Answer
See below
Work Step by Step
Given:
$a_0P_0+a_1P_1+a_2P_2+a_3P_3$$=\frac{1}{2}a_3(5x^3-3x)+\frac{1}{2}a_2(3x^2-1)+a_1x+a_0\\
=\frac{5a_3x^3}{2}+\frac{3a_2x^2}{2}+(a_1-\frac{3a_3}{2})x+a_0-\frac{1}{2}a_2\\
=2x^3+x^2+5$
Compare cofficients on both sides:
$a_0-\frac{a_2}{2}=5\\
a_1-\frac{3}{2}a_3=0\\
\frac{3}{2}a_2=1\\
\frac{5}{2}a_3=2$
Consequently,
$a_0=\frac{16}{3}\\
a_1=\frac{6}{5}\\
a_2=\frac{2}{3}\\
a_3=\frac{4}{5}$
Therefore,
$2x^3+x^2+5=\frac{16}{3}P_0+\frac{6}{5}P_1+\frac{2}{3}P_2+\frac{4}{5}P_3$
