Chapter A.4, Problem 11AYU

To determine

To calculate: The quotient and remainder when the polynomial $4x^6-3x^4+x^2+5$ is divided by $x-1$ with help of synthetic division.

Answer to Problem 11AYU

The quotient is $4x^5+4x^4+x^3+x^2+2x+2$ and reminder is $7$ .

Explanation of Solution

Given information:

The polynomial $4x^6-3x^4+x^2+5$ 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 $4x^6-3x^4+x^2+5$ and a factor of it $x-1$ .

To divide the polynomial $4x^6-3x^4+x^2+5$ by $x-1$ follow the steps provided below,

Here dividend is $4x^6+0\cdot x^5-3x^4+0\cdot x^3+x^2+0\cdot x+5$ and divisor is $x-1$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $4,0,-3,0,1,0,5$ .

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,

  $1\overline{)\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}}$

Step 3: Bring down $4$ to in third row,

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\\ \begin{array}{ccccccc}4& & & & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& & & & & \end{array}\\ \begin{array}{ccccccc}4& & & & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& & & & & \end{array}\\ \begin{array}{ccccccc}4& 4& & & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& 4& & & & \end{array}\\ \begin{array}{ccccccc}4& 4& & & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& 4& & & & \end{array}\\ \begin{array}{ccccccc}4& 4& 1& & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& 4& 1& & & \end{array}\\ \begin{array}{ccccccc}4& 4& 1& & & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& 4& 1& & & \end{array}\\ \begin{array}{ccccccc}4& 4& 1& 1& & & \end{array}\end{array}}$

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

  $1\overline{)\begin{array}{l}\begin{array}{ccccccc}4& 0& -3& 0& 1& 0& 5\end{array}\\ \begin{array}{ccccccc}& 4& 4& 1& 1& 2& 2\end{array}\\ \begin{array}{ccccccc}4& 4& 1& 1& 2& 2& 7\end{array}\end{array}}$

Now, the first six 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 $4x^5+4x^4+x^3+x^2+2x+2$ and reminder is $7$ .

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

  $4x^6-3x^4+x^2+5=\left(4x^5+4x^4+x^3+x^2+2x+2\right)\left(x-1\right)+7$

To verify multiply the terms on the right hand side,

  $\begin{array}{c}\left(4x^5+4x^4+x^3+x^2+2x+2\right)\left(x-1\right)+7=4x^6-4x^5+4x^5-4x^4+x^4-x^3+x^3-x^2+2x^2-2x+2x-2+7\\ =4x^6-3x^4+x^2+5\end{array}$

Thus, the quotient is $4x^5+4x^4+x^3+x^2+2x+2$ and reminder is $7$ when the polynomial $4x^6-3x^4+x^2+5$ is divided by $x-1$ .