Find algebraically the zeros for p (x) = x^3+x^2-4x-4

One zero is - 1

because p (-1) = (-1) ^3 + (-1) ^2 - 4 (-1) - 4 = 0

So one factor of p (x) is x + 1

Dividing p (x) by x + 1 gives the quotient

x^2 - 4

x - 2) (x + 2) = 0

x = 2, - 2

The zeroes are - 2,-1 and 2