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If a polynomial $\mathrm{P}(x)$ given by $P(x)=2 x^4+a x^3+b x^2+c x+d$ is such that $\mathrm{P}(1)=4, \mathrm{P}(2)=7, \mathrm{P}(3)=12$ and $\mathrm{P}(4)=19$, then $\mathrm{P}(5)=$
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76
$\mathrm{P}(x)=2 x^4+a x^3+b x^2+c x+d$
$\begin{aligned} & \mathrm{P}(1)=4=1^2+3 \\ & \mathrm{P}(2)=7=2^2+3 \\ & \mathrm{P}(3)=12=3^2+3 \\ & \mathrm{P}(4)=19=4^2+3 \\ & \therefore \mathrm{P}(x)=x^2+3\end{aligned}$
But to make it degree 4 polynomial with leading coefficient $=2$
$\begin{aligned} & \mathrm{P}(x)=2(x-1)(x-2)(x-3)(x-4)+x^2+3 \\ & \mathrm{P}(5)=2(5-1)(5-2)(5-3)(5-4)+\left(5^2+3\right) \\ & =2 \cdot 4 \cdot 3 \cdot 2 \cdot 1+28=76\end{aligned}$
$\begin{aligned} & \mathrm{P}(1)=4=1^2+3 \\ & \mathrm{P}(2)=7=2^2+3 \\ & \mathrm{P}(3)=12=3^2+3 \\ & \mathrm{P}(4)=19=4^2+3 \\ & \therefore \mathrm{P}(x)=x^2+3\end{aligned}$
But to make it degree 4 polynomial with leading coefficient $=2$
$\begin{aligned} & \mathrm{P}(x)=2(x-1)(x-2)(x-3)(x-4)+x^2+3 \\ & \mathrm{P}(5)=2(5-1)(5-2)(5-3)(5-4)+\left(5^2+3\right) \\ & =2 \cdot 4 \cdot 3 \cdot 2 \cdot 1+28=76\end{aligned}$
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