Finding solutions for 5th-degree polynomial functions  TOPIC_SOLVED

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Finding solutions for 5th-degree polynomial functions

Postby honest_denverco09 on Wed Apr 22, 2009 2:13 am

Please help me to solve this function : y = x^5 - 6x^4 + 20x^3 - 60x^2 + 99x - 54
I know the answer are x = 1, x = 2, x = 3, x = 3i, x = -3i . Thanks!
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Re: Finding solutions for 5th-degree polynomial functions

Postby Martingale on Wed Apr 22, 2009 4:40 am

honest_denverco09 wrote:Please help me to solve this function : y = x^5 - 6x^4 + 20x^3 - 60x^2 + 99x - 54
I know the answer are x = 1, x = 2, x = 3, x = 3i, x = -3i . Thanks!


if you are given this polynomial and you don't know any of the roots then you can start my using the rational root theorem to see if there are any rational roots to your polynomial.

otherwise if you can factor it, because you just don't see how, you could always try a numerical technique.
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Postby stapel_eliz on Wed Apr 22, 2009 11:26 am

honest_denverco09 wrote:Please help me to solve this function : y = x^5 - 6x^4 + 20x^3 - 60x^2 + 99x - 54

To learn the set of techniques used for solving polynomials, please try here. :wink:
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Re: Finding solutions for 5th-degree polynomial functions

Postby honest_denverco09 on Thu Apr 23, 2009 10:50 pm

Thank you, guy a lot!

Here is some steps that i did :
We have y = x^5 - 6x^2 + 11x - 6
1st :Using calculator to find the visible x-intercepts by looking at the graph, we will see that this function has 3 x-intercepts, 1, 2, 3
Or (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6
2nd :But we know that this is the fifth-degree one, so it must have 2 more non-real zeros. So, we set up :
(x^3 - 6x^2 + 11x - 6)(factor) = x^5 - 6x^2 + 11x - 6
3rd :Using the polynomial quotient method to find the factor in the equation above, you will get factor x^2 + 9
Eventually, we get y = (x - 1)(x - 2)(x - 3)(x - 3i)(x + 3i)
The x-intercepts are also the zeros! ^^!

Now i got how to solve the higher degree polynomial function by Factor Theorem, but i hear that there is one more Theorem to find the answer for this problem is Rational Root Theorem. So, can some one tell me what is the differences between these two theorems ? And when should we use one of them ?Thank!
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Re: Finding solutions for 5th-degree polynomial functions  TOPIC_SOLVED

Postby stapel_eliz on Thu Apr 23, 2009 11:07 pm

To learn about the techniques for solving polynomials, including how to use the Rational Roots Test, try following the link provided in an earlier reply.

The lesson at that link will also demonstrate how much easier it is to divide out known factors, rather than guessing factors from the picture and then attempting to reconstruct the original factorization. :shock:

For instance, you'll learn how to divide out the x - 1, thus proving, when the remainder is zero, that x - 1 is actually a factor. Then, working from the now-simpler and lower-degree quotient, divide out the next factor. And so forth. Once you get down to a quadratic, you apply the Quadratic Formula.

Study the various examples in the lesson. They make the process fairly straightforward. :D
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