- Thread starter MathSlot
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Can you factorize the numerator?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?

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

+/- 1 is also a possible rational number solution. You need to check the factors ofFirst, 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.

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.

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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 0First, a fraction is zero if and only if its numerator is 0.

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Ummmm.... \(\displaystyle 6x^2 - 12x - 18 = 6(x - 1)(x - 2)\)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.

so x - 2

-Dan

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Would you consider going beyond retracting your comment to putting a

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

-Dan

thumbs-up on post # 5?

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No problem.Would you consider going beyond retracting your comment to putting a

thumbs-up on post # 5?

-Dan