If $\mathrm{P}(x)=x^5+a x^4+b x^3+c x^2+d x+e$ is a polynomial such that $\mathrm{P}(0)=1, \mathrm{P}(1)=2,…
If $\mathrm{P}(x)=x^5+a x^4+b x^3+c x^2+d x+e$ is a polynomial such that $\mathrm{P}(0)=1, \mathrm{P}(1)=2, \mathrm{P}(2)=5, \mathrm{P}(3)=10$ and $\mathrm{P}(4)=17$, then $\mathrm{P}(5)=$
$26$
$146$
$126$
$76$
Solution
$P(x)=x^5+a x^4+6 x^3+c x^2+d x+e$
$P(x)-x^2-1$ is a polynomial of degree 5 and has 5 roots $0,1,2,3,4$
$P(x)-x^2-1=(x-0)(x-1)(x-2)(x-3)(x-4)$
$P(x)=x(x-1)(x-2)(x-3)(x-4)+x^2+1$
$P(5)=5!+25+1=146$