Answer
$P(a)=0$
Work Step by Step
Using the synthetic division shown below, the remainder (bottom-right number) of $(
x^3-7x^2+15x-9
)\div(
x-3
),$ is
\begin{align*}
0
.\end{align*}
Substituting $x=
3
$ in $P(x)=
x^3-7x^2+15x-9
,$ then by the Remainder Theorem, the remainder when $P(x)$ is divided by $
x-3
$ is
\begin{align*}
P(3)&=
(3)^3-7(3)^2+15(3)-9
\\&=
27-7(9)+15(3)-9
\\&=
27-63+45-9
\\&=
0
.\end{align*}
The two solutions above show that the remainder, $P(a),$ is
\begin{align*}
P(a)=0
.\end{align*}