|
|
|
|
|
|
|
||
|
|
|
|
|
Factoring Polynomials (page 2 of 2) Since factors and zeroes are so closely related, you can use solving techniques to factor. About the only differences are that, for factoring, you have to keep track of any numbers that you divide off (like the 2 that I divided out of the quadratic in the previous problem) and you may need to use the difference- or sum-of-cubes formula, but they generally give you less messy factors, without the complex and square-rooted numbers.
All the coefficients are even, so I can factor out a 2. I need to keep in mind, though, that since I'm finding factors rather than zeroes, I can't just divide off the 2; I'm going to have to remember to include the 2 in my final factored form. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved Even after dividing out the 2, I'm still stuck with some pretty big numbers: 15x5 – 83x4 – 271x3 + 1419x2 + 760x – 400. Since the solutions to polynomials are usually pretty close to the middle, I won't bother trying x = 400 or something else big like that in my synthetic division, at least not initially. I'll keep my first guesses small. And I'll check the graph right away to see which guesses would be most likely. (If you don't have a graphing calculator, you've got a rough road ahead of you. Apply the Rational Roots Test. Start testing the smaller whole-number values it gives you (usually factors of the constant term), and work out from there. In this example, you'd probably want to start with values like x = ±1, ±2, and ±5.. Be sure to use the synethetic division trick for knowing when to stop moving to higher or lower test values.)
I can tell from the graph that x = 5 looks like it'll be a zero, and a repeated one at that, since the graph just touches the axis instead of going through it. So if x = 5 divides out evenly, then the first thing I'll do is try x = 5 again.
This means that x – 5 is a factor twice, and the division leaves me with 15x3 + 67x2 + 24x – 16. (I've taken out two zeroes -- two factors -- so the degree of the polynomial has gone from 5 down to 3.) I'll do another quick graph to see what I'm left working with now:
Judging from the graph, I'll try x = –4.
Taking out the x + 4 factor leaves me with 15x2 + 7x – 4. I can use regular factoring methods to finish the problem: 15x2 + 7x – 4 = (5x + 4)(3x – 1). Remembering the initial 2 that I factored out, the complete factorization is: 2(x – 5)(x – 5)(x + 4)(5x + 4)(3x – 1) = 2(x – 5)2(x + 4)(5x + 4)(3x – 1) The factors don't have to be listed in any particular order, by the way, since order doesn't matter for multiplication. But any constant term (the "2", in this case) should go in front. And many books prefer repeated factors to be written in exponential form, like the "(x – 5)2" in the second form of the final answer above. A variant of this sort of question is where they give you a factor or a zero (usually something messy), and want you to find the rest of the factors or zeroes. There are examples of how this works in the last page of the synthetic division lesson. << Previous Top | 1 | 2 | Return to Index >>
|
|
|
|
Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
|
|
|
|
|
|
