# 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

Question:

Let $\mathrm{P}(\mathrm{x})$ be a real polynomial of degree 3 which vanishes at $x=-3$. Let $P(x)$ have local minima at $\mathrm{x}=1$, local maxima at $\mathrm{x}=-1$ and

$\int_{-1}^{1} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=18$, then the sum of all the

coefficients of the polynomial $\mathrm{P}(\mathrm{x})$ is equal to____________

Solution:

Let $p^{\prime}(x)=a(x-1)(x+1)=a\left(x^{2}-1\right)$

$\mathrm{p}(\mathrm{x})=\mathrm{a} \int\left(\mathrm{x}^{2}-1\right) \mathrm{d} \mathrm{x}+\mathrm{c}$

$=a\left(\frac{x^{3}}{3}-x\right)+c$

Now $\mathrm{p}(-3)=0$

$\Rightarrow a\left(-\frac{27}{30}+3\right)+c=0$

$\Rightarrow-6 a+c=0$

Now $\int_{-1}^{1}\left(a\left(\frac{x^{3}}{3}-x\right)+c\right) d x=18$

$=2 c=18 \Rightarrow c=9$        .......(2)

$\Rightarrow$ from (1) \& (2) $\Rightarrow-6 a+9=0 \Rightarrow a=\frac{3}{2}$

$\Rightarrow p(x)=\frac{3}{2}\left(\frac{x^{3}}{3}-x\right)+9$

sum of coefficient

$=\frac{1}{2}-\frac{3}{2}+9$

$=8$