Question

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

Answer 1

The answer is `x in ] -oo,-2[uu] -2,1 [`

Explanation:

Let's rewrite the inequality

`x^3+3x^2-4<0`

We must factorise the LHS

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

The domain of `f(x)` is `D_f(x)=RR`

`f(1)=1+3-4=0`

So, `(x-1)` is a factor

To find the other factors, we do a long division

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

`color(white)(aaaa)` `x^3-x^2` `color(white)(aaaa)` ###color(white)(aaaaaaaa)#`∣` `x^2+4x+4`

`color(white)(aaaa)` `0+4x^2` `color(white)(aaaa)` ###color(white)(aaaaaaaa)#

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

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

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

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

Therefore,

`x^3+3x^2-4=(x-1)(x^2+4x+4)=(x-1)(x+2)^2`

So, we can make the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `-2` `color(white)(aaaaaaaa)` `1` `color(white)(aaaaaaaa)` `+oo`

`color(white)(aaaa)` `(x+2)^2` `color(white)(aaaa)` `+` `color(white)(aaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<0` when `x in ] -oo,-2[uu] -2,1 [`