Originally Posted by

**Barthayn** Today in math I learned new concepts. However, I am unsure if these three questions are correct. This is what I did:

Solve by considering all cases. Show each solution on a number line.

$\displaystyle x^2-2x-15<0$

Steps I did:

$\displaystyle x^2-2x-15<0$

$\displaystyle = (x-5)(x+3)<0$

$\displaystyle = (x-5<0) (x+3<0)$

$\displaystyle = (x<5) (x<-3)$

Therefore: -3<x<5 is a solution.

$\displaystyle x^3-3x^2-5x+6<0$

$\displaystyle = (x-1)(x^2-x-6)<0$

$\displaystyle = (x-1)(x-3)(x+2)<0$

$\displaystyle = (x-1<0) (x-3<0) (x+2<0)$

$\displaystyle = (x<1) (x<3) (x<-2)$

Therefore: x<-2 or 1<x<3 is a solution.

Solve using intervals.

$\displaystyle 5x^3-12x^2-11x+6 \leq 0$

$\displaystyle \Rightarrow (x+1)(5x^2-17x+6) \leq 0$

Used quadratic equation to find the roots of $\displaystyle 5x^2-17x+6$

$\displaystyle = (x+1)(x-3)(x-0.4)$

Therefore: x≤-1 or 0.4 ≤ x ≤ 3 is a solution.

Am I correct with the findings? Is what I did to solve for all cases correct?