Chapter A.4, Problem 13AYU

To determine

To calculate: The quotient and remainder when the polynomial $0.1x^3+0.2x$ is divided by $x+1.1$ with help of synthetic division.

Answer to Problem 13AYU

The quotient is $0.1x^2-0.11x+0.321$ and reminder is $-0.3531$ .

Explanation of Solution

Given information:

The polynomial $0.1x^3+0.2x$ and a factor of it $x+1.1$ .

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 $0.1x^3+0.2x$ and a factor of it $x+1.1$ .

To divide the polynomial $0.1x^3+0.2x$ by $x+1.1$ follow the steps provided below,

Here dividend is $0.1x^3+0\cdot x^2+0.2x+0$ and divisor is $x-\left(-1.1\right)$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $0.1,0,0.2,0$ .

Now, the divisor is of the form $x-c$ , where $c=-1.1$ .

Step 2: To perform the synthetic division put the coefficients in the division sign along with $-1.1$ on the left side,

  $-1.1\overline{)\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}}$

Step 3: Bring down $0.1$ to in third row,

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ \begin{array}{cccc}0.1& & & \end{array}\end{array}}$

Step 4: Multiply $-1.1$ with the first entry of row 3 and place the result in row 2 column 2.

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& & \end{array}\\ \begin{array}{cccc}0.1& & & \end{array}\end{array}}$

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

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& & \end{array}\\ \begin{array}{cccc}0.1& -0.11& & \end{array}\end{array}}$

Step 6: Multiply $-1.1$ with the second entry of row 3 and place the result in row 2 column 3.

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& 0.121& \end{array}\\ \begin{array}{cccc}0.1& -0.11& & \end{array}\end{array}}$

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

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& 0.121& \end{array}\\ \begin{array}{cccc}0.1& -0.11& 0.321& \end{array}\end{array}}$

Step 8: Multiply $-1.1$ with the third entry of row 3 and place the result in row 2 column 4.

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& 0.121& -0.3531\end{array}\\ \begin{array}{cccc}0.1& -0.11& 0.321& \end{array}\end{array}}$

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

  $-1.1\overline{)\begin{array}{l}\begin{array}{cccc}0.1& 0& 0.2& 0\end{array}\\ \begin{array}{cccc}& -0.11& 0.121& -0.3531\end{array}\\ \begin{array}{cccc}0.1& -0.11& 0.321& -0.3531\end{array}\end{array}}$

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 $0.1x^2-0.11x+0.321$ and reminder is $-0.3531$ .

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

  $0.1x^3+0.2x=\left(0.1x^2-0.11x+0.321\right)\left(x+1.1\right)+\left(-0.3531\right)$

To verify multiply the terms on the right hand side,

  $\left(0.1x^2-0.11x+0.321\right)\left(x+1.1\right)-0.3531=0.1x^3+0.2x$

Thus, the quotient is $0.1x^2-0.11x+0.321$ and reminder is $-0.3531$ when the polynomial $0.1x^3+0.2x$ is divided by $x+1.1$ .