P(x) = x^3 + 6x^2 − 32

Question

All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)

P(x) = x^3 + 6x^2 − 32

I get -4,2 and it keep saying i got it wrong. I don't know where i am going wrong.

Answer 1

You should have 3 roots because it's a cubic function, so one of those zeroes is repeated:

P(x) = x³+6x²-32

P(x) = x³+8x²+16x-2x²-16x-32

P(x) = x(x²+8x+16)-2(x²+8x+16)

P(x) = (x-2)(x²+8x+16)

P(x) = (x-2)(x+4)²

Hence, since the factor x+4 has a multiplicity of 2, then the roots are -4, -4, and 2.