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

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Answer 1 · 2016-07-14T16:35:52

Zeros: -2, 10, 4.

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

$x^{3} - 12 x^{2} + 12 x + 80 = (x + 2) (x^{2} - 14 x + 40)$ $x^3-12x^2+12x+80 = (x+2)(x^2-14x+40)$

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

$x^{2} - 14 x + 40 = (x - 10) (x - 4)$ $x^2-14x+40 = (x-10)(x-4)$

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

How do you find all the zeros of x3−12x2+12x+80x^3-12x^2+12x+80 with −2-2 as a zero?