# Thread: Solve for all possible values of x

1. ## Solve for all possible values of x

Solve for all possible values of x Question1. $x^3+3x^2-13x-15=0$ Question 2. $x^3-3x^2+4x-12=0$

2. Originally Posted by oldschool999
Solve for all possible values of x Question1. $x^3+3x^2-13x-15=0$ Question 2. $x^3-3x^2+4x-12=0$
Just use the rational root theorem here, I get, x=-5,-1,3 for the first one.

3. For the second just factor.
$x^2 \left( {x - 3} \right) + 4\left( {x - 3} \right) = \left( {x^2 + 4} \right)\left( {x - 3} \right) = 0.$

4. Originally Posted by Plato
For the second just factor.
$x^2 \left( {x - 3} \right) + 4\left( {x - 3} \right) = \left( {x^2 + 4} \right)\left( {x - 3} \right) = 0.$

5. Originally Posted by oldschool999