A polynomial $P(x)$ of degree $5$ with lead coefficient one,increases in the $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$ A polynomial function $P(x)$ of degree $5$ with leading coefficient one,increases in the interval $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$. Given that $P(0)=4$ and $P'(2)=0$, find the value of $P'(6)$.



I noticed that $1,2,3$ are the roots of the polynomial $P'(x)$.$P'(x)$ is a fourth degree polynomial. So

Let $$P'(x)=(x-1)(x-2)(x-3)(ax+b)$$
Now I expanded $P'(x)$ and integrated it to get $P(x)$ and use the given condition $P(0)=4$, but I am not able to calculate $a,b$.
As $P(x)=x^5+\ldots$, we have $P'(x)=5x^4+\ldots$. This gives us $a=5$.
Moreover, the facts that $P$ is decreasing on $(1,3)$ and $P'(2)=0$ imply that $x=2$ must in fact be a multiple root of $P'$. We conclude that $2a+b=0$, so $b=-10$.
The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-2 find a possible formula for P(x).