1. ## Algebra 2 help..

3x^3 -4x^2-17x+6=0

I'm just told to solve it, it's in a chapter about polynomials. I know the answers, 3,1/3, and -2. I just don't know how to get them.

3x^3 -4x^2-17x+6=0

I'm just told to solve it, it's in a chapter about polynomials. I know the answers, 3,1/3, and -2. I just don't know how to get them.
there are ways how to solve it..

try using rational root test and synthetic division..

3. Another hint.

When x=-2, 3(-2)^3-4(-2)^2-17(-2)+6=0

therefore, 3x^3-4x^2-17x+6 = (x+2)(ax^2+bx+c)

4. But I can't use -2 if I didn't get the answer from the teacher. Like how do I do it, w/o using one of the answers?

5. By trial and error, you have to determine one of the roots first. then try to solve the problem. Let me give you the answer you try to work it out.

(x+2)(3x^2-10x+3)
(x+2)(3x-1)(x-3)

Hope it helps.

6. Originally Posted by tester85
By trial and error, you have to determine one of the roots first. then try to solve the problem. Let me give you the answer you try to work it out.

(x+2)(3x^2-10x+3)
(x+2)(3x-1)(x-3)

Hope it helps.
trial and error doesn't make sense when you are talking on infinitely many numbers to try on.. i mean, it is a waste of time.. try decreasing the number of numbers you are going to try,,

the rational root test tells you that if you have a polynomial of degree $n$ whose leading coefficient is $a_n$ and whose constant term is $a_0$, then the possible roots of the polynomial is given by $\frac{p}{q}$ where $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$

in your example, $a_0 = 6$ and $a_n = 3$

so $p$ can be equal to $\pm 1, \pm 2, \pm 3, \pm 6$
and $q$ can be equal to $\pm 1, \pm 3$

from here, you may start your trial and error by choosing any combination of $\frac{p}{q}$

after getting one, you may do previous poster's suggestion by taking out and factoring the quadratic factor..