Answer
The sum of $a,b,c$ and $d$ is $-9$.
Work Step by Step
The given expression is:-
$(x^3-2x^2+3x+5)\div (x+2)$
The divisor is $x+2$, so the value of $c=-2$.
and on the right side the coefficients of dividend in descending powers of $x$.
Perform synthetic division to obtain:
$\begin{matrix}
&-- &-- &--&--& && \\
-2) &1&-2&3&5& & &\\
& &-2 &8 &-22 && &\\
& -- & -- & --& -- &&& \\
& 1 & -4& 11 &-17 & &&\\
\end{matrix}$
The remainder is $-17$.
The Quotient is $x^2-4x+11$
Check:
$\text{(Divisor)(Quotient)+Remainder}$
$=(x+2)(x^2-4x+11)-17$
$=x^3-4x^2+11x+2x^2-8x+22-17$
$=x^3-2x^2+3x+5$
The solution is
$=\text{Quotient} + \dfrac{\text{Remainder}}{\text{divisor}}$
$=x^2-4x+11 + \dfrac{-17}{x+2}$
From the question we have.
$x^2-4x+11 + \dfrac{-17}{x+2}=ax^2+bx+c+\frac{d}{x+2}$
Thus, the values are $a=1,b=-4,c=11$ and $d=-17$
The sum of $a,b,c$ and $d$ is
$=1-4+11-17$
$=-9$
Hence, the sum of $a,b,c$ and $d$ is $-9$.
