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

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Answer 1 · 2016-08-07T06:04:35

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

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...

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