# Thread: Solve for all possible values of x

1. ## Solve for all possible values of x

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

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

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

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

5. Originally Posted by oldschool999
No! You have to finish the solution.
I, unlike some here, do not give out ready to turn in answers.
Sorry, but I think you should do some work for yourself.

6. Originally Posted by Plato
No! You have to finish the solution.
I, unlike some here, do not give out ready to turn in answers.
Sorry, but I think you should do some work for yourself.