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(B) Given $P(x) = 2x^4 + ax^3 + bx^2 + cx + d$ is a polynomial of degree $4$ with leading coefficient $2$. We observe that for $x = 1, 2, 3, 4$,the va

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$76$

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(B) Given $P(x) = 2x^4 + ax^3 + bx^2 + cx + d$ is a polynomial of degree $4$ with leading coefficient $2$. We observe that for $x = 1, 2, 3, 4$,the values are $P(x) = x^2 + 3$. Let $Q(x) = P(x) - (x^2 + 3)$. Since $P(x)$ is a polynomial of degree $4$ with leading coefficient $2$,$Q(x)$ is also a polynomial of degree $4$ with leading coefficient $2$. Since $P(1)=4, P(2)=7, P(3)=12, P(4)=19$,we have $Q(1)=0, Q(2)=0, Q(3)=0, Q(4)=0$. Thus,$Q(x) = 2(x-1)(x-2)(x-3)(x-4)$. Therefore,$P(x) = 2(x-1)(x-2)(x-3)(x-4) + x^2 + 3$. To find $P(5)$,substitute $x = 5$: $P(5) = 2(5-1)(5-2)(5-3)(5-4) + (5^2 + 3)$ $P(5) = 2(4)(3)(2)(1) + (25 + 3)$ $P(5) = 2(24) + 28 = 48 + 28 = 76$.

If a polynomial $P(x)$ of degree $4$ is given by $P(x) = 2x^4 + ax^3 + bx^2 + cx + d$ such that $P(1) = 4, P(2) = 7, P(3) = 12$,and $P(4) = 19$,then find the value of $P(5)$.

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