Synthetic Division, Factors, and Zeroes

I will set up the synthetic division, and check to see if the remainder is zero. If the remainder is zero, then x = 1 is a zero of x³ − 1.

To do the initial set-up, note that I needed to leave "gaps" for the powers of x that are not included in the polynomial. That is, I followed the practice used with long division, and wrote the polynomial as x³ + 0x² + 0x − 1 for the purposes of doing the division. If you forget to leave "gaps", your division will not work properly!

Step 1:

Step 2:

Step 3:

Step 4:

Since the last value in the bottom row is zero, then the remainder on this division is zero. Since the remainder is zero, then:

(x − 1)​(x² + x + 1)

Comparing the results of the Rational Roots Test to a quick graph, I decide to test x = 2 as a possible zero. I set up the divison:

...and here is the result:

Since the remainder is zero, then x = 2 is indeed a zero of the original polynomial.

To continue on and find the rest of the polynomial's zeroes, should I start over again with x⁴ + x³ − 11x² − 5x + 30? Well, think about when you factor something like 72. After you divide a 2 out and get a 36, do you go back to the 72 to try the next factor, or do you see what will go into the 36? Of course, you try factors into the 36. It is simplest to follow the same procedure here. So I won't return to the original polynomial, but will instead see what divides into my result.

(Recall that synthetic-dividing out x = 2 is the same as long-dividing out x − 2, so the result has a degree that is one lower than what I started with. That is, to continue, I will be dealing not with the original fourth-degree polynomial x⁴ + x³ − 11x² − 5x + 30, but with the third-degree result from the synthetic division: x³ + 3x² − 5x − 15.)

Continuing, and again comparing the Rational Roots Test with a quick graph, I will try x = −3. Iet up the division:

...and here is the result:

Since the remainder is zero, then x = −3 is a zero of the original polynomial. At this point, the final result is a quadratic, (x² − 5), and I can apply the Quadratic Formula or other methods to get the remaining zeroes:

Then all the zeroes are: