McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
5. Dividing Polynomials
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Exercise 38 Page 709

Use the formula for the volume of a prism.

2w+4

Practice makes perfect

Consider the given prism with a triangular base.

The volume of the prism is 10w^3+23w^2+5w-2, where its height is 2w+1 and the height of the triangular base is 5w-1. We will calculate the length of the base of the triangle. To do so, let's first find an expression for the area of the triangular base B by using the formula for the volume of a prism. V=BhLet's substitute V= 10w^3+23w^2+5w-2 and h= 2w+1 into the formula and solve it for B.

V=Bh
10w^3+23w^2+5w-2=B( 2w+1)
10w^3+23w^2+5w-2/2w+1=B
B=10w^3+23w^2+5w-2/2w+1

We will use polynomial long division to calculate the quotient B. All the terms of the dividend must be present and the polynomial must be in standard form. Since there are no missing terms and our polynomial is in descending degree order, we do not need to rewrite the polynomial. Let's divide!

l r 2w+1 & |l 10w^3+23w^2+5w-2
â–¼
Divide

10w^3/2w= 5w^2

r 5w^2 r 2w+1 & |l 10w^3+23w^2+5w-2

Multiply term by divisor

r 5w^2 rl 2w+1 & |l 10w^3+23w^2+5w-2 & 10w^3+5w^2

Subtract down

r 5w^2 r 2w+1 & |l 18w^2+5w-2
â–¼
Divide

18w^2/2w= 9w

r 5w^2+9w r 2w+1 & |l 18w^2+5w-2

Multiply term by divisor

r5w^2+ 9w rl 2w+1 & |l 18w^2+5w-2 & 18w^2+9w

Subtract down

r 5w^2+9w r 2w+1 & |l - 4w-2
â–¼
Divide

- 4w/2w= - 2

r 5w^2+9w-2 r 2w+1 & |l - 4w-2

Multiply term by divisor

r5w^2+9w - 2 rl 2w+1 & |l - 4w-2 & - 4w-2

Subtract down

r 5w^2+9w-2 r 2w+1 & |l 0

The quotient is 5w^2+9w-2. This is the area of the triangle B. We will find the measure of the base of the triangle. To do this, let's use the formula for the area of a triangle. B=1/2bh Let's now substitute B= 5w^2+9w-2 and h= 5w-1 into the formula and solve it for b.

B=1/2bh
5w^2+9w-2=1/2b( 5w-1)
â–¼
Solve for b
2(5w^2+9w-2)=b(5w-1)
10w^2+18w-4=b(5w-1)
10w^2+18w-4/5w-1=b
b=10w^2+18w-4/5w-1

We will again use polynomial long division to calculate the quotient b. Notice that there are no missing terms in the dividend and our dividend is in descending degree order. Therefore, we do not need to rewrite it. Let's divide!

l r 5w-1 & |l 10w^2+18w-4
â–¼
Divide

10w^2/5w= 2w

r 2w r 5w-1 & |l 10w^2+18w-4

Multiply term by divisor

r 2w rl 5w-1 & |l 10w^2+18w-4 & 10w^2-2w

Subtract down

r 2w r 5w-1 & |l 20w-4
â–¼
Divide

20w/5w= 4

r 2w+4 r 5w-1 & |l 20w-4

Multiply term by divisor

r2w+ 4 rl 5w-1 & |l 20w-4 & 20w-4

Subtract down

r 2w+4 r 5w-1 & |l 0

The quotient is 2w+4. This is the measure of the base of the triangle.

Showing Our Work

Long division by hand...
When we are doing long division by hand, it looks a bit different than how we have it in this solution. Here is how yours should look when you are writing it in your notebook. Let's start with the division of 10w^3+23w^2+5w-2 by 2w+1.

Let's now show how to divide 10w^2+18w-4 by 5w-1 by hand.