1. ## solving equation

solve the equation $\displaystyle 6x^3-19x^2+11x+6=0$ so far i got (x-2) as a factor thats as far as i can seem to get

2. Hi

You are right !

So you can factor $\displaystyle 6x^3-19x^2+11x+6=0$ by $\displaystyle x-2$ which will give a second degree polynomial

3. Originally Posted by william
solve the equation $\displaystyle 6x^3-19x^2+11x+6=0$ so far i got (x-2) as a factor thats as far as i can seem to get
Hi william,

When you divide $\displaystyle 6x^3-19x^2+11x+6=0$ by $\displaystyle x-2$, you get $\displaystyle 6x^2-7x-3$ which should be easily solved using the quadratic formula. The quadratic can be factored as well.

4. Originally Posted by masters
Hi william,

When you divide $\displaystyle 6x^3-19x^2+11x+6=0$ by $\displaystyle x-2$, you get $\displaystyle 6x^2-7x-3$ which should be easily solved using the quadratic formula. The quadratic can be factored as well.
i got 1.5 and -0.3 is that correct?

5. Originally Posted by william
i got 1.5 and -0.3 is that correct?
Actually, when you solve by factoring $\displaystyle 6x^2-7x-3=0$, you get:

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

So, the roots are:

$\displaystyle x=-\frac{1}{3}$ and $\displaystyle x=\frac{3}{2}$.

So, one of your roots was correct. You don't want to round off the other one to -.3

Leave it as a fraction.