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Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

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BuyFindarrow_forward

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

For Problems 53-64, use synthetic division to determine the quotient and remainder. (Objective 2)

( 3 x 4 x 3 + 2 x 2 7 x 1 ) ÷ ( x + 1 )

To determine

To Find:

The quotient and remainder of the polynomials by performing the synthetic division.

Explanation

Approach:

Division of a polynomial by binomial is done by using the synthetic division method.

In this method take the coefficeints of the polynomial in the descending powers of variable and if any term is missing than a zero must be used in that place.

Equate the divisor to zero and take the variable value for division.

Bring down the first coefficient of the dividend.

Mutiply the first coefficient of the dividend by the divisor and write that value under the dividend.

Find the result by adding or subtracting.

The obtained results multiplied by the divisor again, and write that value under the dividend.

Repeat the process till the last coefficient and simplify.

Convert the coeficients of the last row to an equation of the quotient by decreasing the variable exponent by 1 and the last value of the row represents the remainder.

Calculation:

Consider the expression (3x4x3+2x27x1)÷(x+1).

First write the coefficients of the polynomial in the descending order.

They are 3,1,2,7,1.

Take the divisor and equate it to zero.

That is,

x+1=0

Add 1 on both sides of the equation.

x+11=01x+11=01x=1

Arrange the coefficients in the descending order and write the divisor in the following way and simplify.

13        1        2     7         1_

The calculation steps are as follows:

Bring down the leading coefficient 3 and multiply it with 1.

13        1        2     7         1                                                  _   3                                                 

The result is 3.

Write the result under the coefficient 1 and add.

The result is 4.

13        1        2     7         1           3                                 _   3        4                                  

Multiply the divisor 1 by 4.

Which is equals to 4, write the value under the coefficient 2.

The result is 6.

13        1        2     7         1           3      4                     _   3        4        6                        

Multiply the divisor 1 by 6

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