Answer
$P(a)=168$
Work Step by Step
Using the synthetic division shown below, the remainder (bottom-right number) of $(
6x^3-x^2+4x+3
)\div(
x-3
),$ is
\begin{align*}
168
.\end{align*}
Substituting $x=
3
$ in $P(x)=
6x^3-x^2+4x+3
,$ then by the Remainder Theorem, the remainder when $P(x)$ is divided by $
x-3
$ is
\begin{align*}
P(3)&=
6(3)^3-(3)^2+4(3)+3
\\&=
6(27)-9+4(3)+3
\\&=
162-9+12+3
\\&=
168
.\end{align*}
The two solutions above show that the remainder, $P(a),$ is
\begin{align*}
P(a)=168
.\end{align*}
