If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ must have least 
degree $3$ polynomial. So Let $P(x)=Ax^3+Bx^2+Cx+D$ and $P'(x)=3Ax^2+2Bx+C.$
Now Given $P(1)=6\Rightarrow A+B+C+D = 6....................(1)$
and Given $P(3)=2\Rightarrow 27A+9B+3C+D=2..............(2)$
and Given $P'(1)=0\Rightarrow 3A+2B+C = 0.........................(3)$
and Given $P'(3)=0\Rightarrow 9A+6B+C = 0.........................(4)$
Now Subtract $(4)-(3)$, We Get $6A+4B=0\Rightarrow 3A+2B=0$
Similarly Sub $(2)-(1)\;,$ We Get $26A+8B+2C=-4\Rightarrow 13A+4B+C=-2$
Is there is any other method my which we can solve the above question in less complex way.
plz explain me, Thanks

Thanks Friends ,I have got it
My Solution:: Let $P'(x) = A\cdot (x-1)\cdot (x-3)\;,$ Then Integrate both side w. r. to $x\;,$ We Get
$\displaystyle P(x) = A\left[\frac{x^3}{3}-2x^2+3x\right]+\mathcal C\;,$ Now for calculation of $A$ and $\mathcal C\;,$ Put $x=1$ and $x=3.$
So $\displaystyle P(1) = \frac{4A}{3}+\mathcal C\Rightarrow 4A+3\mathcal C = 18$ and $\displaystyle P(3) = \mathcal C\Rightarrow \mathcal C = 2.$ Now Solve These
Equation we get $\displaystyle A=3$ and $\displaystyle \mathcal C = 2$. So $P'(0) = 3A = 9\Rightarrow P'(0)=9$ 
