How do you find all the zeros of #x^3-12x^2+12x+80# with #-2# as a zero?

1 Answer
Jul 14, 2016

Answer:

Zeros: -2, 10, 4.

Explanation:

Since we are told that #x=-2# is a zero, this cubic has a factor #(x+2)#:

#x^3-12x^2+12x+80 = (x+2)(x^2-14x+40)#

To factor the remaining quadratic, we can find a pair of factors of #40# with sum #14#. The pair #10, 4# works, hence:

#x^2-14x+40 = (x-10)(x-4)#

So the other two zeros are #x=10# and #x=4#