Finding the values of x in an inequality.

Question

Question: Find the values of x such that $6x^3-x^2-10x-3>0$ Source: James S. Rickards Fall Invitational (Algebra II Individual)

I tried substituting $x^2$ for $y$ and solving, but I ended up with $\sqrt{y}$ and it seemed like that approach wouldn't work. So, I tried factoring the expression and got $(x^2-5/3)(6x-1)-4/3>0$, but this way also seems like a dead end. Is there an efficient way to solve this inequality?

Thanks

Answer 1

-1 is a root.

Divide by x+1 or use the rational root theorem to find the other roots.

Then see where the terms are positive or negative.

Answer 2

$$6x^3-x^2-10x-3=6x^3+6x^2-7x^2-7x-3x-3=$$ $$=(x+1)(6x^2-7x-3)=(x+1)(2x-3)(3x+1).$$ Can you end it now?

I got: $$\left(-1,-\frac{1}{3}\right)\cup\left(\frac{3}{2},+\infty\right).$$

Answer 3

Use that $$6x^3-x^2-10x-3=(2x-3)(3x+1)(x+1)$$