# section 1.2 solutions ```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
Think of this as (x2 + 9x) + (4x + 36)
Factor out an x from the first parenthesis and a 4 from the second parenthesis.
= x(x+9) + 4(x+9)
Now factor out a (x+9)
Solution: (x+9)(x+4)
3
Chapter 1: Factoring and Quadratic Equations
7) x2 – 9x – 4x + 36
Think of this as
(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)(y+3)
4
Chapter 1: Factoring and Quadratic Equations
11) y2 – 10y – 3y + 30
Think of this as
(y2 – 10y) + (-3y + 30) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an y from the first parenthesis and -3 from the second
= y(y – 10) + -3(y – 10) (notice what I did with the signs in the second parenthesis when I
factored out a negative)
This can be written better.
= y(y – 10) – 3(y – 10)
Now factor out a (y – 10)
Solution: (y – 10)(y – 3)
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)
5
Chapter 1: Factoring and Quadratic Equations
15) 5x2 – 5x – 6x+ 6
Think of this as
(5x2 – 5x) + (-6x + 6) notice I put a plus in front of the second parenthesis. This does equal the
original problem.
Factor out an 5x from the first parenthesis and -6 from the second
=5x(x-1) + (-6)(x- 1) (notice what I did with the signs in the second parenthesis when I factored
out a negative)
This can be written better.
= 5x(x – 1) – 6(x – 1)
Now factor out a (x-1)
Solution: (5x-6)(x-1)
17) 4y2 + 3y + 4y + 3
Think of this as
(4y2 + 3y) + (4y + 3)
Factor out an y from the first parenthesis and 1 from the second
=y(4y + 3) + 1(4y + 3)
Now factor out the GCF of 4y + 3
Solution: (4y + 3)(y + 1)
6
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 y 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
= (2z2 – 2z) + (7z – 7)
= 2z(z – 1) + 7(z – 1)
Solution: (z – 1)(2z – 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)
7
Chapter 1: Factoring and Quadratic Equations
25) x3 + 2x2 + 6x + 12
= (x3 + 2x2) + (6x + 12)
= x2(x + 2) + 6(x + 2)
Solution: (x + 2)(x2 + 6)
27) x3 + 6x + 3x2 + 18
Think of this as (x3 + 6x) + (3x2 + 18)
Factor out an x2 from the first parenthesis and a 3x from the second parenthesis.
= x(x2+6) + 3x(x2+6)
Now factor out a (x2+6)
Solution: (x+3)(x2+6)
29) y3 + 4y2 + y + 4
= (y3 + 4y2) + (y + 4)
= y2(y + 4) + 1(y + 4)
Solution: (y + 4)(y2 + 1)
31) z3 + 5z2 – z – 5
8
Chapter 1: Factoring and Quadratic Equations
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.
33) 5x3 + 4x2 + 10x + 8
= (5x3 + 4x2) + (10x + 8)
= x2(5x + 4) + 2(5x + 4)
Solution: (5x + 4)(x2 + 2)
35) 7t3 + 5t2 + 21t + 15
= (7t3 + 5t2) + (21t + 15)
= t2(7t + 5) + 3(7t + 5)
Solution: (7t + 5)(t2 + 3)
9
```