Question

How do you solve `x^ { 3} - 6x ^ { 2} + 16x - 16= 0`?

Answer 1

This cubic has roots:

`x = 2" "` and `" "x = 2 +- 2i`

Explanation:

Given:

`x^3-6x^2+16x-16 = 0`

By the rational roots theorem, any rational roots of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-16` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-2, +-4, +-8, +-16`

In addition, note that the signs of the coefficients are in the pattern `+ - + -`. With `3` changes of sign, Descartes' Rule of Signs tells us that this cubic polynomial has `3` or `1` positive real zeros. Reversing the signs of the coefficients on the terms of odd degree we get the pattern `- - - -`. With no change of sign, we can tell that this cubic has no real negative zeros.

So the only possible rational roots are:

`1, 2, 4, 8, 16`

We find:

`(color(blue)(2))^3-6(color(blue)(2))^2+16(color(blue)(2))-16 = 8-24+32-16 = 0`

So `x=2` is a root and `(x-2)` a factor:

`0 = x^3-6x^2+16x-16`

`color(white)(0) = (x-2)(x^2-4x+8)`

The remaining quadratic has non-real complex zeros that we can find by completing the square:

`0 = x^2-4x+8`

`color(white)(0) = x^2-4x+4+4`

`color(white)(0) = (x-2)^2+2^2`

`color(white)(0) = (x-2)^2-(2i)^2`

`color(white)(0) = ((x-2)-2i)((x-2)+2i)`

`color(white)(0) = (x-2-2i)(x-2+2i)`

Hence:

`x = 2+-2i`