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 December 27th, 2006, 08:38 AM #1 Member   Joined: Nov 2006 From: BIG APPLE Posts: 63 Thanks: 0 two roots If one root of x^3 - 2x^2 - 9x + 18 = 0 is 2, give the other two roots.
 December 27th, 2006, 09:15 AM #2 Newbie   Joined: Dec 2006 Posts: 2 Thanks: 0 Divide by x-2.
 December 27th, 2006, 09:21 AM #3 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 Since we know that one of the roots is x=2, (x-2) must be a factor of this equation. Thus we can divide the equation by x-2 to give us a 2nd-degree equation, from which it is easy to find the other two roots. I can tell you without doing any work that one of the other roots will a be positive integer and the other a negative one, and that both will be factors of 18. If you know synthetic division, the following work will make sense to you. If you don't, please ask, and I will try to explain it. Synthetic division is a very powerful tool for solving this type of problem. Code:  1 -2 -9 18 2 | 2 0 -18 _______________ 1 0 -9 0 We can now write the equation as (x-2)(x^2-9), from which it is easy to see that the other equations are 3 and -3. Do not worry if there is something that you do not understand beyond this point as your teacher will probably cover it, and likely do a better job than I will. As to how I knew some information about the roots before doing any of the work, an any polynomial function with integer coefficients, all rational roots will have absolute values that are of the form p/q, where p is an integer factor of the constant term and q is an integer factor of the left-most coefficient. Also, for any polynomial function f(x) with integer coefficients, the number of positive rational roots will be equal to the number of changes of signs in f(x), minus some whole multiple of 2, where a change of sign is when one coefficient is positive and the next negative, or one coefficient is negative and the next positive. The number of negative rational roots will be equal to the number of changes in sign in f(-x), again minus some whole multiple of 2. Allowing for the subtraction of whole multiples of two takes into account the possibility of irrational or imaginary roots, which will come in pairs. Using these two methods will allow you to find the roots of equations relatively efficiently even if there are no roots given. Also remember that in any polynomial function with integer coefficients, if a + sqrt(b) is a factor, then a - sqrt(b) must also be a factor, and that if a + bi is a factor, than a - bi must also be a factor, where i is defined to be sqrt(-1). I know this is a lot of information to work through; don't worry if you don't understand all of it.
January 19th, 2007, 02:52 PM   #4
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Re: two roots

Quote:
 Originally Posted by Soha If one root of x^3 - 2x^2 - 9x + 18 = 0 is 2, give the other two roots.
Since 2 is a root, it's very easy to spot that the polynomial factorizes as
(x - 2)(x² - 9) = (x - 2)(x - 3)(x + 3)
and it's only after factorizing that it can be seen that the remaining roots are rational, namely 3 and -3.

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