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I'm solving this rational expression question and I'm stuck. What should I do next?

My work is below.

Thank you!

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Jyrki Lahtonen
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TheFloatingMan
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    What happens to $x^3$ at step #2 VI)? And in step #2 I), what was the operation you wanted to perform? Else, the rest seems ok. – Berci Jan 07 '15 at 23:32
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    You correctly wrote $(x-1)(x+2)$ but later changed it to $(x-2)(x+2)$. ${}\qquad{}$ – Michael Hardy Jan 07 '15 at 23:37
  • @turkeyhundt & MichaelHardy Thank you, that's a typo. As you can see I factored it in step ii) and then rewrote it incorrectly in v)... – TheFloatingMan Jan 07 '15 at 23:40
  • Retagging, because this had nothing whatsoever to do with either division-algebras or regular-expressions. Artemisveras, when considering adding a tag, mouseover it. That shows a tag excerpt, which is a brief description. As a rule of thumb: If that description contains words you are not familiar with, it is nearly certain that you should not use that tag. Most of the tags involve concepts from more advanced math only understood by advanced undergrads or graduate students. They may contain familiar sounding words, but that is misleading. – Jyrki Lahtonen Jan 08 '15 at 08:40
  • @JyrkiLahtonen I'll be more careful next time, sorry about that. – TheFloatingMan Jan 08 '15 at 15:35

2 Answers2

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I think you just made a copying error on step 1v.

You wrote $(x-2)$ instead of $(x-1)$. If you fix that you will get an additional pair of terms that cancel out and lead to a 2nd degree polynomial on top and bottom of the fraction.

turkeyhundt
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  • Is everything else correct? – TheFloatingMan Jan 07 '15 at 23:41
  • The other methods look good. You might want to put a set of parentheses around your $+x$ to avoid confusion that you are adding in step 2ii. And, you won't have an $x^3$ term once you correct that mistake, but for future reference, you cannot just cancel out a term from the fraction that is added, as you did in step 2vi when you struck through the $x^3$ terms. You can only cancel out terms which are multiplied through the entire expression. – turkeyhundt Jan 07 '15 at 23:44
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If you don't know yet, verify yourself that $(x-a)(x-b)=x^2-(a+b)x+ab$ which means that the roots $a,b$ of $x^2-Ux+V$ will just satisfy $U=a+b$ and $V=ab$.

Knowing this, you can easily find integer roots by simple tries for a quadratic with small coefficients, e.g. for $x^2+x-2$ we look for $a,b$ such that $a+b=-1$ and $ab=-2$, so it is $-2$ and $1$, hence $x^2+x-2=(x+2)(x-1)$.

You can do it, but you need to pay more attention. E.g. the fractions (after taking the reciprocal of the divisor) has to be multiplied, but in your solution, the nominators were added.

And, at step #2 VI), you seem to cancel $x^3$, please forget it, because it is a summand of both the nominator and the denumerator, and not a multiplier.

Berci
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