1. The problem statement, all variables and given/known data When we divide the polynom P(x) with x-a, we receive reminder [tex]\alpha[/tex], and if we divide P(x) with x-a, we receive reminder [tex]\beta[/tex]. What will be the reminder, if we divide P(x) with (x-a)(x-b)? 2. Relevant equations 3. The attempt at a solution [tex]P(x)=(x-a)f(x) + \alpha [/tex] [tex]P(x)=(x-b)c(x) + \beta [/tex] [tex]P(x)= (x-b)(x-a)g(x)+ R(x) [/tex] We need to find R(x)
Physicsissuef, you were supposed show us what you'd already done, in your original post. Also the next thing you tried, and how it didn't work. That's what 3. The attempt at a solution is for. If you don't, other PF members don't know where to start, and it wastes our time. Start again …
ok, your R(x) = px + q, for some constants p and q (because R(x) can't contain x^2, since it's a remainder after dividing by x^2 + …). So you should be able to write two equations involving (px + q) and alpha and beta. What are they? …
Remainders must be smaller Actually, "after dividing by x^2 + …" If you divide by (x^2 plus anything smaller), then the remainder has to be smaller. If the remainder was, say, 3x^2 + 5, then you'd just subtract 3 more lots of (x^2 + …), to give you a new remainder with only x and 1. Generally, the remainder after dividing by any polynomial beginning with x^n will be a polynomial beginning with x^(n-1) or less.
It's (x-a)(x-b), which is x^2 - (a+b)x + ab; so when you divide any polynomial by it, the remainder will always be either linear or constant.
And how do you know that the polynom P(x) is in this form: [tex]ax^3+bx^2+cx+d[/tex], so you can write [tex]R(x)=px+q[/tex] ??
Sorry, Physicsissuef, you've completely lost me. Where did [tex]ax^3+bx^2+cx+d[/tex] come from? Your own equation, [tex]P(x)= (x-b)(x-a)g(x)+ R(x) [/tex], where R(x) is a remainder, shows that R(x) can't have anything higher than x.
You said that [tex]R(x)=px+q[/tex], right? So when I divide some polynom with x^3 with x^2, I will receive R(x)=px+q, right? I asked how did you know that it is x^3? Maybe it is x^4?
Physicsissuef, divide anything with x^2 + …, you will receive R(x)=px+q. It could be x^3 or x^4 or x^307 … it doesn't matter … the remainder will always be of the form px +q (possibly, of course, with p or q = 0).
I've lost the plot now … ah … we have to go back to post #6: Hint: divide (px + q) by (x - a) … what is the remainder?
You've got the principle, but you've got rather confused writing it. From post #1, P(x) = f(x)(x-a) + alpha. And so alpha equals pa + q (not px + q), from your post #17, because, as you say, alpha is the remainder. Right! Nearly there! alpha = pa + q. Now what does beta equal?