## Polynomial equation

For discussing Olympiad Level Algebra (and Inequality) problems
jagdish
Posts: 38
Joined: Wed Jan 19, 2011 2:21 pm
Location: India (Himanchal Pradesh)

### Polynomial equation

Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$
jagdish

nfstiham
Posts: 10
Joined: Wed Mar 09, 2011 1:23 pm

### Re: Polynomial equation

probably 1

jagdish
Posts: 38
Joined: Wed Jan 19, 2011 2:21 pm
Location: India (Himanchal Pradesh)

### Re: Polynomial equation

Now How can I Calculate other Roots.
jagdish

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

### Re: Polynomial equation

I think with $x$ excessing a limited value,$(x+1)(x+2)(x+3)(x+4)>120x^2$ so we will find the limit of $x$ and then finding the $x$ 's is not at all hard.(for the first part ,you have to expand the L.H.S. and do some argument like, for $x>a,x^3>ax^2$)
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: Polynomial equation

jagdish wrote:Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$
But you didn't say what is $x$?
$x\in \mathbb{N}$ or else?Since posted in Algebra fora,$x$ can be anything.
Whatever,you may limit the values of $x$ using inequalities as the previous post says.
One one thing is neutral in the universe, that is $0$.

$(x^2+5x+4)(x^2+5x+6)=120x^2$
$\Longrightarrow (x^2+5x+5)^2-1=120x^2$
$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$
$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$
so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.