Let f(x)=P(x)−(x2+1)
Then, x=1,3,5 are roots of f(x).
Therefore, we can write f(x) as A(x−1)(x−3)(x−5)
⇒P(x)=A(x−1)(x−3)(x−5)+(x2+1)
Also, P(x) should be of degree 3.
We have,
limx→∞x3P(x)x6+3x2+7=2
⇒A=2
⇒P(x)=2(x−1)(x−3)(x−5)+x2+1
Now, P(2)=11,P(0)=−29
∴|P(0)|+P(2)10=4