If P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e i | vedclass

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(B) Given $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$. Notice the pattern in the values: $P(x) = x^2 + 1$ for $x = 0, 1, 2, 3, 4$. Let $Q(x) = P(x) - (

Vedclass · Accepted Answer

$146$

Vedclass · Answer

(B) Given $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$. Notice the pattern in the values: $P(x) = x^2 + 1$ for $x = 0, 1, 2, 3, 4$. Let $Q(x) = P(x) - (x^2 + 1)$. Since $P(x)$ is a polynomial of degree $5$ and $x^2 + 1$ is of degree $2$,$Q(x)$ is a polynomial of degree $5$ with roots $0, 1, 2, 3, 4$. Thus,$Q(x) = k(x)(x-1)(x-2)(x-3)(x-4)$ for some constant $k$. Comparing the coefficient of $x^5$ in $P(x)$,we see $k = 1$. So,$P(x) = x(x-1)(x-2)(x-3)(x-4) + x^2 + 1$. To find $P(5)$,substitute $x = 5$: $P(5) = 5(5-1)(5-2)(5-3)(5-4) + 5^2 + 1$ $P(5) = 5 	imes 4 	imes 3 	imes 2 	imes 1 + 25 + 1$ $P(5) = 120 + 26 = 146$.

If $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial such that $P(0) = 1, P(1) = 2, P(2) = 5, P(3) = 10$ and $P(4) = 17$,then $P(5) =$

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