To give feedback about these practice question solutions, please fill out this survey.

# Solutions for Lesson 11 Practice Questions

To see the solution to a question, hover your mouse over the empty area beneath the question.  The answer should then magically appear!

Contents

## Factoring

Factor each expression completely.

a) x^2+4x-21

x^2+4x-21=(x+7)(x-3)

b) x^2+5x+4

x^2+5x+4=(x+4)(x+1)

c) 2x^2+7x-15

2x^2+7x-15 = (2x-3)(x+5)

d) x^2-3x-18

x^2-3x-18=(x-6)(x+3)

e) -x^2-14x-40

-x^2-14x-40 = -(x^2+14x+40) = -(x+10)(x+4)

f) 4x^2-12x+9

4x^2-12x+9=(2x-3)^2

g) 7x^2-63x+56

7x^2-63x+56=7(x^2-9x+8)=7(x-8)(x-1)

h) -2x^2+14x+36

-2x^2+14x+36=-2(x^2-7x-18)=-2(x-9)(x+2)

i) x^2+17x+66

x^2+17x+66=(x+6)(x+11)

j) 4x^2-24x

4x^2-24x=4x(x-6)

## Extending Ideas

Let's look at taking some of the ideas you've met so far and apply them to problems where they might not at first glance appear to apply.

a) Factor w^2-9

w^2-9=(w-3)(w+3)

b) The question above is an example of the difference of two squares since both the terms are squares.  We're going to use your answer to part a) to factor (x-1)^2-9 .  First, what is w ?

w=x-1

c) Use your answers to a) and b) to fully factor (x-1)^2-9 .  Simplify each of the expressions in the parentheses.

(x-1)^2-9=[(x-1)-3][(x-1)+3]=(x-4)(x+2)

d) By applying the same method, completely factor 16-(x+3)^2

16-(x+3)^2=[4-(x+3)][4+(x+3)]=(1-x)(7+x)=-(x-1)(x+7)

## More Extending Ideas

You saw in the lesson that you can factor expressions like 6x^2+5x+1 by reversing distribution.  You'd split the term in x and then factor the terms in pairs.
6x^2+5x+1=6x^2+3x+2x+1=3x(2x+1)+(2x+1)=(3x+1)(2x+1)
This technique can be applied more widely, starting from where you can factor the terms in pairs.  Try these examples.  Factor fully.

a) 6x^3+3x^2+4x+2

6x^3+3x^2+4x+2 = 3x^2(2x+1)+2(2x+1)=(2x+1)(3x^2+2)

b) 12x^5+18x^3+10x^2+15

12x^5+18x^3+10x^2+15 = 6x^3(2x^2+3)+5(2x^2+3) = (2x^2+3)(6x^3+5)

c) 7x^3+x-14x^2-2

7x^3+x-14x^2-2 = 7x^3-14x^2+x-2 = 7x^2(x-2)+(x-2) = (x-2)(7x^2+1)

## Length of a Room

We have a room that has one side 8 feet longer than the other side.  If we want the total area to be 240 square feet, how long should the shorter side be?

Let's call the shorter side of the room x.  Then the longer side will be x+8 .  We can find the area of the room by multiplying the side lengths together.
x(x+8)=240
x^2+8x=240
x^2+8x-240=0
(x+20)(x-12)=0
x+20=0 or x-12=0
x=-20 or x=12
We cannot have x=-20 as a solution, since x is supposed to be a measure of length, which shouldn't be negative.
That means that x=12, so the shorter side is 12 feet.

## Andy's Backyard

Andy is designing the backyard of his new house.  He wants to place a garden around the reflecting pool, which is 40 feet by 10 feet.  If he has 1000 square feet for the entire project, what will the width of the garden be?

If we call the width of the garden w , then the new length of his backyard will be 40+2w , and the new width will be 10+2w .  We need to add 2w to each dimension since we want the garden to be on both ends of each.  We know that the area of the yard will be the product of these new dimensions.
(40+2w)(10+2w)=1000
400+100w+4w^2=1000
100+25w+w^2=250
w^2+25w+100-250=0
w^2+25w-150=0
(w+30)(w-5)=0
w+30=0 or w-5=0
w=-30 or w=5
Since w is a length, we want it to be a positive number.  This means that the solution we will use is w=5.  The width of the garden will be 5 feet.

## Solve by Factoring

a) 2x^2+6x+4=0

2x^2+6x+4=0
x^2+3x+2=0
(x+2)(x+1)=0
x+2=0 or x+1=0
x=-2, -1

b) x^2-6x-16=0

x^2-6x-16=0
(x-8)(x+2)=0
x-8=0 or x+2=0
x=8, -2

c) 16x^2-56x+49=0

16x^2-56x+49=0
(4x-7)(4x-7)=0
4x-7=0
4x=7
x=\frac {7}{4}

d) 16x^2-24x+9=0

16x^2-24x+9=0
(4x-3)(4x-3)=0
4x-3=0
4x=3
x=\frac {3}{4}

## Completely Factor 1

Completely factor each expression.
a) x^2-4x+4

x^2-4x+4 = (x-2)(x-2) = (x-2)^2

b) x^3-9x

x^3-9x = x(x^2-9) = x(x-3)(x+3)

c) (z-2)^2-16

(z-2)^2-16 = [(z-2)-4][(z-2)+4] = (z-6)(z+2)

d) a^4-a^3+2a-2

a^4-a^3+2a-2 = a^3(a-1)+2(a-1) = (a-1)(a^3+2)

## Completely Factor 2

Completely factor each expression.
e) y^4-\frac {1}{16}

y^4 - \frac {1}{16}=(y^2 - \frac {1}{4})(y^2 + \frac {1}{4})=(y-\frac {1}{2})(y+\frac {1}{2})(y^2 + \frac {1}{4})

f) 4x^2-16x+16

4x^2-16x+16 = 4(x^2-4x+4) = 4(x-2)(x-2) = 4(x-2)^2

g) 49 - (x+3)^2

49 - (x+3)^2 = 49-(x^2+6x+9) = 49-x^2-6x-9 = -x^2-6x+40 = -(x^2+6x-40) = -(x+10)(x-4)

h) 2e^5+4e^4+2e+4

2e^5+4e^4+2e+4 = 2(e^5+2e^4+e+2) = 2[e^4(e+2)+(e+2)]=2(e+2)(e^4+1)