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

1 Answer
Aug 7, 2016

Answer:

#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...