Question

How do you solve `x^3+3x^2+x+3<=0`?

Answer 1

`x<=3`

Explanation:

`x^3+3x^2+x+3<=0`

`hArrx^2(x+3)+1(x+3)<=0`

or `(x^2+1)(x+3)<=0`

As `x^2+1` is always positive,

dividing by `(x^2+1)`, we get `x+3<=0` or `x<=-3`