How to solve the inequality $x^3 + 2x^2 - 2x - 1 < 0$?

Question

What will be the solution to the following inequality?

$$x^3 + 2x^2 - 2x - 1 < 0$$

I can't solve this using the table of signs method. Please include an explanation as to how it was solved.

Answer 1

HINT: your inequality is equivalent to $$(x-1)(x^2+3x+1)<0$$ solve $$x^2+3x+1=0$$ for $x$ and factor the equation $$x_{1,2}=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}$$

Answer 2

The first derivative of the LHS

$$3x^2+4x-2$$

has two real roots (as the discriminant is positive), so that the function has two local extrema and potentially three real roots. You can't escape from computing them, but you can also establish that the maximum is positive and the minimum negative, so that the three roots are there.

The LHS will be negative before the smallest root and between the two largest (which frame the minimum).

Answer 3

We need to solve $$x^3-1+2x(x-1)<0$$ or $$(x-1)(x^2+x+1+2x)<0$$ or $$(x-1)(x^2+3x+1)<0$$ or $$(x-1)\left(\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\right)<0$$ or $$(x-1)\left(x+\frac{3-\sqrt5}{2}\right)\left(x+\frac{3+\sqrt5}{2}\right)<0,$$ which by the intervals method gives the answer: $$\left(-\infty,\frac{-3-\sqrt5}{2}\right)\cup\left(\frac{-3+\sqrt5}{2},1\right).$$ Done!