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 September 11th, 2012, 05:56 PM #1 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Is this dividing by zero? If you want to find the factors f(x) You dive by (x-a). You get some new function q(x) +remainder/(x-a). If (x-a) is a factor of f(x), then the remainder is zero. Also, f(x)=(x-a)q(x)+remainder, and f(a)=(0)q(x) + remainder or just f(a)= remainder. But Isn't it quasi dividing by zero If you divide f(x) by (x-a), while x=a? I'm thinking f/a=q+r/a, then f=a(q)+r, except when a = zero.
 September 12th, 2012, 07:18 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Is this dividing by zero? What you are saying is that as long as $x\ne a$, $\frac{f(x)}{x- a}= q(x)+ \frac{\text{remainder}}{x- a}$. And, as long as $x\ne a$, $f(x)= q(x)(x- a)+ \text{remainder}$. You say "functions", but all of this only makes sense for polynomials (as you imply when you say "find the factors"). All polynomials are continuous for all x so taking the limit as x goes to a of that formula gives f(a)= q(a)(0)+ remainder= remainder.

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