Adding polynomials 

We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really,  all we have to do is we have to combine like terms-- terms that have x raised to the same  power. And the first thing we can do, we can actually get rid of these parentheses right here,  because we have this whole expression, and then we're adding it to this whole expression. The  parentheses really don't change our order of operations here. So let me just rewrite it once  without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a  minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x  plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start  with the x squared terms. So you have a 5x squared term here and you have a 2x squared term  right there. So 5 of something plus 2 of that same something is going to be 7 of that something. So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right  there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7  of something, you're just going to have 1 of that something. And then if you add 14 of that  something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry.  You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x  minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you  want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a  constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've  simplified it as far as we can go. We are done. 

Subtracting polynomials 

Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you  subtract an entire expression, this is the exact same thing as having 16x plus 14. And then  you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared  plus x minus 9. Or another way to think about it is you can distribute this negative sign along all  of those terms. That's essentially what we're about to do here. We're just adding the negative  of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change  it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So  negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x  because that's positive 1x. Negative 1 times negative 9-- remember, you have to consider this  negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9.  Negative times a negative is a positive. So then we have positive 9. And now we just have to  combine like terms. So what's our highest degree term here? I like to write it in that order. We  have only one x squared term, second-degree term. We only have one of those. So let me write  it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of  just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract  an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view  that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both  being multiplied by x to the 0. 14 plus 9 is 23. And we're done. Negative 3x squared plus 15x plus 23. 

Polynomial subtraction

We're asked to subtract negative two x squared plus four x minus one from six x squared plus  three x minus nine, and like always I encourage you to pause the video and see if you can give it  a go. All right, now let's work through this together. So I could rewrite this as six x squared plus  three x minus nine minus, minus this expression right over here, so I'll put that in parentheses,  minus negative two x squared, negative two x squared plus four x minus one. Now what can we  do from here? Well, we can distribute this negative side, we can distribute this negative side,  and then if we did that, we would get the six x squared plus three x minus nine won't change so  we still have that. Six x squared plus three x minus nine, but if I distribute the negative side, the  negative of negative two x squared is positive two x squared. So that's going to be positive  two... Get a little more space. Positive two x squared, and then subtract, and then the negative  of positive four x is... I'm going to subtract four x now, and then the negative of negative one, or the opposite of negative one is going to be positive one. So I've just distributed the negative  side, and now I can add terms that have the same degree on our x... The same degree terms, I  guess you could say, so I have an x squared term, here it's six x squared, here I have a two x  squared term, so I can add those two together, six x squared plus two x squared. If I have six x  squareds and then I have another two x squareds, how many x squareds am I now going to  have? I'm now going to have eight x squareds, eight x squareds, or six x squared plus two x  squared. We add the coefficients, the six and the two to get eight, eight x squared. Then we can add the x terms. You could view these as the first-degree terms, three x... We have three x and  then we have minus four x, so three x minus four x, if I have three of something and I take away  four of them, I'm now going to have negative one of that thing, or you could say that the  coefficients, three minus four would be negative one. So I now have negative one x. I could  write it as negative one x, but I might as well just write it as negative x. That's the same thing as  negative one x. And then finally, I can deal with our constant terms. I'm subtracting a nine and  then I'm adding a one. So you could say, "Well, what's a negative nine plus one?" Well, that's  going to be negative eight. That's going to be negative eight and we are all done. And one thing that you might find interesting is I had a polynomial here and from that I subtracted another  polynomial, and notice, I got a polynomial, and this is actually always going to be the case. If  you think about the set of all polynomials, if you just think about the set... Let me do this in a  neutral color. So if we think about the set of all polynomials right over here, and if you take one  polynomial, which you could imagine this magenta polynomial. So this is a polynomial right  over here, let's call this p of x. So this is p of x right over there, p of x, and then you have another polynomial, this one right over here, let's call this, I don't know, we can call this q of x, q of x,  just for kicks. So that's q of x, just like that. And if you apply the... In this case, we applied the  subtraction operator. If we apply subtraction... So you took these two, you took these two...  Let me see how I could depict this well. So we took p of x and you subtracted from that q of x.  We still get a polynomial, so that's going to give us... We stay in the set of polynomials. At any  time you have a set of things, and you might be more used to talking about this in terms of  integers, or number sets, but you can talk about this in general. Here we're talking about the  set of polynomials, and we just saw that if we start with two polynomials, two members of the  set of polynomials... Let me be clear, this is polynomials, polynomials, right over here. You take  two members of the set and you perform the subtraction operation, you're still going to get a  member of the set. And when you have a situation like this, I could call this one, I don't know, 

I'm running out of letters. Well let me just call this one f of x, so we've got f of x here. When you  have this situation where you take two members of a set, you apply an operator on them, or  you take a certain number of members of a set, you apply an operator on them, and you still  get a member of the set, we would say that this set is closed under that operation. So we could  say that the set of polynomials, set of polynomials, polynomials closed, closed... I won't even  put it in quotes. Closed under, under subtraction. And I didn't prove it here, I just did one  example where I subtracted two polynomials and I got another one, and there's clearly more  rigorous proofs that you can do. But this is actually the case, as long as you have two  polynomials, you apply subtraction, you're going to get another polynomial. And the fancy way of saying that is the set of polynomials is closed under subtraction. This notion of closure  sometimes seems like this very fancy mathematical idea, but it's not too fancy. It's just you take two members of a set, you apply an operation, if you still get a member of the set after that  operation, then that set is closed under that operation.



Последнее изменение: четверг, 7 апреля 2022, 09:49