I see that you don’t have a part 2 up yet, but in that part are you going to also mention the remainder theorem also works when the divisor is not monic/linear. I remember a contest problem like that, where the polynomial was divided by x^2-1 and is asked for R(x). It was so easy once you did it for x=1 and x=-1, because then you can find the line that passes through it and that will be the remainder. Great explanation btw!
That's because you can factor x^2 - 1 into (x+1)(x-1). Notice how if both of these terms can divide into the original polynomial then the polynomial is divisible by x^2-1. However, the remainder theorem usually only works for monomials, so I wouldn't recommend using it for dividing any other polynomials UNLESS you can factor them into monomials.
Jacuplex I probably overstated myself a bit. If it asked for the remainder when dividing by x^2+2x+9 it would not work because the quotient term would still be relevant, as the divisor would not be zero for any x. And the unique quotient (assuming the original polynomial was of degree n) would be a polynomial of n-2 (in this particular case since we are dividing by a polynomial of degree 2), which would require n-1 points, and then there is other stuff to deal with, and you may as well divide. So I would personally say it still “works”, but it isn’t useful
It should be but it’s just how you define it. For me it goes : P(x)/D(x) = Q(x) + reminder/D(x) . So our definition is that the reminder is reminder/D(x), at least that’s what i’ve been taught in the u.k. And then multiplying that equation by D(x) : P(x) = q(x)d(x) + r(x) which is on the right side of his whiteboard.
Greetings Mr. Woo, your lecture's wonderfully delivered as usual! However, there's an error that you made which was perhaps just a slip of the tongue, i.e. at minute 4:50 you stated that "If a divisor D(x) is of degree n, then the remainder R(x) is of degree (n-1)", which is not true in general. A simple counterexample would be the following: Take P(x)= x^4 + 2x + 3, divide it by D(x)=x^3+1, we will get R(x)=x+3 as our remainder. Note that D(x) is cubic, yet R(x) is not quadratic, but linear. A correct statement would be "If a divisor D(x) is of degree n, then the remainder R(x) is a polynomial of degree *at* *most* (n-1)".
@@dextro9322 eddie just said that the degree of the remainder is 1 less than that of the divisor. So, in @Mystery Kid's example, the remainder's degree should've been 2, but it was actually 1. So Mystery Kid is right.
I'm a little confused, would the remainder be -70/x+3 at 3:58? Or is it -70 because the remainder is always one degree less than the divisor, which in this case was a degree of one. Even if that's the case, I've seen examples where it would write the answer as -70/x+3 because the Dividend was dividing by x+3, and if the remainder is -70/x+3 it would have a reminder with degree negative one since (-70)(x+3)^-1= -70/x+3 thanks to anyone who answers
I think the remainder at 3:58 is just -70 You might want to write this on a paper because it's hard to read in this format: x^3 - 4x^2 + 2x -1 = (x+3)(x^2 - 7x + 23) - 70 is also (x^3 - 4x^2 + 2x -1)/(x+3) = x^2 - 7x + 23 - 70/(x+3) where you would write the -70/x+3 Do you see it now? ((Hope this helps
p(a)=(a-a)Q(x)+R p(a)=(0)Q(x)+R (0)Q(x) is going to be equal to 0, so 0+R is going to be equal to P(x) P(x) in this form is not being divided. It is on the other side of the equal sign and is unaffected by multiplying by zero
