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Simplifying Rational Functions with Factoring and GCFs  Problem 4
Alissa Fong
Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
This rational function needs to be simplified and I need to find the excluded values. So the first thing I’m going to do is factor the top and factor the bottom and look for any factors they have in common. On top 3 multiplies into both of those terms.
So I’m going to write the top like this. And the bottom I’m going to have to do some rewriting because this is in standard form I’m going to rewrite it with 6x in the middle, 6x in the middle. Now it’s going to be easier for me to factor.
Notice how we have a common factor on the bottom. 3 multiplies into all those terms so I’m going to factor out the 3 and I’m left with this okay now this is going to be easier to factor I’m going to have 3 and then let’s see x and x are going to be my first terms.
Again this factoring is like unfoiling my last terms need to be 3 and 1 they have to add up to negative excuse me multiply to 3 and add up to 2. So those are my two factors. Okay so this is my completely factored form and it looks like the only things that can cancel out are those 3s.
Let me rewrite it and we’ll see what we’ve got to work with. Right now we at three take away x on top and on bottom we have x minus 3 times x plus 1. Many students will think that this is totally simplified but we have tricky little thing called opposite factors. Look at how these are super close I’m going to show you how these are opposite factors
On top here instead of 3 minus x I’m going to factor out the 1. I’m going to come over here so I can write it without messing up too much my problem. 3 minus x if I factor out 1 that would be 3 plus x, rearrange that and I’ll have 1 times x minus 3.
That is the same as that guy in the denominator. So going back to my problem I can rewrite this top, instead of 3 minus x I’m going to write it as 1 times the quantity x minus 3 that’s what I got to do here on the bottom I have x minus 3 times the x plus 1. I’m almost done.
The x minus 3 factor is cancelled and here is my most simplified answer. 1/x plus 1.Who know that big nasty thing with all those 3’s and 9’s and 6’s could be simplified with this, it’s pretty cool.
Okay so now that I have my most simplified form what I need to do is look excluded values. Again excluded values are anything that would make my denominator equal to 0. Well here is my factored form of the denominator and that’s handy because I want to set that equal to 0. I’m looking for factors, oh excuse me, the x values that would make this somehow equal to 0.
If I use the 0 products property I’ll see this doesn’t give me a solution it gives me an excluded value x minus 3 equals 0 tells me my x value could not be 3 because x were 3 by denominator will be bottom my denominator will be 0. And the other excluded value is going to be x cannot be 1 okay so that was kind of tricky I had to do lots of steps. This might be a video you want to rewind because whole opposite factor business.
Let me back up and talk you through it one more time and if you want to you can watch this video again. The first thing I did was factor the top no problem then I had to rewrite the bottom into standard form so I can factor it here. Then one the 3s were eliminated I noticed that these two factors are almost exactly the same the only difference is where the minus sign showed up.
So what I did was looking at the top I factored out a 1 it’s like I divided this guy by 1 and divided that guy by 1 to get this and the reason why this is good Is because x minus 3 shows up in my denominator.
So I could cancel those guys out to get my final simplified form of 1/x plus 1. These problems are tricky you guys, these are like advanced level simplification problems if you can do this you are on your way to an A+. Stick with it, don’t give up and go talk to your tutor or your teacher if you have some questions.
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Alissa Fong
M.A. in Secondary Mathematics, Stanford University
B.S., Stanford University
Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.
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