Friday, 6 February 2009

Synthetic Division When the Linear Coefficent is NOT One

I was hoping this would be the last one on synthetic division, I have a problem I promised Al Harmon from Misawa, Japan I would answer, and I wanted to talk about the Law of Sines (and how much I hate when people write it upside down)... So I will defer the responses to those of you who asked about the history of synthetic divison for a later day. Today I wanted to answer the challenge to post the way to divide by ax+b without factoring anything out first.

Apparently my ex-students never throw away their notes hoping to catch me if i ever do something a different way, as I did a couple of days ago with dividing both divisor and dividend before I did the division so as to eliminate the leading coefficient of the divisor..... so in case there are other teachers who teach this out there...here is a modified synthetic division approach that will work for 3x+5 and such and still give the right remainder...

I will use the same example 6x2 + 16x + 16 by 3x+5 . To accomodate the 3 in the leading coefficent, a new line is added below the usual bottom line. After each column is summed, (except for the remainder) the result is divided by the linear coefficient, in this case 3.
Notice that after we add the first two numbers in a column, we divide that sum by three to get the actual value of the quotient for that term.... but we do NOT divide the remainder.

I also wondered (I had never tried it) if you could do the same, or something similar with the higher degree divisors. I had previously divide x5+ 6x4+4x3-39x2-122x-120 by the cubic x3+3x2-10x-24, so I thought I would try 6x5+ 6x4+4x3-39x2-122x-120 by the cubic 2x3+3x2-10x-24, which Should give a quotient of 3x2 - 3x/2 +77/4 with a remainder of 159/4 x2 + 69x/2 + 342 and see what happens.....


Once more we create an extra line to divide by the coefficient of the term with the highest degree (in this case 2) and proceed as we do in the non-linear cases by shifting over one column as we go down the list of coefficients (or the negatives of them, actually).


And once more, the result emerges rather effortlessly... Ok, we have some way ugly fractions in there... but you would get them even with all the mess you create with long division...
Call that a wrap...
Post a Comment