EBK ELEMENTARY AND INTERMEDIATE ALGEBRA
EBK ELEMENTARY AND INTERMEDIATE ALGEBRA
6th Edition
ISBN: 9780321877949
Author: Johnson
Publisher: YUZU
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Chapter C, Problem 1YT
To determine

To calculate: The value of (x3+2x217x+6)÷(x3) using synthetic division.

Expert Solution & Answer
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Answer to Problem 1YT

Solution:

The value of (x3+2x217x+6)÷(x3) is x2+5x2_.

Explanation of Solution

Given Information:

The expression is (x3+2x217x+6)÷(x3).

Calculation:

The constant term of the divisor is 3 with the opposite sign is written to left.

The coefficients of the dividend 1, 2, 17, 6 are written to the right.

Evaluate the value of (x3+2x217x+6)÷(x3) using synthetic division as follows.

3     1   2  17    6                               _        1          

Multiply 1 by 3 and add 2 to the result.

3     1   2  17    6            3                  _        1  5

Multiply 3 by 5 and add 17 to the result.

3     1   2  17    6            3     15        _        1   5   2               

Multiply 3 by 2 and add 6 to the result.

3     1   2  17    6            3     15 6_        1   5   2     0          

Finally, replace the variables in descending order. The maximum degree of the dividend is 3. Thus, the maximum degree of the quotient should be 2.

3     1      2   17    6               3       15 6_        1x2 5x   2      0          

The obtained numbers are the coefficients of a polynomial and the last digit is the remainder.

So, the polynomial with coefficients 1, 5, 2 is x2+5x2.

Therefore, the value of (x3+2x217x+6)÷(x3) is x2+5x2.

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