Exercise 38 Page 709

Consider the given prism with a triangular base.

10 w^{3} + 23 w^{2} + 5 w - 2,

where its height is

2 w + 1

and the height of the triangular base is

5 w - 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 = B h

Let's substitute

V = 10 w^{3} + 23 w^{2} + 5 w - 2

and

h = 2 w + 1

into the formula and solve it for

B .

V = B h

SubstituteII

$V = 10 w^{3} + 23 w^{2} + 5 w - 2$ , $h = 2 w + 1$

10 w^{3} + 23 w^{2} + 5 w - 2 = B (2 w + 1)

DivEqn

$LHS / 2 w + 1 = RHS / 2 w + 1$

\frac{1 0 w ^{3} + 2 3 w ^{2} + 5 w - 2}{2 w + 1} = B

RearrangeEqn

Rearrange equation

B = \frac{1 0 w ^{3} + 2 3 w ^{2} + 5 w - 2}{2 w + 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!

2 w + 1 10 w^{3} + 23 w^{2} + 5 w - 2

▼

Divide

$\frac{1 0 w ^{3}}{2 w} = 5 w^{2}$

5 w^{2} 2 w + 1 10 w^{3} + 23 w^{2} + 5 w - 2

Multiply $term$ by $divisor$

5 w^{2} 2 w + 1 10 w^{3} + 23 w^{2} + 5 w - 2 10 w^{3} + 5 w^{2}

Subtract down

5 w^{2} 2 w + 1 18 w^{2} + 5 w - 2

▼

Divide

$\frac{1 8 w ^{2}}{2 w} = 9 w$

5 w^{2} + 9 w 2 w + 1 18 w^{2} + 5 w - 2

Multiply $term$ by $divisor$

5 w^{2} + 9 w 2 w + 1 18 w^{2} + 5 w - 2 18 w^{2} + 9 w

Subtract down

5 w^{2} + 9 w 2 w + 1 - 4 w - 2

▼

Divide

$\frac{- 4 w}{2 w} = - 2$

5 w^{2} + 9 w - 2 2 w + 1 - 4 w - 2

Multiply $term$ by $divisor$

5 w^{2} + 9 w - 2 2 w + 1 - 4 w - 2 - 4 w - 2

Subtract down

5 w^{2} + 9 w - 2 2 w + 1 0

The quotient is

5 w^{2} + 9 w - 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 = \frac{1}{2} b h

Let's now substitute

B = 5 w^{2} + 9 w - 2

and

h = 5 w - 1

into the formula and solve it for

b .

B = \frac{1}{2} b h

SubstituteII

$B = 5 w^{2} + 9 w - 2$ , $h = 5 w - 1$

5 w^{2} + 9 w - 2 = \frac{1}{2} b (5 w - 1)

▼

Solve for

b

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

2 (5 w^{2} + 9 w - 2) = b (5 w - 1)

Distr

Distribute $2$

10 w^{2} + 18 w - 4 = b (5 w - 1)

DivEqn

$LHS / 5 w - 1 = RHS / 5 w - 1$

\frac{1 0 w ^{2} + 1 8 w - 4}{5 w - 1} = b

RearrangeEqn

Rearrange equation

b = \frac{1 0 w ^{2} + 1 8 w - 4}{5 w - 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!

5 w - 1 10 w^{2} + 18 w - 4

▼

Divide

$\frac{1 0 w ^{2}}{5 w} = 2 w$

2 w 5 w - 1 10 w^{2} + 18 w - 4

Multiply $term$ by $divisor$

2 w 5 w - 1 10 w^{2} + 18 w - 4 10 w^{2} - 2 w

Subtract down

2 w 5 w - 1 20 w - 4

▼

Divide

$\frac{2 0 w}{5 w} = 4$

2 w + 4 5 w - 1 20 w - 4

Multiply $term$ by $divisor$

2 w + 4 5 w - 1 20 w - 4 20 w - 4

Subtract down

2 w + 4 5 w - 1 0

The quotient is

2 w + 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 $10 w^{3} + 23 w^{2} + 5 w - 2$ by $2 w + 1 .$

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