Chapter A.4, Problem 10AYU

To determine

To calculate: The quotient and remainder when the polynomial $x^4+x^2+2$ is divided by $x-2$ with help of synthetic division.

Answer to Problem 10AYU

The quotient is $x^3+2x^2+5x+10$ and reminder is $22$ .

Explanation of Solution

Given information:

The polynomial $x^4+x^2+2$ and a factor of it $x-2$ .

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 $x^4+x^2+2$ and a factor of it $x-2$ .

To divide the polynomial $x^4+x^2+2$ by $x-2$ follow the steps provided below,

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

Step 1: List down the coefficients of dividend in the descending powers of x, that are $1,0,1,0,2$ .

Now, the divisor is of the form $x-c$ , where $c=2$ .

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

  $2\overline{)\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& & & & \end{array}\\ \begin{array}{ccccc}1& & & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& & & \end{array}\\ \begin{array}{ccccc}1& & & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& & & \end{array}\\ \begin{array}{ccccc}1& 2& & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& 4& & \end{array}\\ \begin{array}{ccccc}1& 2& & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& 4& & \end{array}\\ \begin{array}{ccccc}1& 2& 5& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& 4& 10& \end{array}\\ \begin{array}{ccccc}1& 2& 5& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& 4& 10& \end{array}\\ \begin{array}{ccccc}1& 2& 5& 10& \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}1& 0& 1& 0& 2\end{array}\\ \begin{array}{ccccc}& 2& 4& 10& 20\end{array}\\ \begin{array}{ccccc}1& 2& 5& 10& 22\end{array}\end{array}}$

Now, the first four 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 $x^3+2x^2+5x+10$ and reminder is $22$ .

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

  $x^4+x^2+2=\left(x^3+2x^2+5x+10\right)\left(x-2\right)+22$

To verify multiply the terms on the right hand side,

  $\begin{array}{c}\left(x^3+2x^2+5x+10\right)\left(x-2\right)+22=x^4-2x^3+2x^3-4x^2+5x^2-10x+10x-20+22\\ =x^4+x^2+2\end{array}$

Thus, the quotient is $x^3+2x^2+5x+10$ and reminder is $22$ when the polynomial $x^4+x^2+2$ is divided by $x-2$ .