# Find zero

#### MathSlot

##### New member
Hello, I am currently trying to find the zeros of this equation: (3x^3-6x^2+x-2)/(6x^2-12x-18) = 0
Can sb. help?

#### Subhotosh Khan

##### Super Moderator
Staff member
Hello, I am currently trying to find the zeros of this equation: (3x^3-6x^2+x-2)/(6x^2-12x-18) = 0
Can sb. help?
Can you factorize the numerator?

#### HallsofIvy

##### Elite Member
First, a fraction is zero if and only if its numerator is 0. So you are really asking for zeroes of the cubic polynomial $$\displaystyle 3x^3- 6x^2+ x- 2$$. There is a general formula for solutions to a cubic equation but it is rather complicated so first lets hope for a simpler solution. By the "rational roots theorem" the only possible rational number solutions are $$\displaystyle \pm 2$$, $$\displaystyle \pm\frac{1}{3}$$, and $$\displaystyle \pm\frac{2}{3}$$. Right off, we see that $$\displaystyle 3(2^3)- 6(2^2)+ 2- 2= 24- 24+ 2- 2= 0$$. So x= 2 is a solution! That means (x- 2) is a factor: $$\displaystyle 3x^3- 6x^2+ x- 2= (x- 2)(3x^2+ 1)$$. That quadratic term has no real zeroes but two complex zeroes.

#### Dr.Peterson

##### Elite Member
The numerator can be factored "by grouping". Have you tried that yet?

#### lookagain

##### Senior Member
First, a fraction is zero if and only if its numerator is 0. So you are really asking for zeroes of the cubic polynomial $$\displaystyle 3x^3- 6x^2+ x- 2$$. . . . . . By the "rational roots theorem" the only possible rational number solutions are $$\displaystyle \pm 2$$, $$\displaystyle \pm\frac{1}{3}$$, and $$\displaystyle \pm\frac{2}{3}$$. Right off, we see that $$\displaystyle 3(2^3)- 6(2^2)+ 2- 2= 24- 24+ 2- 2= 0$$. So x= 2 is a solution! That means (x- 2) is a factor: $$\displaystyle 3x^3- 6x^2+ x- 2= (x- 2)(3x^2+ 1)$$. That quadratic term has no real zeroes but two complex zeroes.
+/- 1 is also a possible rational number solution. You need to check the factors of
that denominator, too, because if there is a match of a factor between the denominator and the numerator, then there will be one fewer zero. It turns out that
(x - 2) is not also a factor in the denominator. Had it been, then the expression (the rational fraction)
would have been undefined at x = 2.

#### Jomo

##### Elite Member
First, a fraction is zero if and only if its numerator is 0.
I am sorry but I do not accept that statement. I like to say that a fraction is 0 iff the numerator is 0 AND the denominator is not 0

#### topsquark

##### Full Member
It turns out that
(x - 2) is not also a factor in the denominator. Had it been, then the expression (the rational fraction)
would have been undefined at x = 2.
Ummmm.... $$\displaystyle 6x^2 - 12x - 18 = 6(x - 1)(x - 2)$$
so x - 2 is a factor in the denominator!

-Dan

#### Dr.Peterson

##### Elite Member
As I tell students, always check your factoring by multiplying it out, because factoring is an error-prone process ...

#### Jomo

##### Elite Member
Dan,
As Dr Peterson hinted, maybe you should check your factoring (I don't like hinting too much!)

#### topsquark

##### Full Member

Comment retracted! (Too much root beer last night.)

-Dan

#### lookagain

##### Senior Member

Comment retracted! (Too much root beer last night.)

-Dan
Would you consider going beyond retracting your comment to putting a
thumbs-up on post # 5?

#### topsquark

##### Full Member
Would you consider going beyond retracting your comment to putting a
thumbs-up on post # 5?
No problem.

-Dan