## Polynomial equation

### 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 $x>a,x^3>ax^2$)

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Nur Muhammad Shafiullah | Mahi

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

Nur Muhammad Shafiullah | Mahi

### Re: Polynomial equation

But you didn't say what is $x$?jagdish wrote:Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$

$x\in \mathbb{N}$ 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 $0$.

- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### Re: Polynomial equation

$(x^2+5x+4)(x^2+5x+6)=120x^2$

$\Longrightarrow (x^2+5x+5)^2-1=120x^2$

$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$

$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$

so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.

$\Longrightarrow (x^2+5x+5)^2-1=120x^2$

$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$

$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$

so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.