Let P(x) be a real polynomial of degree 3 whi | vedclass

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(B) Let $P'(x) = k(x - 1)(x + 1) = k(x^2 - 1)$ for some constant $k$.Integrating $P'(x)$,we get $P(x) = k(\frac{x^3}{3} - x) + C$.Given $P(-3) = 0$,we

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$8$

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(B) Let $P'(x) = k(x - 1)(x + 1) = k(x^2 - 1)$ for some constant $k$.Integrating $P'(x)$,we get $P(x) = k(\frac{x^3}{3} - x) + C$.Given $P(-3) = 0$,we have $k(\frac{-27}{3} - (-3)) + C = 0$,which simplifies to $k(-9 + 3) + C = 0$,so $-6k + C = 0$,or $C = 6k$.Given $\int_{-1}^{1} P(x) dx = 18$,we calculate $\int_{-1}^{1} (k(\frac{x^3}{3} - x) + C) dx = 18$.Since $k(\frac{x^3}{3} - x)$ is an odd function,its integral over $[-1, 1]$ is $0$. Thus,$\int_{-1}^{1} C dx = 18$,which gives $2C = 18$,so $C = 9$.Using $C = 6k$,we find $9 = 6k$,so $k = \frac{9}{6} = \frac{3}{2}$.Thus,$P(x) = \frac{3}{2}(\frac{x^3}{3} - x) + 9 = \frac{1}{2}x^3 - \frac{3}{2}x + 9$.The sum of the coefficients of $P(x)$ is $P(1) = \frac{1}{2}(1)^3 - \frac{3}{2}(1) + 9 = \frac{1}{2} - \frac{3}{2} + 9 = -1 + 9 = 8$.

Let $P(x)$ be a real polynomial of degree $3$ which vanishes at $x = -3$. Let $P(x)$ have local minima at $x = 1$,local maxima at $x = -1$,and $\int_{-1}^{1} P(x) dx = 18$. Then the sum of all the coefficients of the polynomial $P(x)$ is equal to ....... .

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