0.1x^3-0.4x^2-3.1x+7<0

Question

How do I break this down

Answer 1

Multiply through by 10: x³-4x²-31x+70<0.

Now look for rational zeroes of the cubic. 70=2×5×7 so 2, 5 or 7 could be a zero or their negatives.

First, let x=2: 8-16-62+70=78-78=0, so x=2 is a zero. By dividing by this zero we can reduce the cubic to a quadratic (synthetic division):

2 | 1 -4 -31   70

     1  2  -4  | 70

     1 -2 -35 |  0 = x²-2x-35=(x-7)(x+5).

So there are three zeroes: 2, 7, -5.

Now we have to consider the inequality for (x+5)(x-2)(x-7). The product of these three factors must be negative.

x<-5: (-)(-)(-)<0 so x<-5 is one solution;

-5<x<2: (+)(-)(-)>0 so no solution here;

2<x<7: (+)(+)(-)<0 so 2<x<7 is another solution;

x>7: (+)(+)(+)>0 so no solution here.

SOLUTION: x<-5 or 2<x<7.