How do you use synthetic division and the Remainder Theorem to find #P(a)#, if #P(x) = 6x^4+19x^3-2x^2-44x-24# and #a=-2/3#?

Question

Benjamin Achtenberg · Accepted Answer

#P(-2/3) = 0#

The Reaminder Theorem tells us that if you wish to find #P(a)#, or the value of the function #P(x)# when #x = a#, you can divide by the related factor #(x-a)# and examine the coefficient of the remainder term to determine #P(a)#.

In Synthetic Division (SD), this would correspond to the final value on the row of numbers beneath the line as we solve the problem using synthetic division. As a "bonus", we already have the "starting number" for SD: #-2/3#, or the value we're evaluating #P(x)# at.

Write the value #-2/3# in a little box off to the left edge of the paper, and then write out all of the coefficients of the terms in the function, in descending order from the highest power down to the constant term (being sure to write a 0 in for any missing terms), in a row across the page. Leave a little space beneath that row of numbers and draw a horizontal line, like an addition problem. Finally, copy down the leading coefficient beneath the line to set up the SD workspace:

#-2/3__|color(white)("aaaa")6color(white)("aaaaa")19color(white)("aaaaa")-2color(white)("aaaaa")-44color(white)("aaaaa")-24# #color(white)("aaaaaaaa|")underline(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"))# #color(white)("aaaaaaaaa")6#

Now, perform the following steps repeatedly until you run out of columns to work with:

Multiply the last number written underneath the line by the number in the box to the upper-left, and write that product just above the line in the column to the right.
Add the topmost coefficient number with the product you just wrote, and write that sum just under the line in the same column. This should be written just to the right of the rightmost number in the bottom row under the line.

Keep working in this manner until you run out of columns/numbers. I've color coded the sets of numbers in the worked out solution below:

#-2/3__|color(white)("aaaa")6color(white)("aaaaa")19color(white)("aaaaa")-2color(white)("aaaaa")-44color(white)("aaaaa")-24# #color(white)("aaaaaaaa|")underline(color(white)("aaaaaa")color(blue)(-4)color(white)color(white)("aaaa")color(green)(-10)color(white)("aaaaaaaaa")color(purple)(8)color(white)("aaaaaaa|")color(red)(24))# #color(white)("aaaaaaaaa")color(blue)(6)color(white)("aaaaa")color(green)(15)color(white)("aaaaa")color(purple)(-12)color(white)("aaaaa|")color(red)(-36)color(white)("aaaaaaaa")color(red)(0)#

The last number on the bottom row is 0, which is the value of #P(-2/3)# per the Remainder Theorem. (It also conveniently tells us that #(x+2/3)# is a factor of #P(x)#, by the way.)

To check (and completely defeating the purpose of using SD in the first place):

#{:(P(-2/3),=,6((-2)/3)^4+19((-2)/3)^3-2((-2)/3)^2-44((-2)/3)-24),(,=,6(16/81)-19(8/27)-2(4/9)+88/3-24),(,=,96/81-152/27-8/9+88/3-24),(,=,96/81-456/81-72/81+2376/81-1944/81),(,=,0/81=0):}#

As you can see, SD calculations are far, far easier to do - sometimes even in your head!