## Rapalje's Intermediate Alg: simplifying polynomial fraction

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rf0
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Joined: Sat Mar 09, 2013 1:32 pm
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### Rapalje's Intermediate Alg: simplifying polynomial fraction

I have been working through Dr. Rapalje's Algebra texts regarding multiplying and dividing fractions:
http://www2.seminolestate.edu/rrapalje/IntermediateAlgebra/Int%20One%20Step%20Ch%202/2204_Mult_Div_Frac.htm
I'm having difficulty reducing problem 38 to the solution. The expression starts as:

$\frac{x^3 + 27}{x^2 + 6x + 9} * \frac{x^2 - 9}{x^2 - 3x + 9}$

and I factor the components:

$x^3 + 27 =$ (unable to reduce)
$x^2 + 6x + 9 = (x + 3)(x + 3)$
$x^2 - 9 = (x + 3)(x - 3)$
$x^2 - 3x + 9 =$ (unable to reduce)

and then reassemble the expression and simplify:

$\frac{x^3 + 27}{(x + 3)(x + 3)} * \frac{(x + 3)(x - 3)}{x^2 - 3x + 9}$

...cancel out one of the (x + 3) from the denominator of the first fraction and
one from the numerator of the second fraction...

$\frac{x^3 + 27}{(x + 3)} * \frac{(x - 3)}{x^2 - 3x + 9}$

...then cancel out the x^3 and the x^2 leaving just x in the numerator of the
first fraction and then divide the 27 by the 9 leaving an expression:

$\frac{x + 3}{(x + 3)} * \frac{(x - 3)}{- 3x}$

The first fraction (x + 3) cancels altogether leaving:

$\frac{(x - 3)}{- 3x}$

but this isn't correct, as it should simply be

$(x - 3)$

Somewhere I'm missing a step necessary to eliminate the denominator value entirely.

rf0
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Joined: Sat Mar 09, 2013 1:32 pm
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### Re: Rapalje's Intermediate Alg: simplifying polynomial fract

Hi,

I just got sorted this out.
The problem is that I omitted a step when factoring the components:

$x^3 + 27 =$ (unable to reduce)
$x^2 + 6x + 9 = (x + 3)(x + 3)$
$x^2 - 9 = (x + 3)(x - 3)$
$x^2 - 3x + 9 =$ (unable to reduce)

The first one is incorrect that it doesn't reduce, it does, it reduces to

$x^3 + 27 = (x + 3)(x2 - 3x + 9)$

(I don't think "reduce" is the proper word there, perhaps factor?)
Subsequently, reassembling produces an expression:

$\frac{(x + 3)(x2 - 3x + 9)}{(x + 3)(x + 3)} * \frac{(x + 3)(x - 3)}{x^2 - 3x + 9}$

...and then canceling:

* cancel (x + 3) in the numerator and denominator of the first fraction leaving:
$\frac{(x2 - 3x + 9)}{(x + 3)} * \frac{(x + 3)(x - 3)}{x^2 - 3x + 9}$

* cancel (x + 3) in the numerator of the second fraction and denominator of the first fraction leaving:
$\frac{(x2 - 3x + 9)}{1} * \frac{(x - 3)}{x^2 - 3x + 9}$

* cancel (x2 - 3x + 9) in the numerator of the first and denominator of the second leaving:
$\frac{(x - 3)}{1}$

...and I'm left with the correct solution of $(x - 3)$.

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Joined: Sun Feb 22, 2009 11:12 pm

### Re: Rapalje's Intermediate Alg: simplifying polynomial fract

rf0 wrote:$\frac{x^3 + 27}{(x + 3)} * \frac{(x - 3)}{x^2 - 3x + 9}$

...then cancel out the x^3 and the x^2 leaving just x in the numerator of the
first fraction...

You can't do this. They explain why here. Read down to the part about "ripping off arms and legs".

rf0 wrote:...and then divide the 27 by the 9...

You can't do this either, and for the same reason. Make sure you understand why not, cos teachers like putting stuff on tests that'll tempt you to do this exact thing to mess you up.

rf0
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Joined: Sat Mar 09, 2013 1:32 pm
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### Re: Rapalje's Intermediate Alg: simplifying polynomial fract

Thanks for taking the time to respond and point out my error. I know it doesn't count but had actually read that section before but had forgotten it.
Without the factorization of $x^3 + 27$ I was struggling to find the next step. Evidently the information in that section at has not been reinforced enough to stick in my mind. I'll go back and study those concepts more. I really just want learn this stuff, again. At this point in my life if I have to take a math test, it'll be a weird thing indeed...