Precalculus with Limits
Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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Chapter A.7, Problem 72E

a.

To determine

Simplify the equations.

a.

Expert Solution
Check Mark

Answer to Problem 72E

  2(2x−3)32(x+15)

Explanation of Solution

Given information:

You are taking a course in calculus, and for one of the homework problems you obtain the following answer.

  110(2x−1)52+16(2x−1)32

The answer in the back of the book is

  115(2x−1)32(3x+1)

Show how the second answer can be obtained from the first. Then use the same technique to simplify each of the following expressions.

  23x(2x−3)32−215(2x−3)52

Calculation:

Here, the answers given to us are:

  110(2x−1)52+16(2x−1)32

  115(2x−1)32(3x+1)

Now, we will simplify the first answer to get the second:

  110(2x−1)52+16(2x−1)32=(2x−1)32(110(2x−1)+16)

  =(2x−1)32(x5−110+16)

  =(2x−1)32(6x−3+530)

  =(2x−1)32(6x+230)

  =(2x−1)32(3x+115)

Now, we will simplify the equation:

  23x(2x−3)32−215(2x−3)52

  =2(2x−3)32(x3−2x−315)=2(2x−3)32(5x−2x+315)=2(2x−3)32(3x+315)=2(2x−3)32(x+15)

Hence, the answer is 2(2x−3)32(x+15) .

b.

To determine

Simplify the equations.

b.

Expert Solution
Check Mark

Answer to Problem 72E

  8(4+x)32(x−115)

Explanation of Solution

Given information:

You are taking a course in calculus, and for one of the homework problems you obtain the following answer.

  110(2x−1)52+16(2x−1)32

The answer in the back of the book is

  115(2x−1)32(3x+1)

Show how the second answer can be obtained from the first. Then use the same technique to simplify each of the following expressions.

  23x(4+x)32−215(4+x)52

Calculation:

Now, we will simplify the equation:

  23x(4+x)32−215(4+x)52

  =23(4+x)32−215(4+x)52=2(4+x)32(x3−4+x15)=2(4+x)32(5x−4−x15)=2(4+x)32(4x−415)=8(4+x)32(x−115)

Hence, the answer is 8(4+x)32(x−115) .

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Chapter A.7, Problem 71E

Chapter A Solutions

Precalculus with Limits

Ch. A.1 - Prob. 1ECh. A.1 - Prob. 2ECh. A.1 - Prob. 3ECh. A.1 - Prob. 4ECh. A.1 - Prob. 5ECh. A.1 - Prob. 6ECh. A.1 - Prob. 7ECh. A.1 - Prob. 8ECh. A.1 - Prob. 9ECh. A.1 - Prob. 10E
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