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### Course: Algebra 2>Unit 4

Lesson 1: Dividing polynomials by x

# Dividing polynomials by x (no remainders)

Discover how to divide polynomials by 'x' in two different ways. Finding the quotient (x⁴-2x³+5x)/x is the same as asking "what should we multiply by x to get x⁴-2x³+5x?" First, we distribute '1/x' to each term in the polynomial. Second, we factor out an 'x' from each term. Both methods simplify complex expressions,.

## Want to join the conversation?

• So basically you just factor out an x?
• Yeah, that is essentially what he did. I think that Sal did it this way to base the intuition for dividing a polynomial by x.
• When I search up "what is polynomial long division useful for?" I see the definition of it and not an actual real-world use. Is it safe to say this is a useless thing to just have in your head? I'd like to be proven wrong, or for someone to say that I've just not been scrolling enough but Sal never really tells us what we'll ever use this for. I'm still going to learn it though because, because... yeah, because.
• It is so crazy that most of algebra, trigonometry, or geometry is just going straight out the door once we go into the work world
• In this case, x can't be zero, right?
• why isn't it 1x instead of 1/x wouldn't that be easier?
• They are 2 different things. Like for example 1/5*5 is the same as 5/5 since 5 is the same as 5/1 and you are multiplying 1/5*5. 1x is 1*x and 1/x is 1 divided by x. Same thing for the equation: x^4-2x^3+5x/x is the same as 1/x(x^4-2x^3+5x). The reason of why you are multiplying 1/x with the equation is because we start of with division and when you are dividing the numerator with itself it gets replaced with a one so then you still have to make it true so u multiply the the numerator with 1/x. 1/x * x^4-2x^3+5x/1=x^4-2x^3+5x/x
• If there were a term in the polynomial that was not divisible by x would that be considered to be a remainder?
• Let's say you want to divide 3x+2 by x right,
x | 3x +2

A striking feature is that only one term is divisible by x and 2 is not. So yes that is correct, and later on you'll be learning abt how to incorperate remainders into your answer.

hopefully that helps !
• So basically you just factor out an x?
• Pretty much. He is just showing different ways to do it, which may or may not come in handy for other problems.
• I have a question:
Say for example that the equation was

4x to the third power + x squared
----------------------------------
x

and my other equation was:

3x to the fourth power - 6x squared - x
----------------------------------------
x

why can't x2* (* squared)
--
x

x
---
x

because x2* (* squared)
--
x
answr is x. Why is that?
(1 vote)
• maybe t will help with numbers. 3/3 = 1 for instance, but 9/3 = 3 and 9 = 3*3. So we just replace that with variables. x/x=1 then x*x = x^2 so (x^2)/x = x.

Also, you say x=1 and I think there may be some confusion. When you are using variables it basically means it can be any number. so x could be 1, could be 234 could be a million, anything. x/x means some number divided by that same number which is always 1 except when x = 0, that's a special case. (x^2)/x means some number squared divided by just that number, so 9/3 for instance, which is just that original number.

Does that help?