Steve C. answered • 06/03/15
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Steve C. Math & Chemistry Tutoring
Assuming that the functions are:
(x2 - 4x) / (x3 - 4x2 + 4x) and
(x2 + 6x + 8) / (x2 - 16),
The first function can be factored as:
((x)(x - 4)) / ((x)(x2 + 4x + 4)), and this reduces to. (x - 4) / (x + 2)2.
The zero of this reduced function is x=4.
The second function can be factored as:
((x + 4)(x + 2)) / ((x + 4)(x - 4)). This reduces to:
(x + 2) / (x - 4).
The zero of this reduced function is x = -2.
Neither zero is extraneous.
Michael J.
One more thing to add.
This means that in the first f(x), x=0 is a solution because its in the numerator, but it also makes f(x) undefined because it is in the denominator. x=0 is rather a discontinuity, or in graphical terms, a hole.
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06/03/15
Steve C.
I agree that in certain cases, it's wrong to simplify an equation too much, such as dividing this equation by x:
x2 - 2x = x, then obtaining only x = 2 as a solution. But for the given problems, removing common factors between numerator and denominator does not eliminate any valid zeros. The submitter didn't ask for "holes" in the graph.
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06/03/15
Andrew M.
Steve
The first function actually reduces to
x(x - 4)/x(x - 2)2 You accidentally switched the signs in your initial factoring.
Michael is correct that the undefined value is at x=2 and at x = 0 even though x = 0
looks like it should be a zero of the function.
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06/04/15
Andrew M.
Steve,
Sorry. The sign you dropped was when you initially transcribed the function prior to doing the factoring. You corrected it in your work.
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06/04/15
Michael J.
06/03/15