Chapter A.4, Problem 27AYU

To determine

To calculate: The sum of values of a, b, c and dfor the expression $\frac{x^3-2x^2+3x+5}{x+2}=a{x}^2+bx+c+\frac{d}{x+2}$ with help of synthetic division.

Answer to Problem 27AYU

The sum of values of a, b, c and d is $-9$ .

Explanation of Solution

Given information:

The expression $\frac{x^3-2x^2+3x+5}{x+2}=a{x}^2+bx+c+\frac{d}{x+2}$ .

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 expression $\frac{x^3-2x^2+3x+5}{x+2}=a{x}^2+bx+c+\frac{d}{x+2}$ .

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

Here dividend is $x^3-2x^2+3x+5$ and divisor is $x-\left(-2\right)$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $1,-2,3,5$ .

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,

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ \begin{array}{cccc}& & & \end{array}\end{array}}$

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ \begin{array}{cccc}1& & & \end{array}\end{array}}$

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& -2& & \end{array}\\ \begin{array}{cccc}1& & & \end{array}\end{array}}$

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

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

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& -2& 8& \end{array}\\ \begin{array}{cccc}1& -4& & \end{array}\end{array}}$

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& -2& 8& \end{array}\\ \begin{array}{cccc}1& -4& 11& \end{array}\end{array}}$

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& -2& 8& -22\end{array}\\ \begin{array}{cccc}1& -4& 11& \end{array}\end{array}}$

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

  $-2\overline{)\begin{array}{l}\begin{array}{cccc}1& -2& 3& 5\end{array}\\ \begin{array}{cccc}& -2& 8& -22\end{array}\\ \begin{array}{cccc}1& -4& 11& -17\end{array}\end{array}}$

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 $x^2-4x+11$ and reminder is $-17$ .

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

  $\begin{array}{c}{x}^3-2x^2+3x+5=\left(x^2-4x+11\right)\left(x+2\right)+\left(-17\right)\\ \frac{x^3-2x^2+3x+5}{x+2}=x^2-4x+11-\frac{17}{x+2}\end{array}$

Compare the expression $\frac{x^3-2x^2+3x+5}{x+2}=a{x}^2+bx+c+\frac{d}{x+2}$ with the above,

Here $a=1,b=-4,c=11,d=-17$ .

The sum of the values is,

  $\begin{array}{c}a+b+c+d=1-4+11-17\\ =-9\end{array}$

Thus, the sum of values of a, b, c and d is $-9$ for the expression $\frac{x^3-2x^2+3x+5}{x+2}=a{x}^2+bx+c+\frac{d}{x+2}$ .