Question

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

Answer 1

`x in [-4, 1]`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 = (x-2)(x+4)`

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

At `x = 0 in (-4, 1)`,, 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...