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.

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
= class="AM">`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 class="AM">`(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.

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sscr="-5,5,-5,10,1,1,1,1,1,300,200,func,2x^3-x^2-8x+4,null,1,1, ,
,black,1,none"/>

Thus your answer is class="AM">`(-2 <= x <=
1/2)uu(x>2)`