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Synthetic Division & Finding Zeroes (page 3 of 4)

Sections: Introduction, Worked examples, Finding zeroes, Factoring polynomials

Once you know how to do synthetic division, you can use the technique as a shortcut to finding factors and zeroes of polynomials. Here are some examples:

• Use synthetic division to determine whether x = 1 is a zero of x³ – 1.

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!









Since the remainder is zero, then x = 1 is a zero of x³ – 1.

Since x = 1 is a zero of x³ – 1, then x – 1 is a factor, so the polynomial x³ – 1 factors as
(x – 1)(x² + x + 1). Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Use synthetic division to find all the zeroes of x⁴ + x³ – 11x² – 5x + 30.

Comparing the results of the Rational Roots Test to a quick graph, I decide to test x = 2 as a possible zero. 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 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. Follow the same procedure here. 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. Set 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: 

This is how synthetic division is usually used. You try various zeroes until one works, and then you try zeroes on the result until something works, and you keep on going like this until you get down to a quadratic, at which point you use the Quadratic Formula or other methods to get the last two zeroes.

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