Answer
$P(a)=10$
Work Step by Step
Using the synthetic division below, the remainder (bottom-right number) of $(
2x^3-x^2+10x+5
)\div\left(
x-\dfrac{1}{2}
\right)$ is
\begin{align*}
10
.\end{align*}
Substituting $x=\dfrac{1}{2}$ in $P(x)=
2x^3-x^2+10x+5
,$ then by the Remainder Theorem, the remainder is
\begin{align*}
P\left(\dfrac{1}{2}\right)&=
2\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2+10\left(\dfrac{1}{2}\right)+5
\\\\&=
2\left(\dfrac{1}{8}\right)-\left(\dfrac{1}{4}\right)+10\left(\dfrac{1}{2}\right)+5
\\\\&=
\dfrac{1}{4}-\dfrac{1}{4}+5+5
\\\\&=
10
.\end{align*}
Both solutions above show that the remainder, $P(a),$ is
\begin{align*}
P(a)=10
.\end{align*}
