We multiply this sum by the “test root” and place the product beneath the entry −7:
(^210) − 7 − 6
24#
12##
We add the numbers in the fourth column, writing the sum in the bottom row:
(^210) − 7 − 6
24#
(^12) − 3 #
Do you see the pattern? Next, we multiply the “test root” by −3, and write the product under
the entry −6:
210 − 7 − 6
(^24) − 6
12 − 3 #
Adding the numbers in the last column yields the final result of our test. Let’s write this last
number, in the bottom row:
(^210) − 7 − 6
24 − 6
12 − 3 − 12
This final number is called the remainder. What does this particular result mean? Well, the
news is not good. We want the remainder to be 0! That is always the ultimate goal of synthetic
division when we’re looking for a binomial factor of a cubic equation. We must try another
“test root” and go through this ritual again.
We try, we fail, we learn
Let’s try 4 as our “test root” this time. We put a bold numeral 4 in the top left slot. Then the pro-
cess goes along with the new numbers, in the same way as before. Here’s the sequence of steps:
410 − 7 − 6
Polynomial Equation of Third Degree 425