How do you divide #(5x^3+11x^2+26x+26)div(5x+6)# using synthetic division?

1 Answer
Oct 1, 2017

Answer:

#x^2+x+4+2/(5x+6)#

Explanation:

To perform synthetic division, first find the divisor and set it equal to 0 to determine the number we will use in our synthetic division for the multiplication steps:

#5x + 6 = 0#

#5x = -6#

#x = -6/5#

Write this number at the left edge of your paper, and put a slight dividing box next to it to isolate it from the next numbers we will write.

Next to that number, spaced out a bit, write down all of the coefficients of the dividend polynomial, starting with the highest degree term, and proceeding all the way down to the constant term.

Note: If there is a term of #x# missing, be sure to write a 0 for the coefficient! (This is not the case in this problem, though.)

#-6/5__|color(white)("aaaaaa")5color(white)("aaaaaa")11color(white)("aaaaaaa")26color(white)("aaaaaaa")26#

Leave some vertical space and draw a horizontal line, much like down for addition problems:
#-6/5__|color(white)("aaaaaa")5color(white)("aaaaaa")11color(white)("aaaaaaa")26color(white)("aaaaaaa")26#
#color(white)("aaaaaaaaaa")#
#color(white)("aaaaaaaaaa")ul(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaaaaa")#

Lastly, copy down the first coefficient (5) and write it below the line:

#-6/5__|color(white)("aaaaaa")5color(white)("aaaaaa")11color(white)("aaaaaaa")26color(white)("aaaaaaa")26#
#color(white)("aaaaaaaaaa")#
#color(white)("aaaaaaaaaa")ul(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaaaaa")#
#color(white)("aaaaaaaaaaa")5#

From here on out, you perform three steps repeatedly until you finish the last column of numbers:

1) Multiply the last number written below the line by the number in the box to the upper-left.

2) Write that product in the next column to the right, just above the line.

3) Add the two numbers in the next column and write the sum underneath the line.

#-6/5__|color(white)("aaaaaa")5color(white)("aaaaaa")11color(white)("aaaaaaa")26color(white)("aaaaaaa")26#
#color(white)("aaaaaaaaaa")color(blue)(-6)color(white)("aaaaa")color(green)(-6)color(white)("aaaaa")color(blue)(-24)#
#color(white)("aaaaaaaaaa")ul(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaaaaa")#
#color(white)("aaaaaaaaaaa")5color(white)("aaaaaaa")color(blue)(5)color(white)("aaaaaaa")color(green)(20)color(white)("aaaaaaa")color(red)(2)#

As you can see, we began with 5, and then multiplied that by #-6/5# to get #-6#, which we wrote under the #11#. The sum #11 + (-6)# is #5#, which we write under the line. We then repeat again and again until we write the final value #2# (colored in red here for emphasis).

All that remains is writing the result. The last value (in red) represents the remainder of the division, and should be written over the divisor #5x+6#.

The other numbers under the line are all coefficients of the "whole" part of the resulting polynomial. There is one trick remaining: Since the original divisor was #5x#, each of these values are currently 5 times their final values. Reading from right to left these represent the constant term, #x# term, and #x^2# term, all of which should be divided by 5 before being written in the final result. Thus:

#x^2+x+4+2/(5x+6)#