TITLE: Let $P$ be a 4-th degree real polynomial with 5 conditions given. How to compute $P(4)$?
QUESTION [5 upvotes]: Yesterday I was math tutoring a 18-years old girl. And she asked me for the following problem: given $P\in\Bbb R[X]_4$, i.e. $P$ a real polynomial of degree exactly $4$, such that:

$P(1)=0$
It has a relative extrema in the points $x=2,3$, which value is $3$.

compute $P(4)$.
Now the second condition tells us that $P'(2)=P'(3)=0$ and $P(2)=P(3)=3$. Thus in total I have $5$ linear conditions on the $5$ real coefficients which define $P$, once we write it as
$$
P(x)=ax^4+bx^3+cx^2+dx+e.
$$
I.e. I have a linear system of $5$ equations in $5$ variables, which has (provided the conditions are all indipendent one each other) one solution: thus I'd have identified uniquely my polynomial, hence I could compute easily $P(4)$ and conclude my exercise.
My problem is: this girl doesn't know matrices, Gauss elimination and all the linear algebra tool which help to solve quickly this kind of problems, thus in order to solve such a system she should do it by subsitutions and so on, which is really tedious and not instructive (to me, at least), and it seems weird that her teacher gave her such an exercise to solve.
Moreover, what is asked is to compute $P(4)$, NOT to determine the polynomial $P$.
So I am asking myself: is there another way to do it? A way which avoids all that calculation?
I tried to write $P$ as
$$
P(x)=a(x-x_0)(x-x_1)(x-x_2)(x-x_3)
$$
but nothing good came out. Any idea?

REPLY [2 votes]: Consider the polynomial $q(x)=p(x)-3$. Then, $q(2)=q(3)=q'(2)=q'(3)=0$ and $q(1)=-3$. So, $$q(x)=a(x-2)^2(x-3)^2$$ And $q(1)=-3$ implies that $a=\frac{-3}{4}$ and from here: $p(4)=q(4)+3=-3$.