If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Simplifying rational expressions: higher degree terms

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.
• 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!
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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!
• 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.