Question

How do you solve `x^3-6x^2-4x+24>0` using a sign chart?

Answer 1

The answer is `x in ] -2,+2 [ uu ] 6, +oo[`

Explanation:

Let `f(x)=x^3-6x^2-4x+24`

As a polynomial, the domain of `f(x)` is `D_f(x)=RR`

Then, by trial and error,

`f(2)=8-24-8+24=0`

Therefore,

`(x-2)` is a factor

To find the other factors, we do a long division

`color(white)(aaaa)` `x^3-6x^2-4x+24` `color(white)(aaaa)` `∣` `x-2`

`color(white)(aaaa)` `x^3-2x^2` `color(white)(aaaaaaaaaaaaa)` `∣` `x^2-4x-12`

`color(white)(aaaa)` `0-4x^2-4x`

`color(white)(aaaaaa)` `-4x^2+8x`

`color(white)(aaaaaaa)` `-0-12x+24`

`color(white)(aaaaaaaaaa)` `-12x+24`

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

Therefore,

`x^2-4x-12=(x+2)(x-6)`

and `x^3-6x^2-4x+24=(x-2)(x+2)(x-6)`

Now, we can establish the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaa)` `2` `color(white)(aaaaa)` `6` `color(white)(aaaa)` `+oo`

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

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

`color(white)(aaaa)` `x-6` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaa)` `+`

Therefore,

`f(x)>0` when `x in ] -2,+2 [ uu ] 6, +oo[`