Question on how this function was factored...

[samalex]

Hi everyone,

I'm going through examples on this site to prep myself for a College Algebra class I'm about to start next week, but thus far most of the examples under http://freemathhelp.com/ make great since and are easily followed. The one exception thus far is this one: http://freemathhelp.com/finding-asymptotes.html

.. specifically the part where \(\displaystyle y=\frac{x^2 - x - 6}{x^2 - 9} \) is factored down to \(\displaystyle y=\frac{x + 2}{x + 3}\).

I see how \(\displaystyle \frac{-6}{- 9} \) is factored down to \(\displaystyle \frac{2}{3}\) but I'm not sure how \(\displaystyle \frac{x^2 - x}{x^2}\) is factored to just \(\displaystyle \frac{x}{x} \). Does the \(\displaystyle x^2\) in both the denominator and numerator cancel each other out? If so wouldn't that just leave -x? That's where I'm not sure what's next, if I'm even remotely correct at this point...

Sorry if this is an elementary question I should know, I'm still rusty but trying to get caught-up on the basics before my College Algebra class begins next week.

Take care,

Sam

 

[Deleted member 4993]

Hi everyone,

I'm going through examples on this site to prep myself for a College Algebra class I'm about to start next week, but thus far most of the examples under http://freemathhelp.com/ make great since and are easily followed. The one exception thus far is this one: http://freemathhelp.com/finding-asymptotes.html

.. specifically the part where \(\displaystyle y=\frac{x^2 - x - 6}{x^2 - 9} \) is factored down to \(\displaystyle y=\frac{x + 2}{x + 3}\).

I see how \(\displaystyle \frac{-6}{- 9} \) is factored down to \(\displaystyle \frac{2}{3}\) but I'm not sure how \(\displaystyle \frac{x^2 - x}{x^2}\) is factored to just \(\displaystyle \frac{x}{x} \). Does the \(\displaystyle x^2\) in both the denominator and numerator cancel each other out? If so wouldn't that just leave -x? That's where I'm not sure what's next, if I'm even remotely correct at this point...

Sorry if this is an elementary question I should know, I'm still rusty but trying to get caught-up on the basics before my College Algebra class begins next week.

Take care,

Sam

Those are absolutlely WRONG concept.

[x²-x-6]/[x²-9] = [(x-3)(x-2)]/[(x-3)(x+3)] = (x-2)/(x+3)

If you have trouble understanding the steps above, I suggest a face-to-face tutor.

 

[samalex]

I did some reading on factoring binomials and trinomials, and yeah I was WAY off in my first post... but I see how it's done now. This stuff is slowly but surely coming back...

Thanks

Sam