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## Integrated math 3

### Course: Integrated math 3 > Unit 13

Lesson 1: Cancelling common factors- Reducing rational expressions to lowest terms
- Intro to rational expressions
- Reducing rational expressions to lowest terms
- Simplifying rational expressions: common monomial factors
- Reduce rational expressions to lowest terms: Error analysis
- Simplifying rational expressions: common binomial factors
- Simplifying rational expressions: opposite common binomial factors
- Simplifying rational expressions (advanced)
- Reduce rational expressions to lowest terms
- Simplifying rational expressions: grouping
- Simplifying rational expressions: higher degree terms
- Simplifying rational expressions: two variables
- Simplify rational expressions (advanced)

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# Simplifying rational expressions: higher degree terms

CCSS.Math:

Sal simplifies & states the domain of (x⁴+8x²+7)/(3x⁵-3x).

## Want to join the conversation?

- As the exercise is about 'simplifying' the rational expression, would it not have been better to put in the denominator 3x(x+1)(x-1)? It is the equivalent of Sal's two examples, but more 'factored out' if you will.(15 votes)
- You are right. But it might be more simplified in the last form. I did notice Sal missed a step. Sal might have wanted to avoid the long process of multiplying 3x(x+1)(x-1) which equals to 3x(x^2-1). See, it would lead us to the same step. In conclusion, we simplify the denominator to lead us no factors left behind. Hopefully it helps!(5 votes)

- In the case of test questions in general, do you expect answers in an expanded form or in a factored form after cancellations? That has confused me and caused some of my answers, which were technically correct, to be counted as incorrect. I do note that on some questions you do specify "expanded" form for answers.(3 votes)
- In my experience, Khan Academy Practice Problems software will accept answers whether they're in expanded form or not; for example, they list both '(x - 2)/4(x-3)' and '(x - 2)/(4x-12)' as answers to the question. So maybe you're not simplifying down to those two basic equations? One thing that kept marking me as wrong (and that I incorrectly assumed was due to the fact of my simplification in the wrong form) was that I did not choose a second option for what x cannot equal; I thought it was not needed to say that x cannot -3 in the equation (x+6)/(x+3), since in previous videos Mr. Khan said it was unnecessary since it was obvious if you just look at the equation; but the Practice Problems want you to select ALL the correct options that x cannot be equal to. Hope this solves some of your problems, Walter.(4 votes)

- Can I simplify x⁴+8x²+7 to x⁴+5x+3x+7?

Then I could regroup x⁴+5x+3x+7 to 3x⁵+5x+7?

I'm doing this in order to simplify (x⁴+8x²+7)/(3x⁵-3x). By regrouping the numerator I'm trying to eliminate the 3x⁵ from both expressions.(2 votes)- You have multiple errors.

1) 5x+3x is not the same as 8x². Addition does not change exponents.

2) Similarly, x⁴+3x does not equal 3x⁵. Addition is not the same as multiplication which is what you would have done to get to 3x⁵

3) Making the first term into 3x⁵ will not allow to you eliminate the 3x⁵ in the denominator. When we reduce fractions, we can only cancel out common**factors**(items being multiplied). You are trying to cancel out**terms**(items being added/subtracted).

To do your problem, you need to completely factor both the numerator and denominator. And then only cancel out common factors.

Hope this helps.(4 votes)

- I need help factoring x³-1. The answer is (x-1)(x²+x-1). I see that the answer checks out when I multiply it together, but I don't understand how to arrive at the answer from the problem. Can anyone explain or point me to a video that will explain? Thought this one might, but I still don't understand my problem, hence my asking this. Thanks!(1 vote)
- You have a difference of 2 cubes. It is factored using a pattern. Search on KA for "factoring difference of cubes" and you should find the video.

FYI - You have a sign error on the last 1, it should be +1.

x³-1 = (x-1)(x²+x+1)

Hope this helps.(4 votes)

- I simplified the rational expression to the simplest form which i found to be:

x^2+7/3x(x+1)(x-1)

Upon reaching the step of simplifying when the equation was:

(x^2+7)(x^2+1)/3x(x^2+1)(x^2-1), I realized x^2-1 as a difference of squares so i simplified it further and canceled out the (x^2+1) ... would my final answer be considered equivalent as well?(1 vote)- Yes, your version would be equivalent to Sal's final version. In fact, it is a little better. It is preferred to have the polynomials completely factored. Yours is. Sal's is not.(4 votes)

- Okay, seeing what would happen if quadratics or sum of squares was in the denominator is amazing. But what about inverse of square functions, what about square roots in the denominator for all real numbers and also complex ones, how would the domain look then? Am guessing x is defined only for numbers greater than 0 since we cannot allow 0 to be in the denominator or anything lower than 0 since we want to exclude complex numbers?(1 vote)
- In the exercises later they ask: "Simplify the following rational expression and express in expanded form."

What does mean "express in expanded form"?(1 vote)- Don’t leave something in factored form, like 2(x+3). You want to multiply everything out like this: 2x+6. Hope this helps!(2 votes)

- At about 5 minutes into the video, Sal cancelled out x^2+1.

Can someone please explain why this can never be equal to zero?(1 vote)- If x²+1=0, then x²=-1. So multiplying a number by itself yields a negative number, which is impossible in the real numbers. So x²+1 cannot be 0.(2 votes)

- Sorry but this video had a problem in the restrictions. Sal stated that x could not equal 0, 1, -1. However, x could also not equal i (imaginary number), as the (x^2+1) in the denominator will give the x-value to be sqrt{-1} which is essentially i.(1 vote)
- At the high school level, the domain of a function is assumed to be a subset of ℝ unless otherwise stated, especially if the function is not a polynomial (as is the case here). Nonreal numbers like i needn't be considered.(2 votes)

- Are there any videos about simplifying rational expressions like (40x-56)/(8) / (50x-70)/(10x^2) or expressions like (4/5x) / (x/25)?(1 vote)

## Video transcript

- [Voiceover] Let's see if we
can simplify this expression. So pause the video and have a try at it, and then we're gonna do
it together right now. All right, so when you look at this, it looks like both the
numerator and 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 to the fourth, plus eight
x squared, plus seven. At first it might be
a little intimidating, because you have an x to the fourth here. 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 squared plus eight x plus seven, you'd say, oh, well, this
is pretty straightforward to factor. What two numbers add up to eight and when I take their product I get seven? Well, there's only two numbers
where you take their product and you get positive seven
that are going to be positive, and they need to be
positive, if they're going to add up to positive eight,
and that's one and seven. So this would be x plus seven times x plus one. Well, if you just think of, instead of thinking in
terms of x and x squared, if you just think in terms of x squared and x to the fourth, it's going to be the exact same thing. So this thing can be written as x squared plus seven times x squared plus one. If you want, you can do
some type of a substitution saying, saying that a
is equaled to x squared, in which case, so if you said
that a is equal to x squared, then this thing would become a squared plus eight a plus seven, and then you would factor this into a plus seven and a plus one, and then you would undo the substitution, and that's x squared plus seven and x squared plus one. 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. And so you could do a substitution, or you could just recognize, oh okay, instead of dealing with x squared, I'm dealing with x to the fourth. All right, so that's the numerator. Now let's think about, let's think about the denominator. So the denominator, both of these terms are divisible by three x. So let's factor out a three x. So it's three x times, three x times, if you factor out a three x here, three divided by three is one, x to the fifth divided
by x is x to the fourth, and then if you factor out a three x here, you're just gonna get one. And so far this doesn't seem too helpful. I don't see an x to the fourth minus one, or a three x in the numerator, but maybe I can factor this out further, x to the fourth minus one. And that's because it is
a difference of squares. And you might say, wait, I'm always used to recognizing a difference of squares as something like a squared minus one, which you could write as a plus one times a minus one. Well, this would be a squared minus one if you say that a is equal to x squared. Then this would be a squared minus one. So let's rewrite all of this. So let's rewrite. So this is all going to be equal to, same numerator, let's see, let's do it in green. Same numerator: x squared plus seven, can't
factor that out any more, times x squared plus one,
can't factor that out any more, all of that over three x, but this I can view as
a difference of squares. So this is x squared squared, and this is obviously one squared, so this is going to be x squared plus one times x squared minus, times x squared minus one. Now clearly have an x
squared in the numerator, x squared minus one in the numerator, x squared, sorry, x squared
plus one in the numerator, x squared plus one in the denominator, and so I could cancel them out, and I'm going to be left
with, in the numerator, x squared plus seven, over three x times x squared minus one. Now, this looks pretty simple, and we want to be a little careful, because whenever we do this cancelling out we don't want, we want to make sure that we restrict the xes for which the expression's defined, if we want them to be
algebraically equivalent. So this one, would this be, this would obviously be undefined, 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 squared 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 zero, dividing by zero not defined, or if x is plus or minus one, it would make the denominator equal zero. But that is, that comes
straight 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, so x squared plus seven over three x times x squared
is three x to the third minus three x. So these are all, these are
all equivalent expressions, and we are done.