\(P(4)=5, P′(4)=3\), and \(P′′ (4)=3.\)

\(P(4)=5P(4)=5\) gives us

\(a+4b+16c=5\).

Now differentiating P, we get \(P′=b+2cx\). Now putting x=4, we get

\(b+8c=3.

Now differentiating P′, we get \(P''=2cP\). Now putting x=4, we get

\(b+12=3\Rightarrowb=−9.\)

Now putting \(c=\frac{3}{2} and \(b= -9 in {eq1}\) we get

\(a−36+48=5\Rightarrowa=−7.\)

So required second degree polynomial is \(P(x)=−7−9x+(\frac{3}{2})x2\)