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Synthetic Division: The Process (page 1 of 4) Sections: Introduction, Worked examples, Finding zeroes, Factoring polynomials Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor (and only works in this case). It is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. If you are given, say, y = x2 + 5x + 6, you can factor this as y = (x + 3)(x + 2). Then you can find the zeroes of y by setting each factor equal to zero and solving. You will find that x = –2 and x = –3 are the two zeroes of y. You can, however, also work backwards from the zeroes to find the originating polynomial. For instance, if you are given that x = –2 and x = –3 are zeroes of a quadratic, then you know that x + 2 = 0, so x + 2 is a factor, and x + 3 = 0, so x + 3 is a factor. Therefore, you know that the quadratic must be of the form y = a(x + 3)(x + 2). (The extra number "a" in that last sentence is in there because you don't know, when working backwards from the zeroes, which quadratic you're working toward. For any non-zero value of "a", your quadratic will still have the same zeroes. This is a technical point; as long as you see the relationship between the zeroes and the factors, that's all you really need to know for this lesson.) Anyway, the above is a long-winded way of saying that, if x – n is a factor, then x = n is a zero, and if x = n is a zero, then x – n is a factor. And this is the fact you use when you use synthetic division. Let's look again at the quadratic from above: y = x2 + 5x + 6 . From the Rational Roots Test, you know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring above, you know that the zeroes are –3 and –2.) How would you use synthetic division to check the potential zeroes? Well, think about how long divison works. If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x2 + 5x + 6 by x – 1, we would get a zero remainder. Let's check:
As expected (since we know that x – 1 is not a factor), we got a non-zero remainder. What does this look like in synthetic division? Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
Note that we got the same result as when we did the long division:
The result is formatted differently, but you should recognize that, in each format, we arrived at the result "x + 6, with remainder 12".
As you can see above, while the results are formatted differently, the results are, in a sense, the same: In the long division, I divided by the factor x + 3, and arrived at the result of x + 2 with a remainder of zero. This means that x + 3 is a factor, and that x + 2 is left after factoring out the x + 3. Setting the factors equal to zero, I get that x = –3 and x = –2 are the zeroes of the quadratic. In the synthetic division, I divided by x = –3, and arrived at the same result of x + 2 with a remainder of zero. Because the remainder is zero, this means that x + 3 is a factor and x = –3 is a zero. Also, because of the zero remainder, x + 2 is the remaining factor after division. Setting this equal to zero, I get that x = –2 is the other zero of the quadratic. I will return to this relationship between factors and zeroes throughout what follows; the two topics are inextricably intertwined. Top | 1 | 2 | 3 | 4 | Return to Index Next >>
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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