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- Thread starter mathland
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- Aug 30, 2012

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This is a bit high level for you, isn't it? There is no way to solve a quintic equation in general, though there are some tricks you can employ. Graphically we can see how many real solutions there are. (There are three.)Solve for x∈ℤ.

x^5-15x^3-x-60 = 0

How do I get started? I think the solution is not over the real numbers.

You say?

Have you covered the "rational root theorem?" There is a rational (actually integer) solution you can find, but unless you use Excel or something it will be a bit of a hassle to get. Once you get that you can use synthetic division to reduce it to a quartic equation. There

-Dan

- Mar 1, 2012

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1 real & 4 imaginary, or

3 real & 2 imaginary, or

all 5 real

rational root theorem shows there is one real root that is rational

- Jan 30, 2018

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And the set of all integers is a subset of the set of rational number so I would start by using the

"rational root theorem" suggested by TopSquark and Klaas Van Aarsen.

The rational root theorem says that any rational root of the polynomial equation $\alpha_nx^n+ \alpha_{n-1}x^{n- 1}+ \cdot\cdot\cdot+ \alpha_1 x+ \alpha_0= 0$ is of the form $\frac{a}{b}$ where a divides $\alpha_0$ and b divides $\alpha_n$.

Here $\alpha_0= 1$ so the denominator must be 1- any rational root must be an integer. $\alpha_n= 60$ so any integer solution must be a divisor of 60. Such numbers are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 10, -10, 12, -12, 15, -15, 30, -30, 60, and -60. Try those into the equation to see which, if any, of those actually satisfy the equation.

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- Jan 30, 2018

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I get x= 4 as the only integer solution.

I wrote a short program to evaluate the polynomial at the values of x above.

(I accidently left 5 and -5 from the list of possible roots but they are not actually roots.)

I wrote a short program to evaluate the polynomial at the values of x above.

(I accidently left 5 and -5 from the list of possible roots but they are not actually roots.)

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