How do I do this Algebra 2 problem?

Question

Find the four real zeros (including multiplicity) of the polynomial f(x) = x^4 + 4x^3 - 11x^2 -66x - 72. Determine the sum of all the zeros of f(x).
Answer choice; A)-4 B)-1 C)-5 D)None of these

This is a multiple choice question and I chosed D, but they said the answer was A, -4. I kept doing this problem over and over but didn't seem to get the right answer. Can someone show me the step by step on how to figure out this problem? Thanks.

Answer 1

Consider this:
Suppose that the zeros of f(x) are a,b,c, and d.
f(x) = (x-a)(x-b)(x-c)(x-d) = (x^2-(a+b)x+ab)(x^2-(c+d)x+cd) =
x^4-(a+b)x^3+abx^2-(c+d)x^3
+(a+b)(c+d)x^2-ab(c+d)x^2
+cdx^2-cd(a+b)x+abcd.

Give attention to the variable x^3. The index for that variable is -(a+b+c+d). This is what you need. Since the index of x^3 in f(x) is 4, then the answer is -4.

Furthermore, the multiplication of all zeros is given by -72.

Answer 2

Here is one way to do it though there is probably a less cumbersome way:
Let a,b,c,d be constants to be determined so that
(x-a)(x-b)(x-c)(x-d) = x^4+4x^3-11x^2-66x-72
Now if you multiply out the LHS and equate coefficients:
a+b+c+d=-4 choice A) YOU are correct....but if not, please tell me why not