# Find all values of x that satisfy: 2x^3-x^2-8x+4>=0.    I put ^ to tell you that it represents power and >= means equal to or greater than.

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Find all values of x that satisfy `2x^3-x^2-8x+4 >= 0`

First we factor the left hand side:

`2x^3-x^2-8x+4` look at the terms two terms at a time
=`x^2(2x-1)-4(2x-1)` factoring out the common factor.
=`(x^2-4)(2x-1)` apply the distributive law
=`(x+2)(x-2)(2x-1)` factor the perfect square

So we now have `(x+2)(x-2)(2x-1)>=0`

Note that the function is zero at -2,2, and 1/2. We check the value of the function on the intervals x<-2; -2<x<1/2; 1/2<x<2; x>2

For x<-2 the function is negative (plug in a test value like -3)
For  -2<x<1/2 the function is positive (plug in 0)
For 1/2<x<2 the function is negative (plug in 1)
For x>2the function is positive. (plug in 3)

We can also recognize that a cubic with leading coefficient greater than zero rises,falls, then rises and use the graph along with the zeros to find the intervals where the function is positive.