Simplifying rational expressions: higher degree terms | High School Math | Khan Academy

Let's see if we can simplify this expression, so pause the video and have a try at it, and then we're going to do it together right now.

All right, so when you look at this, it looks like both the numerator and the denominator, they might—you might be able to factor them, and maybe they have some common factors that you can divide the numerator and the denominator by to simplify it.

So let's first try to factor the numerator: (x^4 + 8x^2 + 7). At first, it might be a little intimidating because you have an (x^4) here. It's not a—it’s not a quadratic; it’s a fourth degree polynomial. But like any if you like a lot of quadratics that we've seen in the past, it does seem to have a pattern.

For example, if this said (x^2 + 8x + 7), you'd say, "Oh, well this is pretty straightforward to factor." What two numbers add up to eight, and when I take the product, I get seven? Well, there are only two numbers that, where you take their product and you get positive seven, that are going to be positive—and they need to be positive. They're going to add up to positive 8, and that's 1 and 7.

So this would be ((x + 7)(x + 1)). Well, if you just think of— instead of thinking in terms of (x) and (x^2)—if you just think in terms of (x^2), and (x^4) is going to be the exact same thing. So this thing can be written as ((x^2 + 7)(x^2 + 1)).

If you want, you could do some type of a substitution saying that (a) is equal to (x^2), in which case—so if you said that (a) is equal to (x^2), then this thing would become (a^2 + 8a + 7), and then you would factor this into ((a + 7)(a + 1)), and then you would undo the substitution, and that’s (x^2 + 7) and (x^2 + 1).

But hopefully, you see what's going on here. This is the higher order term, and then this is half the degree of that, so it fits this mold. You could do a substitution or you could just recognize, "Well, okay, instead of dealing with (x^2), I’m dealing with (x^4)."

All right, so that's the numerator. Now, let's think about the denominator. The denominator—both of these terms are divisible by (3x), so let's factor out a (3x).

So it’s (3x) times ((x^4 + 1)). If you factor out a (3x) here, (3) divided by (3) is (1), (x^5) divided by (x) is (x^4), and if you factor out a (3x) here, you're just going to get (1).

And so far, this doesn’t seem too helpful. I don’t see an (x^4 - 1) or (3x) in the numerator, but maybe I can factor this out further: (x^4 - 1). That’s because it is a difference of squares.

You might say, "Wait, I'm always used to recognizing a difference of squares as something like (a^2 - 1), which you could write as ((a + 1)(a - 1))." Well, this would be (a^2 - 1) if you say that (a) is equal to (x^2).

So let's rewrite all of this. This is all going to be equal to the same numerator, so let’s do it in green. Same numerator: (x^2 + 7)—can't factor that out anymore—times (x^2 + 1)—can't factor that out anymore—all that over (3x), but this I can view as a difference of squares.

So this is (x^2) squared, and this is obviously (1) squared. So this is going to be ((x^2 + 1)(x^2 - 1)). Now, clear to have an (x^2) in the numerator, (x^2 - 1) in the numerator (x^2)—sorry, (x^2 + 1) in the numerator and (x^2 + 1) in the denominator, and so I could cancel them out.

I’m going to be left with in the numerator: (\frac{x^2 + 7}{3x \times (x^2 - 1)}). Now this looks pretty simple, and we want to be a little careful because whenever we do this canceling out, we want to make sure that we restrict the (x)'s for which the expression is defined if we want them to be algebraically equivalent.

So this one would—this would obviously be undefined if, so (x) cannot be equal to zero. (x) cannot be equal to plus or minus one—positive or negative one would make this expression right over here equal zero.

So it cannot be equal to zero; (x) cannot be equal to—I’ll write plus or minus one—that would make this part zero. But this right over here, this one—unless we're assuming we’re dealing only with real numbers—this one can’t ever equal zero if you're dealing with real numbers, because (x^2) is always going to be non-negative, and you're adding it to a positive value.

And so this part, this factor, would have never made the entire thing undefined. So we can actually just factor it out or cancel it out without worrying much about it, and so this is actually algebraically equivalent to what we had originally.

Now, we could write these constraints on it if we want. If someone were to ask me, “You know, for what (x) is this expression not defined?” Well, it’s clear it’s not defined for (x) that would make the denominator equal zero.

Dividing by zero is not defined, or if (x) is plus or minus one, it would make the denominator equal (0). But that comes straight out of this expression, so this expression and our original expression are algebraically equivalent.

Now, if you wanted to, you could expand the bottom out a little bit; you could multiply it out if you like. That’s equivalent to (\frac{x^2 + 7}{3x \times (x^2)}) which is (3x^3 - 3x). So these are all equivalent expressions, and we are done.