Question

How do you solve `2x^3-3x^2-32x+48>=0` using a sign chart?

Answer 1

The answer is `x in [-4,3/2] uu [4, +oo[`

Explanation:

Let `f(x)=2x^3-3x^2-32x+48`

Then

`f(4)=128-48-128+48=0`

So, `(x-4)` is a factor of `f(x)`

To find the other factors, we do a long division

`color(white)(aaaa)` `2x^3-3x^2-32x+48` `color(white)(aaaa)` `∣` `color(red)(x-4)`

`color(white)(aaaa)` `2x^3-8x^2` `color(white)(aaaaaaaaaaaaaa)` `∣` `color(blue)(2x^2+5x-12)`

`color(white)(aaaaaa)` `0+5x^2-32x`

`color(white)(aaaaaaaa)` `+5x^2-20x`

`color(white)(aaaaaaaaaa)` `+0-12x+48`

`color(white)(aaaaaaaaaaaaaa)` `-12x+48`

`color(white)(aaaaaaaaaaaaaaaa)` `-0+0`

Therefore,

`f(x)=2x^3-3x^2-32x+48=(x-4)(2x^2+5x-12)`

`=(x-4)(2x-3)(x+4)`

Now, we can make the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaaa)` `3/2` `color(white)(aaaa)` `4` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+4` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `2x-3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-4` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaa)` `+`

Therefore,

`f(x)>=0` when `x in [-4,3/2] uu [4, +oo[`