How do you divide #(m^3-20)div(m-3)# using synthetic division?

1 Answer
Oct 1, 2017



Synthetic division for polynomials that have missing terms is a little more complicated than usual, but not difficult.

As always, begin by setting the divisor equal to 0 and solving for the variable. This number will go on the outside in a "box":

#m-3=0 :. m=3#

Write that number and surround it with a partial box to keep it off to the side. Next, you must write down the coefficients of all terms in the dividend. Begin with the leading term and write down all of the coefficients in order from the highest degree down to the constant term. If any term is missing, you must put a 0 in its place. Write these numbers out in a line, like this:

#3__|color(white)("aaaaaa")1color(white)("aaaaaa")0 color(white)("aaaaaa")0 color(white)("aaaaaa")-20#

Leave a bit of vertical space, and then draw a line as though you were doing an addition problem. Underneath the line, directly below the 1, copy that value of 1, as you can see here:

#3__|color(white)("aaaaaa")1color(white)("aaaaaa")0 color(white)("aaaaaa")0 color(white)("aaaaaa")-20#
#color(white)("aaaaaaaa")ul(color(white) ("aaaaaaaaaaaaaaaaaaaaaaaaaa")#

Now perform 2 steps repeatedly until you run out of columns:

1) Multiply the last number you wrote under the line by the number in the "box". Write that product just above the line in the next column over, under the top row of numbers.

2) In the next column over, add the top row number to the number you just wrote, and place that sum under the line in the same column.

#3__|color(white)("aaaaaa")1color(white)("aaaaaa")0 color(white)("aaaaaa")0 color(white)("aaaa")-20#
#color(white)("aaaaaaaa")ul(color(white) ("aaaaaaaaaaaaaaaaaaaaaaaaaa")#

Now we can write the polynomial answer. The last number in the bottom row (which I colored in red) is the remainder, which you should write as the numerator in a fraction with the divisor as the denominator. Moving towards the left, you will see the coefficients in order from the constant, to the #m# term, to the #m^2# term.

The final answer is: #m^2+3m+9+7/(m-3)#