# How do you solve x^3+x^2-10x+8>=0?

Aug 7, 2016

$x \in \left[- 4 , 1\right]$and $x = 2.$ ..

#### Explanation:

Sum of the coefficients = 0. So, x-1 is a factor of the cubic. The other

quadratic factor is ${x}^{2} + x - 8 = \left(x - 2\right) \left(x + 4\right)$

The cubic (y) is 0, at x = -4, 1. or 2#.

At $x = 0 \in \left(- 4 , 1\right)$,, y = 10 > 0,, and so, y > 0, in this interval.

Also, this implies that the cubic has to be < 0 beyond, before

reaching the third zero x = 2, to become positive once again...