A Quintic Equation?

Question

Determine all possible solutions to the undernoted quintic equation:

x^5 - 64*(x^3) - 108x + 864 = 0.

Answer 1

Quintic equations are tricky... but possible.

First try all possible divisors (positive and negative) of 864 as possible solutions. You'll find out that 8 is a solution. As such, you can factor the polynomial as

(x - 8)(x^4 + 8 x^3 - 108) = 0

No other integer solutions exist. Then, try to split the 4-th degree polynomial into two quadratic polynomials. Some trial and error should bring you to

x^4 + 8 x^3 - 108 = (x^2 + 2 x + 6)(x^2 + 6 x - 18)

Now it's just a matter of finding the roots of each of these two quadratic polynomials.

The first (x^2 + 2 x + 6) has roots -1 (+/-) i*sqrt(5), and the second (x^2 + 6 x - 18) has roots -3 (+/3) 3*sqrt(3).

Please double-check for yourself.

Answer 2

Since the highest exponent is 5, there are 5 possible roots (or solutions).
First, decompose the equation into factors in such a way that each factor has no exponent higher than 2.
Secondly, resolve all factors, some of them as first degrees, others as quadratic equations.

Answer 3

634

Answer 4

=x^3(x^2-64)-108(x-8)=x^3(x-8)(x+8)-108(x-8)=(x-8)(x^3(x+8)-108)=
=(x-8)(x^4+8x^3-108)=(x-8)(x^4-x^3+9x^3-108)=(x-8)(x^3(x-4)+4(x^3-27))=...