# How do you solve x^3 -3x^2 +16x -48 = 0?

Feb 16, 2017

$3 \mathmr{and} \pm i 4$. See the Socratic graph of the cubic, making x-intercept 3..

#### Explanation:

From sign changes in the coefficients, the equation has utmost 3

positive roots,There are no changes in sign,

when x is changed to $- x$. And so, there are no negative roots.

The cubic is 0 at x = 0. So, it becomes

$\left(x - 3\right) \left({x}^{2} + 16\right)$

The other solutions are from by ${x}^{2} + 16 = 0$, giving $x = \pm i 4$

graph{(x-3)(x^2+16) [2, 4, -500, 500]} 0, -500, 500]}

Not to scale. It is large y vs small x, for approximating the solution.