Psion Posted August 3, 2005 Posted August 3, 2005 I don't know the latex symbol to get long division symbols but hopefully someone can figure this out for me. I've been doing pretty good until I came across this problem: (x^3 + 1 + 2x) / (x^5 + 3x + 2) Answer is: [hide]x^2 - 2 + [(-x^2 + 7x + 4) / (x^3 + 2x + 1)][/hide] I was going to go in depth about this, however, I don't see the latex symbol that closely looks like the square root symbol for division.... It's very obvious to me that I need long division. I just didn't know how to solve the problem, that's all. I don't understand the long division processes used in this. Long division in plain numbers is somewhat easy, however, this polynomial division isn't. I was wondering if someone could give me detailed steps on how to take on this problem.
matt grime Posted August 3, 2005 Posted August 3, 2005 polynomial division is exactly as hard as dvision of integers if explained properly i think you made a typo and it ought to be [math]\frac{x^5 + 3x + 2}{x^3 + 2x+1}[/math] you are simplyfying. the easiest way is, perhaps, this. we have an x^3+2x+1 on bottom and x^5 on top. how many times does the denominator go into x^5 and what is the remainder? clearly x^5 = x^2(x^3) now if i add and subtrac cleverly then x^5 = x^2(x^3+2x+1) - 2x^3 - x^2 this is exactly the same as saying, well how many times does 14 go into 57? 57=4*14+1 so the answer is 4 times remainder 1 or 57/14= 4+1/14. back to polys. given how we can rewrite x^5, we see that the numerator is exactly x^2(x^3+2x+1) - 2x^3 - x^2+3x+2 so the whole thing simplfies to: [math]\frac{x^5 + 3x + 2}{x^3 + 2x+1}=x^2 + \frac{- 2x^3 - x^2+3x+2}{x^3 + 2x+1}[/math] now you can repeat for the higest power of x remaining
Psion Posted August 4, 2005 Author Posted August 4, 2005 no. the answer is: x^2 - 2 + (-x^2 + 7x + 4) / (x^3 + 2x + 1) I didn't type anything wrong. I don't know how to get to the answer, though.
DQW Posted August 4, 2005 Posted August 4, 2005 Psion, What you've shown in the picture IS [math]\frac{x^5 + 3x + 2}{x^3 + 2x+1}[/math] If you follow matt's procedure and go one step further, you will get the required answer. If you're not convinced, look at [imath]x^5/x^3[/imath] and [imath]x^3/x^5[/imath]. Which one gives you a quotient of [imath]x^2[/imath] ?
Klaynos Posted August 4, 2005 Posted August 4, 2005 Follow matt's above instructions as Dak said, work it through again using the same method using: x3=(x3+2x+1-2x-1) You get your given answer.
Psion Posted August 4, 2005 Author Posted August 4, 2005 Hmm... I actually don't understand any of that. ;_; i know x^5 / x^3 = x^2 && 2/1 = 2 I don't know where people are obtaining a -2 from though.. Here's the problem worked out.. I know I'm screwing up somewhere though..
DQW Posted August 4, 2005 Posted August 4, 2005 When you arrive at a remainder of x^2 - x, you must stop, because the order of the remainder polynomial is smaller the the order of the divisor. In fact, if you look at your next step, you will find your mistake. How did you multiply (x^3 + 2x + 1) by (-x) and get (x^2 - 2x - x) ?
matt grime Posted August 4, 2005 Posted August 4, 2005 iff you SUBTRACT the x^5+2x^2+x^2, why do you end up with a +2x^3 on the next line. in your forst post you wrote pseudo-ascii-math that means to divide the left hand poly by the right hand one not the other way round ie 3/5 does not mean divide 5 by 3. sorry for the confusion, it looks like a fraction. try [math] x^3+2x+1 \ | \overline{x^5+3x+2}[/math] as a makeshift solution
Psion Posted August 4, 2005 Author Posted August 4, 2005 It was all about the 2x^3.. i like totally came to realization about this. i have to remember to imagine a - sign and be all like - (x^5 + 2x^3 ...) kool. I now have come to peace with this problem. i think it really helped my by putting it up on the computer instead of paper so i could mix the numbers around in my head and on screen.
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