Synthetic Division when Coefficent of Linear Term is Not 1Date: 12/15/2005 at 21:33:24 From: Seth Subject: Synthetic division with a linear coefficient not =1 I'm trying to use synthetic division to do 4x^3 - 7x^2 - 11x + 5 divided by 4x + 5. I can do this problem with regular long division, but can't figure out how to do it with synthetic division because the coefficient of the linear term in the divisor is not equal to 1. I only know how to use synthetic division with a linear coefficient equal to 1, but I'm told that there is a way to do this kind of division, too. Date: 12/15/2005 at 23:04:00 From: Doctor Peterson Subject: Re: Synthetic division with a linear coefficient not =1 Hi, Seth. You have to rewrite the divisor in the right form, and then adjust: 4x^3-7x^2-11x+5 4x^3-7x^2-11x+5 4x^3-7x^2-11x+5 --------------- = --------------- = --------------- * 1/4 4x+5 4(x+5/4) x+5/4 So you'd divide by x + 5/4 synthetically, then divide all coefficients by 4. Equivalently, you can divide both the divisor and the dividend by 4 before using synthetic division: 4x^3-7x^2-11x+5 [4x^3-7x^2-11x+5]/4 x^3-7/4x^2-11/4x+5/4 --------------- = ------------------- = -------------------- 4x+5 [4x+5]/4 x+5/4 Let's try it that way: -5/4 | 1 -7/4 -11/4 5/4 | -5/4 15/4 -5/4 +----------------------- 1 -3 1 0 That worked pretty well! It won't always give whole numbers, of course, but you're likely to use it only where it will. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/