Integrated math 3
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,.
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- So basically you just factor out an x?(14 votes)
- 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.(10 votes)
- 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.(6 votes)
- 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(4 votes)
- why isn't it 1x instead of 1/x wouldn't that be easier?(3 votes)
- 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(4 votes)
- If there were a term in the polynomial that was not divisible by x would that be considered to be a remainder?(2 votes)
- 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 !(5 votes)
- So basically you just factor out an x?(3 votes)
- Pretty much. He is just showing different ways to do it, which may or may not come in handy for other problems.(2 votes)
- I have a question:
Say for example that the equation was
4x to the third power + x squared
and my other equation was:
3x to the fourth power - 6x squared - x
why can't x2* (* squared)
can be the same answer* (answer is 1) as
because x2* (* squared)
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?(5 votes)
- Can't you just skip to the distributed version as opposed to rewriting the numerator? That part confused me and I could see factoring out the x easier than the rewritten version.(3 votes)
- [Instructor] What I'd like to do in this video is try to figure out what x to the fourth minus two x to the third plus five x divided by x is equal to. So pause this video and see if you can have a go at that before we work through this together. All right, so if we're saying what is this top expression divided by this bottom expression, another way to think about it is, what do I have to multiply, so I'm going to multiply something, I'll put that in parentheses. If I multiply that something times x, I should get x to the fourth minus two x to the third plus five x. Now how do I approach that? Well there's two ways that I could tackle it. One way is I could just rewrite this expression as being, and I will just make this x in yellow so I can keep track of it. I could just rewrite this as one over x times, times x to the fourth minus two x to the third plus five x. And then I can distribute the one over x, and so what is that going to be equal to? Well it's going to be equal to x to the fourth. Let me do this, x to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of these going to be equal to? X to the fourth divided by x, if I have four x's that I'm multiplying together and then I divide by x, that's going to be equivalent to x to the third power. So this right over here is equal to x to the third. You could also get there from your exponent properties. In the denominator, you have an x to the first power, and so you would subtract the exponents. You have the same base here, so that's x to the third. And then, in this part right over here, what would that equal to? Well it's going to be minus two x to the third divided by x to the first. Well by the same property, that's going to be x squared. And then last but not least, if you take five x's and then you divide by x, you are just going to be left with five. And you can verify that this, indeed, if I were to multiply it by x, I'm gonna get x to the fourth minus two x to the third plus five x. Let me do that. If I put x to the third minus two x squared plus five times x, what I could do is distribute the x. X times x to the third is x to the fourth. X times negative two x squared is negative two x to the third. X times five is five x. Now I mentioned there's two ways that I could do it. Another way that I could try to tackle it is I could look at this numerator and try to factor an x out. I would try to factor out whatever I see in the denominator. So if I do that, actually let me just rewrite the numerator. So I can rewrite x to the fourth as x times x to the third. And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could write this five x as being equal to plus x times five. And then I'm gonna divide everything by x, divide everything by x. I just rewrote the numerator here, but for each of those terms, I factored out an x. And now I can factor out x out of the whole thing. So I sometimes think of factoring out an x out of the whole thing as reverse distributive property. So if I factor out this x out of every term, what am I left with? I'm left with an x times x to the third minus two x squared plus five; I ended up doing that in the same color, but hopefully you're following, plus five, and then all of that is divided by x. And as long as x does not equal zero, x divided by x is going to be equal to one, and we're left with what we had to begin with, or the answer that we had to begin with. So these are two different approaches. Nothing super sophisticated here. When you're dividing by x, you're just like hey, that's the same thing as multiplying every term by one over x, or you can factor out an x out of the numerator, and then they cancel out.