Answer
$x+3$ is not a factor of the given polynomial.
Work Step by Step
The given expression is:-
$(-4x^3+5x^2+8)\div (x+3)$
Rewrite as descending powers of $x$.
$(-4x^3+5x^2+0x+8)\div (x+3)$
The divisor is $x+3$, so the value of $c=-3$.
and on the right side the coefficients of dividend in descending powers of $x$.
Perform synthetic division to obtain:
$\begin{matrix}
&-- &-- &--&--& \\
-3) &-4&5&0&8& & \\
& &12 &-51 &153 && \\
& -- & -- & --& -- && \\
& -4 & 17& -51 &161 & \\
\end{matrix}$
The remainder is $161$.
The Quotient is $-4x^2+17x-51$
Check:-
$\text{(Divisor)(Quotient)+Remainder}$
$=(x+3)(-4x^2+17x-51)+162$
$=-4x^3+17x^2-51x-12x^2+51x-153+161$
$=-4x^3+5x^2+8$
The remainder is not zero.
Hence, $x+3$ is not a factor of the given polynomial.
