$P(x)$ is a polynomial of least degree with a local maximum of $6$ at $x=1$ and minimum of $2$ at $x=3$. Find $P^\prime(0)$. 
Let $P(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and minimum at $x=3$.If $P(1)=6$ and $P(3)=2$, then find $P'(0)$. 

Method:  Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be 
$$P(x) = ax^3+ bx^2+cx+d$$
Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?
 A: Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
A: HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6, P(3)=2, P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
A: If you use lulu's idea, then you should get the answer quicker because you will be solving a simultaneous system in two variables, instead of four variables.  First, if $P(x)$ is of lowest degree, then we expect $P(x)$ to be a cubic polynomial.  Since $P'(x)$ is a quadratic polynomial with two roots $x=1$ and $x=3$, we have
$$P'(x)=3a(x-1)(x-3)=3ax^2-12ax+9a$$
for some constant $a$ (the factor $3$ is there simply to make the integration easier).  Thus,
$$P(x)=ax^3-6ax^2+9ax+c$$
for some constant $c$.  Since $P(1)=6$ and $P(3)=2$, we get
$$4a+c=6\text{ and }c=2\,,$$
so $a=1$ and $c=2$.  That is,
$$P(x)=x^3-6x^2+9x+2\,,$$
with 
$$P'(x)=3(x-1)(x-3)=3x^2-12x+9\text{ and }P''(x)=6(x-2)\,.$$
This makes $P''(1)=-6<0$ and $P''(3)=6>0$, so $x=1$ and $x=3$ are the local maximum and the local minimum, respectively.  Hence, $P(x)$ is indeed the desired polynomial.
