Question

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

Answer 1

The naswer is `x in [-4,-1] uu [1, +oo]`

Explanation:

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

Then, `f(1)=1+4-1-4=0`

Therefore, `(x-1)` is a factor of `f(x)`

To find the other factors, let's do a long division

`color(white)(aaaaa)` `x^3+4x^2-x-4` `color(white)(aaaaa)` `∣` `x-1`

`color(white)(aaaaa)` `x^3-x^2` `color(white)(aaaaaaaaaaaaa)` `∣` `x^2+5x+4`

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

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

`color(white)(aaaaaaaaaaaa)` `0+4x-4`

`color(white)(aaaaaaaaaaaaaa)` `+4x-4`

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

Therefore, `f(x)=(x-1)(x+1)(x+4)>=0`

Let's do the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `+1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore `f(x)>=0` if `x in [-4,-1] uu [1, +oo]`

graph{x^3+4x^2-x-4 [-20.28, 20.27, -10.13, 10.13]}