Polynomial Division Compared with Long DivisionDate: 02/01/2009 at 18:48:15 From: Rashad Subject: Dividing polynomials I can't seem to grasp the concept! Date: 02/02/2009 at 08:04:11 From: Doctor Ian Subject: Re: Dividing polynomials Hi Rashad, Suppose we have a problem like _____ 23 ) 4899 We can use "long division" like so: 213 _____ 23 ) 4899 46 --- 29 23 --- 69 69 -- 0 Does that look familiar? I think it's a little easier to see what's going on if we don't hide all the zeros: 3 10 200 _____ 23 ) 4899 4600 <- 23 * 200 --- 299 230 <- 23 * 10 --- 69 69 <- 23 * 3 -- 0 Now, what if we write it like this? 2*100 + 1*10 + 3*1 ____________________________ 2*10 + 3 ) 4*1000 + 8*100 + 9*10 + 9*1 4*1000 + 6*100 -------------- 2*100 + 9*10 2*100 + 3*10 ------------ 6*10 + 9*1 6*10 + 9*1 ---------- 0 Same thing as before, but now it's more explicit about showing what's going on. It's even more explicit if we write it this way: 2*10^2 + 1*10^1 + 3*10^0 __________________________________ 2*10^1 + 3 ) 4*10^3 + 8*10^2 + 9*10^1 + 9*10^0 4*10^3 + 6*10^2 -------------- 2*10^2 + 9*10^1 2*10^2 + 3*10^1 --- ------------ 6*10^1 + 9*10^0 6*10^1 + 9*10^0 --------------- 0 So far, so good? Now, suppose that, instead of 10 as the base of our exponents, we just have some unknown number x: 2*x^2 + 1*x^1 + 3*x^0 _____________________________ 2*x^1 + 3 ) 4*x^3 + 8*x^2 + 9*x^1 + 9*x^0 4*x^3 + 6*x^2 -------------- 2*x^2 + 9*x^1 2*x^2 + 3*x^1 --------------- 6*x^1 + 9*x^0 6*x^1 + 9*x^0 ------------- 0 But that's just a polynomial division, isn't it? If we leave out the explicit * symbols, and ^1's, and x^0's, we have something that should look like the kind of problem you're trying to solve: 2x^2 + 1x + 3 _____________________ 2x + 3 ) 4x^3 + 8x^2 + 9x + 9 4x^3 + 6x^2 ----------- 2x^2 + 9x 2x^2 + 3x ------------- 6x + 9 6x + 9 ------ 0 But as you can see, it's really just long division, using x instead of 10 as the base. There are some subtleties that occur with polynomials (e.g., you can have negative coefficients), but can you at least "grasp the concept" now? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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