Synthetic Division when Coefficent of Linear Term is Not 1

Date: 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/