Question

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

Answer 1

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`