Let $P$ be a polynomial with integer coefficients satisfying $P(0)=1, P(1)=3, P'(0)=-1, P'(1)=10$. What is the minimum possible degree of $P$?

Question

I am preparing for GRE MATH SUBJECT TEST, I have reviewed many problems, some of these problems are not in GRE Math Practice books, but they are in some other books. I found the following problem, but not sure if it can be a GRE problem or no. And I do not even know how to start;

Let $P$ be a polynomial with integer coefficients satisfying $P(0)=1, P(1)=3, P'(0)=-1, P'(1)=10$. What is the minimum possible degree of $P$?

(A) $3$

(B) $4$

(C) $5$

(D) $6$

(E) No such $P$ exists.

Answer 1

Since you are given two points of $P'(x)$ , start with a linear function for $P'(x)$, let it be $ax+b$, plug the values to get $P'(x)=11x-1$, now integrate it to get $P(x)=\dfrac{11x^2}{2}-x+C$, you have $P(0)=1$, so $C=1$, but at $x=1$, we are getting $P(1)=\dfrac{11}{2}\ne 3$, so this suggests start with a second degree equation for $P'(x)$, and since this will contain one extra variable, this will give you the answer.

Answer 2

As a polynomial of degree $3$ $p(x) = a x^3+b x^2 + c x +d$ has 4 coefficients, this may be sufficient providing that the equations induced by the conditions are compatible. Namely, that would be the case if the determinant

$$\begin{vmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1\\ 3 & 2 & 1 & 0 \end{vmatrix}$$ doesn't vanishes. As this determinant is equal to one, a polynomial of degree $3$ is solution and answer (a) is the right one.