Chapter 6.7, Problem 38E

To determine

To find: the value of $k$.

Answer to Problem 38E

The value of k is 4.

Explanation of Solution

Given information:

  $\begin{array}{l}\text{dividend}=4x^4+x^3-7x^2+3x+k\\ \text{divisor}=x-1\\ \text{remainder}=5\end{array}$.

Formula used:

  $\text{Dividend=Quotient×Divisor+Remainder}$

Calculation:

We have ,

  $\begin{array}{l}\text{dividend}=4x^4+x^3-7x^2+3x+k\\ \text{divisor}=x-1\\ \text{remainder}=5\end{array}$

Now, using polynomial division

Step 1.

  $x-1\begin{array}{l}\begin{array}{c}\hfill 4x^3\\ \hfill \overline{)\begin{array}{l}4x^4+x^3-7x^2+3x+k\\ 4x^4-4x^3\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $5x^3-7x^2+3x+k$

Step 2.

  $x-1\begin{array}{l}\begin{array}{c}\hfill 4x^3+5x^2\\ \hfill \overline{)\begin{array}{l}5x^3-7x^2+3x+k\\ 5x^3-5x^2\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $-2x^2+3x+k$

Step 3.

  $x-1\begin{array}{l}\begin{array}{c}\hfill 4x^3+5x^2-2x\\ \hfill \overline{)\begin{array}{l}-2x^2+3x+k\\ -2x^2+2x\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $x+k$

Step 4.

  $x-1\begin{array}{l}\begin{array}{c}\hfill 4x^3+5x^2-2x+1\\ \hfill \overline{)\begin{array}{l}x+k\\ x-1\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $k+1$

  $=4x^4+x^3-7x^2+3x+k=(x-1)(4x^3+5x^2-2x+1)+(k+1)$

Since, remainder is 5.

  $\begin{array}{l}=k+1=5\\ =k=5-1\\ =k=4\end{array}$

Hence, the value of k is 4.