Question

Solve the equation:\[(x+1)(x+2)(x+3)(x+4)(x+5)=(x-5)(x+6)(x+7)(x+8)(x+9)\]

Answer 1

Okay, did you try just expanding it and then finding roots?

Answer 2

\[
{(x+5)!\over x!}=(x-5){(x+9)!\over(x+5)!}
\]

Answer 3

And then?

Answer 4

is that an (x-5) on the right?

Answer 5

@wio lol, gears are rolling

Answer 6

Yeah, I got no idea.
I mean I know a \(5\)th degree polynomial is a dragon, but a factorial equation is even more of a dragon for me.

Answer 7

\[
(x+5)!^2-x(x-5)(x+9)!=0\\
(x+5)!\left[(x+5)!-x(x-5){(x+9)!\over(x+5)!}\right]=0\\
\text{1)}\;(x+5)!=0\implies \boxed{x=1,2,3,4,5}
\]
if (x+5)!=0, then the second paranthesis cannot exist!
so,

Answer 8

How can \(y!=0\) for any \(y\)?

Answer 9

@wio so, the polynomial cannot be expressed as a factorial?

Answer 10

http://www.wolframalpha.com/input/?i=(x%2B1)(x%2B2)(x%2B3)(x%2B4)(x%2B5)%3D(x-5)(x%2B6)(x%2B7)(x%2B8)(x%2B9)

If you just cheat it, it turns out to be a \(4\)th degree polynomial with 2 real roots and 2 imaginary roots.

Answer 11

@wio lol.. was doing that when you entered this post .. :D

Answer 12

it was obvious that the x^5 term would cancell off and leave us with a fourth degree polynomial.

Answer 13

the only other way seems to be to expand it and use the "Rational roots" methos

Answer 14

It doesn't actually have rational roots. The only way to do this is with something like newton's method or bisection, or some other method of approximating roots.

Answer 15

it seems to have "2" real and 2 imaginary.,

Answer 16

i guess numerical approach is the only otherway.