Find zero

MathSlot

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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

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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

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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

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

lookagain

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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

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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

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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

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

Jomo

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

topsquark

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🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬

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

-Dan
 

lookagain

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🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬

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

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