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Mathematics Forum
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| Author | Message / Information |
| markai Quote | Reply | | Complex rational roots posted on: 10/31/2006 1:51:25 AM I'm not sure how to do this following problem: f(x)= 6x^8 - 9x^7+ 9x^6 + 15x^5 - 441x^4 - 540x^3 + 324x^2 I have to find all the possible rational roots, along with other things. I have factored the polynomial into the following: f(x)= 3x^2 ( 2x^6 - 3x^5 - 3x^4 + 5x^3 - 147x^2 - 180x + 108) I am at a loss as what to do now...I believe the polynomial to have exactly 8 zeros/roots. The is no constant term in the polynomial (I guess it's implied as "0"?) So I can't use the rational zero test. I am lost/confused by the now factored ploynomial. Please help me, or at least guide me in the right direction. Any help is appreciated. Thank you. |
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Euler
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Complex rational roots
replied on: 10/31/2006 11:38:40 PM I would use the p/q method to factor. You take all the possible factors of the coefficient of the highest degree term, and let that be p, and the possible factors of the constant, and let that be q, then for all of those: p/q gives you possible roots. This will give you all the real roots of the equation, thought not all p/q are roots, so if you have coefficients with many factors this can take a lot of work. It is possible, however, since the degree is even, that this equation has only complex roots, which means this method will be useless. Anyways, in this case (for the sixth degree polynomial you have factored it to) p can be +/-3,+/-2, or +/-1, and q can be +/-1,+/-2,+/-3,+/-4,+/-6, +/-9,or +/-12. Hope this helps! |
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HallsofIvy
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Complex rational roots
replied on: 4/6/2007 1:40:53 PM Yes, the "no constant term" is the same as saying it is 0. However, since EVERY number evenly divides 0, that doesn't really help. Do just what you have done: factor out the 3x^2 so you do have one. Now, the denominator of any rational root must be 1, -1, 2, or -2 and the numerator a divisor of 108. Apparently Euler has already enumerated those so I don't have to1 |
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