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Video: Addition and Subtraction of More Complex Rational Expressions

Tim Burnham

Catch on adding and subtracting rational expressions with polynomial expressions up to order 3 in numerators and/or denominators.

18:01

Video Transcript

We did an introduction to addition and subtraction of rational expressions in a previous video. But in this one, we’re gonna be looking at some slightly more difficult examples of adding and subtracting rational expressions, and learn a few techniques to look out for in simplifying the results.

So the first one here, we’re gonna do a subtraction, three 𝑥 over 𝑥 minus one minus 𝑥 over 𝑥 plus three.

Well there are no common factors in either of these denominators. So we’re gonna have to find equivalent fractions for each of these, in order to get a common denominator. And the way that we’re gonna do that is, we’re gonna take this first denominator here and we’re gonna multiply top and bottom of the second fraction by that denominator. And then we’re gonna take the second denominator and do the same for the first fraction. So now both of the denominators are basically 𝑥 plus three times 𝑥 minus one. Now what you’ll remember is that 𝑥 plus three over 𝑥 plus three is just one. So we’ve taken that first fraction and we’ve multiplied it by one. We’ve taken the second fraction and multiplied it by 𝑥 minus one over 𝑥 minus one, which again of course is just one. So multiplying something by one doesn’t change it. So we’ve still got the same question here, but we’ve now got it in a format where both of those fractions have got the same denominators.

So again, you might remember from before that it’s a really good idea to put those denominators and all these little terms in brackets, when they’re supposed to be together. And then we combine the two terms into one term with that common denominator. So that was three 𝑥 lots of 𝑥 plus three for our first term, and 𝑥 lots of 𝑥 minus one for our second term. And now we’ve gotta multiply those out. So three 𝑥 lots of 𝑥, and three 𝑥 lots of positive three. But it’s negative 𝑥 lots of 𝑥, and negative 𝑥 lots of minus one. So three 𝑥 lots of 𝑥 is three 𝑥 squared, and three 𝑥 lots of positive three is positive nine 𝑥. Negative 𝑥 times 𝑥 is negative 𝑥 squared, and negative 𝑥 times negative one is positive 𝑥. And now we’ve gotta simplify that numerator. So we’ve got three 𝑥 squared and we’re taking away 𝑥 squared, so that gives us two 𝑥 squared. And then we’ve got nine 𝑥 and we’re adding another 𝑥, so that’s gonna give us plus ten 𝑥.

Well now we can factor the numerator. So we’ve got a common factor of two in each of those coefficients, and we’ve also got a common factor of 𝑥 which we can factor out. So that gives us two 𝑥 lots of 𝑥 plus five all over 𝑥 plus three lots of 𝑥 minus one. Now if we’d have been a bit luckier there, that factor there, 𝑥 plus five, would have been the same as one of the brackets on the denominator. We’d have been able to cancel something else out and make it simpler. But as it is, that’s gonna be our answer. Nothing else is gonna cancel down. There we go, two 𝑥 lots of 𝑥 plus five over 𝑥 plus three times 𝑥 minus one.

So the process, just to summarise this, that we’re gonna be using is first of all, we’re looking for a common denominator. Now we’re looking to try and simplify that as much as possible, as soon as possible. But essentially, we’ve got to look for this technique here, to try and get a common denominator. Secondly, we’re looking to combine all of this into a single rational expression, or a single fraction. And then we’re trying to evaluate that numerator, perhaps do some factoring, simplify it down and see if we’re lucky enough if any of the things cancel down in our finar answ- final answer, to simplify that as much as possible.

Right then. In the next question, we’ve got to simplify three over 𝑥 plus two plus two 𝑥 over 𝑥 squared minus four.

Now before we do anything else with that, let’s have a look at this denominator here, 𝑥 squared minus four. We should recognise, that’s the difference of two squares. So we can factor that into 𝑥 minus two lots of 𝑥 plus two. So let’s write out our question with that expressed that way first, and see if that gives us any clues as to how we can move forward. So when we do that, we can see that, look, the denominator here is 𝑥 plus two, but this denominator has also got a factor of 𝑥 plus two. So if we want a common denominator, we can just take this bit here, this factor, 𝑥 minus two, and multiply top and bottom of that first fraction to get a- an equivalent fraction with that same denominator. So remember, 𝑥 minus two over 𝑥 minus two is just one. So we’ve just multiplied that first fraction by one. So we haven’t changed the size of it, the magnitude of it, we’ve just got an equivalent fraction. But now they’ve both got a denominator of 𝑥 minus two times 𝑥 plus two. So let’s combine those into one fraction.

And the first term was three lots of 𝑥 minus two. I’ve just turned those around and called it three lots of 𝑥 minus two. It doesn’t matter which way you multiply those. And then we’ve got two 𝑥, for our second term, on the numerator. So we’re just gonna multiply out that numerator and see if we can simplify it at all. And three lots of 𝑥 is three 𝑥, and three lots of negative two is negative six. So we’ve got three 𝑥 minus six plus two 𝑥 on the top, and we can now say three 𝑥 plus two 𝑥 is five 𝑥. And that leaves us with five 𝑥 minus six over 𝑥 minus two 𝑥 plus two. Looking at that numerator, we haven’t got any common factors. We can’t simplify that down. Nothing cancels top and bottom. So in fact, that is gonna be our final answer.

So the next example we’re gonna look at is simplifying one over 𝑥 squared plus five 𝑥 plus six plus one over 𝑥 squared plus seven 𝑥 plus twelve. Now it’s not gonna be any good just to, sort of, multiply the top and bottom of the second fraction by this denominator here, because we’re gonna end up with incredibly complicated expressions to deal with. But what we can do, first of all, is factorise these two denominators and then pick out if they’ve got a common part, we will multiply by that. If they’ve got other bits, then we’ll multiply by those. So let’s js- let’s just factorise those two denominators first.

Well they’re both quadratics, so we know we’re gonna have 𝑥 here and 𝑥 here because they’ve already got one 𝑥 squared and one 𝑥 squared. So they’re gonna have 𝑥 here and 𝑥 here. And we’ve gotta find numbers in the first case that add to make positive five, and multiply to make positive six. So that’s gonna be plus two and plus three. And in the second case, we’re looking for numbers which add to make seven and multiply to make twelve. So we’re looking at plus four and plus three.

Now before we carry on, let’s just do a quick check. 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times three is three 𝑥, two times 𝑥 is two 𝑥, so that’s two 𝑥 and three 𝑥 is five 𝑥. And then two times three is six. So that one works. And 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times three is three 𝑥, four times 𝑥 is four 𝑥, three 𝑥 and four 𝑥 is seven 𝑥. And four times three is twelve. So that one works as well.

Okay. So both of those denominators have got 𝑥 plus three as a factor. So I’m gonna have to take this missing factor and multiply top and bottom of the second fraction by that. And I’m gonna have to take this missing factor and multiply top and bottom of the first fraction by that. And when we do that, we’ve got a common denominator with three brackets multiplied together, three sets of parentheses multiplied together there. So I’m gonna combine those into a single fraction. I’ve got one lot of 𝑥 plus four plus one lot of 𝑥 plus two. Well of course, one lot of those things is just one lot. So that’s 𝑥 plus four plus 𝑥 plus two. And 𝑥 plus 𝑥 is two 𝑥, and positive four add another two is plus six. And looking at that, on the numerator here, we’ve got a common factor two and six. So two is a common factor; we can factor that to give us two lots of 𝑥 plus three all over 𝑥 plus two times 𝑥 plus three times 𝑥 plus four. Now if we look at that very carefully, we’ve got two times 𝑥 plus three. 𝑥 plus three is a factor on the numerator, but 𝑥 plus three is also a factor on the denominator. So if I divide both of them by 𝑥 plus three, they cancel out, which gives us our final answer of two over 𝑥 plus two times 𝑥 plus four.

Well let’s look at some subtraction ones then. So 𝑥 plus three over 𝑥 plus four minus 𝑥 minus one over 𝑥 plus one. Well there are no common factors in the denominators there. So let’s just write this out and try and get a common denominator.

So first, as we said before, make sure you put those numerators and denominators, cause they’re slightly more complicated numerators now, in brackets to keep those terms together, that need to be kept together. So we’re going to use this denominator here to multiply top and bottom of this fraction. And we’re gonna take this denominator and multiply top and bottom of this fraction by that. So now we have 𝑥 plus one times 𝑥 plus four as our common denominator. We can combine those two terms together into one single fraction. And now we’re gonna have to multiply out these parentheses and these parentheses. Now this is where it gets quite tricky. Because we’ve got a minus sign in here, we’re taking away the whole of that second expression here, from the first expression here. So we’ve gotta be quite careful how we lay that out. So I’m actually gonna multiply out the parentheses and leave them in brackets with a minus in front, and then will evaluate it afterwards.

So multiplying out the first bracket then, we’ve got 𝑥 times 𝑥 is 𝑥 squared. And we’ve got 𝑥 times three is positive three 𝑥, one times 𝑥 is positive one 𝑥. So three 𝑥 and one 𝑥 is four 𝑥. And then we’ve got one times three, both positive, so that’s three. And for the second term here, we’ve got 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times four is plus four 𝑥. Negative one times 𝑥. We’re taking away an 𝑥, so four 𝑥 take away one 𝑥 is three 𝑥. And negative one times positive four is negative four. So we’re gonna be taking away 𝑥 squared. We’re gonna be taking away positive three 𝑥. So take away three 𝑥, negative three 𝑥. And we’re taking away negative four, which is like adding four. So let’s rewrite our numerator.

Well the first few terms here are gonna remain as they are. And then as we said, we’re taking away 𝑥 squared. We’re taking away positive three 𝑥, or taking away three 𝑥. And we’re taking away negative four, which is the same as adding four. So now we can simplify that. We’ve got 𝑥 squared take away 𝑥 squared. Well that’s nothing, so they’re gonna cancel out. We’ve got positive four 𝑥 and we’re taking away three 𝑥, so that’s just one 𝑥. And we’ve got positive three and we’re adding on four, so that’s seven, which leaves us with 𝑥 plus seven all over 𝑥 plus one times 𝑥 plus four. Well we could put brackets around the top if we wanted to, or we could even multiply out the denominator. But nothing else is gonna cancel. That is our final answer.

And the most difficult thing there was the fact that we had quite a lot of working out to do. And taking care of the negative signs here was really the crucial and-and most difficult step. That’s where most people go wrong, if they’re gonna go wrong with that kind of question.

So in our next subtraction of rational expressions, we’ve got three over 𝑥 minus two take away two 𝑥 minus two over two 𝑥 squared minus 𝑥 minus six. And as we’ve seen in some of these questions before, the first thing to try and do is to factor anything that we can. And it looks like the numerator of the second term will factor. And it looks like the denominator will also factor here. So let’s try and do that.

So we’ve put the denominator of the first term in a bracket, because that’s always a good idea. And the numerator on the second term is quite easy to factor. So two is the common factor, so we’ve got two lots of 𝑥 minus one. Now the denominator, slightly trickier, so two 𝑥 squared, so we- we’re gonna have two brackets; it’s a quadratic. One of them is gonna have to be two 𝑥, and one of them is gonna have to be 𝑥, because two 𝑥 times 𝑥 is two 𝑥 squared. And now we’ve gotta sort of play around a bit, to try and work out what the other terms are gonna be.

So we must have two 𝑥 and 𝑥, because two 𝑥 times 𝑥 is 𝑥- two 𝑥 squared. But one of the other terms is gonna be, well they have to multiply together to give negative six and they’re gonna combine together to give this whole expression. And with a bit of trial and error, we see that they’re two 𝑥 plus three and 𝑥 minus two. Well that’s handy because look, we’ve got 𝑥 minus two and we’ve got 𝑥 minus two. So we’ve got a common factor here. So the only thing that would be missing from this denominator, for the first term, would be this term here, two 𝑥 plus three. So I’m gonna multiply top and bottom by two 𝑥 plus three. And that’s an equivalent fraction to the first term. And we’ve now got two terms both with a common denominator of two 𝑥 plus three times 𝑥 minus two. So we can combine those into a single fraction, which is three lots of two 𝑥 plus three for the first term on the numerator, and we’re taking away two lots of 𝑥 minus one for the second term. So we’re gonna have to multiply out these brackets on the numerator. So negative two lots of 𝑥 and negative two lots of negative one. Well three lots of two 𝑥 is six 𝑥, three lots of positive three is positive nine, negative two lots of 𝑥 is negative two 𝑥, and negative two lots of negative one is plus two. So we can now simplify this. We’ve got six 𝑥 take away two 𝑥 is four 𝑥, and nine add two is eleven, positive eleven.

So again, looking at the numerator, there’s no common factors. We can’t factor that down anymore. Nothing cancels on the numerator and the denominator. So there we have it; that would be our answer to this question. So again, once more, we just have to watch what we’re doing in terms of negative signs, when we’re subtracting these. And in this case, we had to do quite a lot of work at the beginning to factor our numerator and denominator, which just made the whole multiplication thing, finding the equivalent fraction with the common denominator, a little bit easier.

Right. Let’s round it off with this quite tricky question here. So in this case, we’ve got to subtract two rational expressions. We’re gonna subtract one from the other. But we’ve gotta show that it equals a particular expression. Now sometimes, when you do your calculation, you might end up with an answer in one format. But in this case, they’re asking for you to present in a very specific format. So let’s go ahead and see what we’re gonna do with this. First of all, it looks like we’re gonna have to do a bit of factorisation on the denominator here. But if we look at that first term and we put a bracket around the numerator, we can see that par- that the numerator, if I divide that by 𝑥 plus five, I get one. If I divide the denominator by 𝑥 plus five, that also becomes one here. So we’ve got one over two minus 𝑥 times three plus 𝑥. So we’ve got a slightly simpler first term. So let’s have a look at what we can do next.

Well in our working out, we can show that cancelling going on and we can show that the first term is equal to one over two minus 𝑥 times three plus 𝑥. And the second term, we can rewrite by factoring the denominator. And to get negative 𝑥 squared, I’ve gotta have a negative 𝑥 times a positive 𝑥. And we’ve just gotta work out what the numbers are, that multiply together to make six and add together to make negative one, which is the coefficient of that 𝑥 term here. So that’s gonna be two minus 𝑥 and three plus 𝑥. So we’ve basically rearranged or reexpressed those first two terms. And it turns out that if we look at the denominators for each of those, they’ve got common denominators. So we’re gonna be able to add them together. So let’s just reexpress that left-hand side. And remember, the first term can be reexpressed as one over two minus 𝑥 times three plus 𝑥. And we’re taking away that second term, which we’ve just reexpressed as 𝑥 over two minus 𝑥 into three plus 𝑥. So we’ve got our common denominator here. So we can combine these two things together into one single fraction.

And that looks incredibly close to what we’ve actually got to show, but it’s not quite the same. Instead of one- instead of 𝑥 minus one, we’ve got one minus 𝑥. And instead of 𝑥 minus two, we’ve got two minus 𝑥. But we’ve got 𝑥 plus three, three plus 𝑥, they’re the same thing. Three plus 𝑥 is the same as 𝑥 plus three. So we’ve got to be a bit creative here about what we’ll do. Well what we could do is multiply the top and the bottom by negative one. And that gives us these expressions, which when I reorder them, give us the answer we were looking for.