second number x + 1
I will replace the “product of the two numbers” with x(x+1)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+1) = 72 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 𝑥 = 72
−72 − 72
__________________
x2 + x – 72 = 0
(x + 9)(x – 8) = 0
x+9=0
x–8=0
x = -9
x=8
x+1 = -8
x+1=9
Solution: There are two sets of answers. -9 and -8 is one set, and 8 and 9 is the other.
43
Chapter 1: Factoring and Quadratic Equations
7) The product of two consecutive even numbers is 24. Find all such numbers.
I will call the first of the two numbers x
first number x
Since they are consecutive even numbers the second number will be called x + 2
second number x + 2
I will replace the “product of the two numbers” with x(x+2)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+2) = 24 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 2𝑥 = 24
−24 − 24
__________________
x2 + 2x – 24 = 0
(x + 6)(x – 4) = 0
x+6=0
x–4=0
x = -6
x=4
x+2 = -4
x+2=6
Solution: There are two sets of answers. -6 and -4 is one set, and 4 and 6 is the other.
44
Chapter 1: Factoring and Quadratic Equations
9) The product of two consecutive even numbers is 63. Find all such numbers.
I will call the first of the two numbers x
first number x
Since they are consecutive even numbers the second number will be called x + 2
second number x + 2
I will replace the “product of the two numbers” with x(x+2)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+2) = 63 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 2𝑥 = 63
−63 − 63
__________________
x2 + 2x – 63 = 0
(x + 9)(x – 7) = 0
x+9=0
x–7=0
x = -9
x=7
x+2 = -7
x+2=9
Solution: There are two sets of answers. -9 and -7 is one set, and 7 and 9 is the other.
45
Chapter 1: Factoring and Quadratic Equations
11) The length of a rectangular bedroom is 2 feet longer than its width. The area of the
bedroom is 120 square feet. Find the dimensions of the room.
Width = W
Length: L = W + 2
The area is found by multiplying the length and the width
This is the equation I need to solve:
Area = 120
or LW = 120
(W + 2)(W) = 120
𝑊 2 + 2𝑊 = 120
−120
− 120
_________________
W2 + 2W – 120 = 0
(W + 12)(W – 10) = 0
W + 12 = 0
W – 10 = 0
W = -12
W = 10
The width can’t be (-12). The answer must be the width is 10 feet. Add 2 feet to get the length
of 12 feet.
Solution: Length 12 feet, Width 10 feet
46
Chapter 1: Factoring and Quadratic Equations
13) A rectangular garden is 4 feet narrower than it is long. The garden has an area of 32 square
feet. Find the dimensions of the garden.
Length = L
Width W = L – 4
Area = LW
32 = L(L-4)
32 = 𝐿2 − 4𝐿
−32
− 32
______________
0 = L2 – 4L – 32
0 = (L + 4)(L – 8)
L+4=0
L–8=0
L = -4
L=8
The length can’t be (-4). The answer must be the width is 8 feet. Subtract 4 feet to get the
width of 4 feet.
Solution: Length 8 feet, Width 4 feet
47
Chapter 1: Factoring and Quadratic Equations
15) The base of a triangle is 2 feet longer than its height. The area of the triangle is 7.5 square
feet. Find the height of the triangle.
Height = h
Base = height plus 2
B=h+2
1
Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡
1
7.5 = 2 (ℎ + 2)(ℎ)
Multiply by 2 to clear the fraction.
1
2*7.5 = 2 ∗ 2 (ℎ + 2)(ℎ)
15 = (h + 2)(h)
15 = ℎ2 + 2ℎ
0 = h2 + 2h – 15
(h + 5)(h – 3) = 0
h+5=0
h–3=0
h = -5
h=3
The height can’t be (-5) so the height must be 3 feet. I am not asked for the base, so I won’t
include it in my answer. I could just add 2 to get the base of 5 feet.
Solution: height = 3 feet
.
48
Chapter 1: Factoring and Quadratic Equations
Section 1.9: Applications that involve factoring
17) The height of a triangle is 2 feet shorter than its base. The area of the triangle is 17.5
square feet. Find the height of the triangle.
Base = B
Height H = B - 2
1
Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡
1
17.5 = 2 𝐵(𝐵 − 2)
1
2*17.5 = 2 ∗ 2 𝐵(𝐵 − 2)
35 = B(B – 2)
35 = B2 – 2B
0 = B2 – 2B – 35
0 = (B+5)(B – 7)
B+5=0
B–7=0
B = -5
B=7
The base can’t be (-5) so the base must be 7 feet. Subtract 2 to find the height of 5 feet.
Solution: Height = 5 feet
49
Chapter 1: Factoring and Quadratic Equations
19) The length of the hypotenuse of a right triangle is 8 inches more than the shortest leg. The
length of the longer leg is 7 inches more than the length of the shorter leg. Find the length of
each side of the triangle.
A = shortest leg
B = longer leg
C = hypotenuse
length of the hypotenuse of a right triangle is 8 inches more than the shortest leg gives
C=A+8
length of the longer leg is 7 inches more than the length of the shorter leg gives
B=A+7
I will use the A2 + B2 = C2 formula, and replace the B and C with the above values.
A2 + (A + 7)2 = (A + 8)2
A2 + (A + 7)(A+7) = (A+8)(A+8)
A2 + A2 + 7A + 7A + 49 = A2 + 8A + 8A + 64
2A2 + 14A + 49 = A2 + 16A + 64
-A2 - 16A - 64 -A2 -16A – 64
A2 – 2A – 15 = 0
(A + 3)(A – 5) = 0
A+3=0
A–5=0
A = -3
A=5
The length can’t be (-3), so A must be 5 inches.
B = 5 + 7 = 12 inches
C = 5 + 8 = 13 inches
Solution: short leg 5 inches, longer leg 12 inches, hypotenuse 13 inches.
50
Chapter 1: Factoring and Quadratic Equations
21) The length of a hypotenuse of a right triangle is 1 foot more than the longer leg. The
length of the shorter leg is 1 foot less than the length of the longer leg. Find the length of each
side of the right triangle.
A = shortest leg
B = longer leg
C = hypotenuse
length of a hypotenuse of a right triangle is 1 foot more than the longer leg gives
C=B+1
length of the shorter leg is 1 foot less than the length of the longer leg gives
A=B-1
I will use the A2 + B2 = C2 formula, and replace the B and C with the above values.
(B – 1)2 + B2 = (B + 1)2
(B – 1)(B – 1) + B2 = (B+1)(B+1)
B2 – 1B – 1B + 1 + B2 = B2 + 1B + 1B + 1
2B2 - 2B + 1 = B2 + 2B + 1
-B2 - 2B - 1 -B2 – 2B – 1
B2 – 4B = 0
B(B – 4) = 0
B=0
B–4=0
B=4
B can’t be 0 feet, so B must be 4 feet.
A = 4 – 1 = 3 feet
C = 4 + 1 = 5 feet
Solution: short leg 3 feet, longer leg 4 feet, hypotenuse 5 feet
51
Chapter 1: Factoring and Quadratic Equations
23) The length of the hypotenuse in a right triangle is 15 inches. The shortest leg is 3 inches
shorter than the length of the longest leg. Find the length of each of the legs.
A = shortest leg B - 3
B = longer leg
C = 15 (the hypotenuse)
(B – 3)2 + B2 = 152
(B – 3)(B – 3) + B2 = 225
B2 – 3B – 3B + 9 + B2 = 225
2B2 – 6B + 9= 225
2B2 – 6B – 216 = 0
This is a trick that will make the work easier. I can divide each number by 2 without changing
the solution.
B2 - 3B – 108 = 0
This is still hard to factor. The lasts will be -12 and 9.
(B– 12)(B+ 9) = 0
B – 12 = 0
B+9=0
B = 12
B = -9
The length can’t be (-9). B must equal 12 inches.
A = 12 – 3 = 9 inches
Solution: short leg 9 inches, long leg 12 inches, hypotenuse 15 inches.
52
Chapter 1: Factoring and Quadratic Equations
25) The length of the short leg in a right triangle is 3 inches. The longest leg is 1 inch less than
the length of the hypotenuse. Find the length of each of the each unknown side.
A = 3 inches
B=C–1
C=C
32 + (C – 1)2 = C2
9 + (C – 1)(C - 1) = C2
9 + C2 – 1C – 1C + 1 = C2
C2 – 2C + 10 = C2
-C2
- C2
-2C + 10 = 0
+2C
+2C
10 = 2C
5 =C
I also need to find B: B = C – 1, so B = 4.
Solution: short side 3 inches, longer side 4 inches, hypotenuse is 5 inches.
53
Chapter 1: Factoring and Quadratic Equations
Chapter 1: Review
ONLY WRITE UP ANSWERS, NO NEED TO WORK OUT SOLUTIONS - I DO THEM ALL IN A VIDEO
1) Completely factor the polynomial. State if a polynomial is prime.
a) x2 + 7x – 18
b) z2 – 13z + 36
c) -y2 +5y + 14
d) -2y3 – 10y2 + 48y
e) 6x2 + 13x + 5
f) x2 + 2x + 3
g) 9n2 + 24n + 15
h) 4x3 – 16x2 – 84x
i) 25b2 – 81
j) x2 + 64
k) x3 + 125
l) 27y3 – 64
m) 5m2 + m – 6
n) 3n2 + 7n +2
o) -3x3 – 21x2 + 54x
p) 25b2 + 30b + 9
q) x2 + 5x + 2x + 10
r) 5x2 + 10x – 3x – 6
2) Solve each equation.
a) 2x(x – 3) = 0
b) (x + 2)(3x – 10)=0
c) x2 – 3x – 10 = 0
d) 5x2 + 16x + 3 =0
e) x(x – 1) = 20
f) 3x(x – 1) – 4 = 14
g) a2 – 49 = 0
h) 25b2 = 16
i) 2x2 + 3x + 1 = 0
j) x(x – 2) = 0
k) y2 + 6y + 9 = 0
l) b2 + 6b = 7
3) The product of two consecutive numbers is 30. Find all such numbers.
4) A rectangular garden is 3 feet narrower than it is long. The garden has an area of 70 square
feet. Find the dimensions of the garden.
5) The length of the hypotenuse in a right triangle is 5 inches. The shortest leg is 1 inch shorter
than the length of the longest leg. Find the length of each of the legs.
54