Question

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

Answer 1

The answer is `x in ]-1, +oo[`

Explanation:

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

Before, we need to find the factors of `f(x)`

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

Therefore, `(x+1)` is a factor

To find the other factors, we perform a long division

`color(white)(aaaa)` `x^3+x^2+4x+4` `color(white)(aaaa)` `|` `x+1`

`color(white)(aaaa)` `x^3+x^2` `color(white)(aaaaaaaaaaaa)` `|` `x^2+4`

`color(white)(aaaaa)` `0+0+4x+4`

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

`color(white)(aaaaaaaaaaaaa)` `0+0`

Therefore,

`f(x)=(x+1)(x^2+4)`

`AA x in RR,(x^2+4)>0`

So, we can build the sign chart

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

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

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

Therefore,

`f(x)>0` when `x in ]-1, +oo[`