Chapter A.4, Problem 5AYU

To determine

To calculate: The quotient and remainder when the polynomial $x^3-x^2+2x+4$ is divided by $x-2$ with help of synthetic division.

Answer to Problem 5AYU

The quotient is $x^2+x+4$ and reminder is $12$ .

Explanation of Solution

Given information:

The polynomial $x^3-x^2+2x+4$ 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 $x^3-x^2+2x+4$ and a factor of it $x-2$ .

To divide the polynomial $x^3-x^2+2x+4$ by $x-2$ follow the steps provided below,

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

Step 1: List down the coefficients of dividend in the descending powers of x, that are $1,-1,2,4$ .

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,

  $2\overline{)\begin{array}{cccc}1& -1& 2& 4\end{array}}$

Step 3: Bring down 1 to in third row,

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ 1\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& & \end{array}\\ 1\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& & \end{array}\\ \begin{array}{cccc}1& 1& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& 2& \end{array}\\ \begin{array}{cccc}1& 1& & \end{array}\end{array}}$

Step7: Now, add the elements of column 3 of row 1 and row 2.

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& 2& \end{array}\\ \begin{array}{cccc}1& 1& 4& \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& 2& 8\end{array}\\ \begin{array}{cccc}1& 1& 4& \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}1& -1& 2& 4\end{array}\\ \begin{array}{cccc}& 2& 2& 8\end{array}\\ \begin{array}{cccc}1& 1& 4& 12\end{array}\end{array}}$

Now, the first three 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 $x^2+x+4$ and reminder is 12.

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

  $x^3-x^2+2x+4=\left(x^2+x+4\right)\left(x-2\right)+12$

To verify multiply the terms on the right hand side,

  $\begin{array}{c}\left(x^2+x+4\right)\left(x-2\right)+12=x^3-2x^2+x^2-2x+4x-8+12\\ =x^3-x^2+2x+4\end{array}$

Thus, the quotient is $x^2+x+4$ and reminder is $12$ when the polynomial $x^3-x^2+2x+4$ is divided by $x-2$ .