Chapter A.4, Problem 15AYU

To determine

To calculate: The quotient and remainder when the polynomial $x^5-1$ is divided by $x-1$ with help of synthetic division.

Answer to Problem 15AYU

The quotient is $x^4+x^3+x^2+x+1$ and reminder is $0$ .

Explanation of Solution

Given information:

The polynomial $x^5-1$ and a factor of it $x-1$ .

Formula used:

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Calculation:

Consider the provided polynomial $x^5-1$ and a factor of it $x-1$ .

To divide the polynomial $x^5-1$ by $x-1$ follow the steps provided below,

Here dividend is $x^5+0\cdot x^4+0\cdot x^3+0\cdot x^2+0\cdot x-1$ and divisor is $x-1$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $1,0,0,0,0,-1$ .

Now, the divisor is of the form $x-c$ , where $c=1$ .

Step 2: To perform the synthetic division put the coefficients in the division sign along with $1$ on the left side,

  $1\overline{)\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}}$

Step 3: Bring down $1$ to in third row,

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& & & & & \end{array}\\ \begin{array}{cccccc}1& & & & & \end{array}\end{array}}$

Step 4: Multiply $1$ with the first entry of row 3 and place the result in row 2 column 2.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& & & & \end{array}\\ \begin{array}{cccccc}1& & & & & \end{array}\end{array}}$

Step 5: Now, add the elements of column 2 of row 1 and row 2.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& & & & \end{array}\\ \begin{array}{cccccc}1& 1& & & & \end{array}\end{array}}$

Step 6: Multiply $1$ with the second entry of row 3 and place the result in row 2 column 3.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& 1& & & \end{array}\\ \begin{array}{cccccc}1& 1& & & & \end{array}\end{array}}$

Step 7: Now, add the elements of column 3 of row 1 and row 2.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& 1& & & \end{array}\\ \begin{array}{cccccc}1& 1& 1& & & \end{array}\end{array}}$

Step 8: Multiply $1$ with the third entry of row 3 and place the result in row 2 column 4.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& 1& 1& & \end{array}\\ \begin{array}{cccccc}1& 1& 1& & & \end{array}\end{array}}$

Step 9: Now, add the elements of column 4 of row 1 and row 2.

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& 1& 1& & \end{array}\\ \begin{array}{cccccc}1& 1& 1& 1& & \end{array}\end{array}}$

Repeat the above steps 8 and 9 for remaining elements of first row,

  $1\overline{)\begin{array}{l}\begin{array}{cccccc}1& 0& 0& 0& 0& -1\end{array}\\ \begin{array}{cccccc}& 1& 1& 1& 1& 1\end{array}\\ \begin{array}{cccccc}1& 1& 1& 1& 1& 0\end{array}\end{array}}$

Now, the first five elements of row 3 are coefficients of quotient in descending powers of x with degree one less than dividend and the last entry is remainder.

Therefore, quotient is $x^4+x^3+x^2+x+1$ and reminder is $0$ .

Now, when a polynomial is divided by its divisor then dividend is the product of divisor and quotient increased by remainder.

  $x^5-1=\left(x^4+x^3+x^2+x+1\right)\left(x-1\right)+0$

To verify multiply the terms on the right hand side,

  $\begin{array}{c}\left(x^4+x^3+x^2+x+1\right)\left(x-1\right)+0=x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x-1\\ =x^5-1\end{array}$

Thus, the quotient is $x^4+x^3+x^2+x+1$ and reminder is $0$ when the polynomial $x^5-1$ is divided by $x-1$ .