Here is the full problem statement:
The polynomial $p(x)=x^4-x^3+x^2-x+1$ has the following values: $(-2,31)$, $(-1,5)$, $(0,1)$, $(1,1)$, $(2,11)$, $(3,61)$.
Find a polynomial $q(x)$ that takes the following values: $(-2,31)$, $(-1,5)$, $(0,1)$, $(1,1)$, $(2,11)$, $(3,-59)$.
I know I can find the polynomial $q(x)$ by the finding the interpolating polynomial using the given data values, but I also know that I should be able to use $p(x)$.
A hint was given saying that this problem can be solved with little work, so how would I use $p(x)$ to solve?
You need to add a polynomial that has $0$'s in the points $-2$ to $2$ and equals $(-120)$ at $3$. This means, to the original polynomial you have to add a multiple of $(x+2)(x+1)x(x-1)(x-2)$. Since $(3+2)(3+1)3(3-1)(3-2) = 120$, the answer is $x^4 - x^3 + x^2 - x^1 + 1 - (x+2)(x+1)x(x-1)(x-2)$. You may recalculate the coefficients yourself.
P.S. The polynomial is not unique, since any polynomial which has zeros at the points $-2$ to $2$ and is nonzero at $3$ would do. All you need is to scale your polynomial accordingly.