# 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 $P$?

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.

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.
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.