One of the solutions to the equation x^3 + x^2 - 16x - 16 = 0 is -4
test the solution to find the full solution set and whether or not they're real or non-real solutions.
To test whether -4 is a solution to the equation x^3 + x^2 - 16x - 16 = 0, we substitute x = -4 into the equation and check if it satisfies the equation:
(-4)^3 + (-4)^2 - 16(-4) - 16 = 0
-64 + 16 + 64 - 16 = 0
0 = 0
Since the result is true, we can conclude that -4 is indeed a solution to the equation.
To find the full solution set, we can use polynomial division or factoring methods to reduce the cubic equation. In this case, we can factor out the common factor of (x + 4) from the equation:
(x + 4)(x^2 - 4x - 4) = 0
Now we can solve for x by setting each factor equal to zero:
x + 4 = 0
x = -4
x^2 - 4x - 4 = 0
Using the quadratic formula, we can solve for x in the second equation:
x = [4 ± √(4^2 - 4(-4))]/2
x = [4 ± √(16 + 16)]/2
x = [4 ± √32]/2
x = [4 ± 4√2]/2
x = 2 ± 2√2
Therefore, the full solution set to the equation x^3 + x^2 - 16x - 16 = 0 consists of x = -4, x = 2 + 2√2, and x = 2 - 2√2.
Regarding the nature of the solutions, we can say that -4 is a real solution since it is a real number. The other two solutions, 2 + 2√2 and 2 - 2√2, are non-real solutions since they involve square roots of negative numbers.