Fifth Degree Equation

[mathland]

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?

 

[topsquark]

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?

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.)

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 is a way to solve these (see the section "solution") but it's pretty hairy. Unless you need to find all of the roots I'd leave it with the integer solution.

-Dan

 

[skeeter]

odd degree says the equation has to at least one real root. could be ...

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

 

[Country Boy]

The problem specifically says that you are to solve for x in Z, the set of integers, so, yes, the solution is not over the real numbers!

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.

 

[mathland]

Thanks everyone.

 

[Country Boy]

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.)