Exercise 42 Page 316

a Let's recall that the area of a rectangle is a product of its width and length. It is given that the area of the rectangle is

2 x^{2} + 7 x + 3

and its length is

2 x + 1 .

Let's use algebra tiles to represent these two measures.

As we can see, the width of the rectangle contains one $x -$ tile and three $1$ -tiles. This allows us to conclude that the width equals $x + 3 .$

b As we already reviewed, the area of a rectangle is the product of its length and width. Hence, to calculate the width, we can divide the rectangle's area by the length.

(2 x^{2} + 7 x + 3) \div (2 x + 1)

c Instead of choosing one method of dividing polynomials, we will solve using each of them.

Synthetic Division

In order to apply synthetic division, the divisor should have a form of

x - r .

In our case, the divisor is

2 x + 1 .

If we divide both the numerator and denominator of the fraction by

2,

the divisor will be in the needed form.

\frac{2 x ^{2} + 7 x + 3}{2 x + 1}

DivFracByFracSameDenom

$\frac{a / 2}{b / 2} = \frac{a}{b}$

\frac{( 2 x ^{2} + 7 x + 3 ) / 2}{( 2 x + 1 ) / 2}

CalcQuot

Calculate quotient

\frac{x ^{2} + 3 . 5 x + 1 . 5}{x + 0 . 5}

Now we are ready to divide!

- 0.5 1 3.5 1.5

Bring down the $first$ $coefficient$

- 0.5 1 3.5 1.5 1

Multiply the $coefficient$ by the $divisor$

- 0.5 1 3.5 - 0.5 1.5 1 3.51.5

Add down

- 0.5 1 3.5 - 0.5 1.5 13 1.5

▼

Repeat the process for all of the coefficients

Multiply the $coefficient$ by the $divisor$

- 0.5 1 3.5 - 0.5 1.5 - 1.5 13 1.5

Add down

- 0.5 1 3.5 - 0.5 1.5 - 1.5 130

The quotient is

x + 3

with a remainder of

0 .

Long Division

Now, let's use the method of polynomial long division. This time we can divide the initial polynomials without changing them.

2 x + 1 2 x^{2} + 7 x + 3

▼

Divide

$\frac{2 x ^{2}}{2 x} = x$

x 2 x + 1 2 x^{2} + 7 x + 3

Multiply $term$ by $divisor$

x 2 x + 1 2 x^{2} + 7 x + 3 2 x^{2} + x

Subtract down

x 2 x + 1 6 x + 3

▼

Divide

$\frac{6 x}{2 x} = 3$

x + 3 2 x + 1 6 x + 3

Multiply $term$ by $divisor$

x + 3 2 x + 1 6 x + 3 6 x + 3

Subtract down

x + 3 2 x + 1 0

Again, we got that the quotient is

x + 3

with no remainder.

Conclusion

In Part A, using algebra tiles, we have found that the width of the rectangle is $x + 3 .$ Here, we provided some algebraic calculations and received the same answer. Therefore, our concrete model checks with the algebraic model.

Subchapter links

Check Your Understanding p.315

Practice and Problem Solving p.315-316

H.O.T. Problems p.316

Standardized Test Practice p.317

Spiral Review p.317

Skills Review p.317