Thread: Is simplification possible (how to know)?

1. Is simplification possible (how to know)?

Is it possible to simplify this expression? I'm hoping that the expression can be written into some quadratic form but I have no idea if that is possible.

$\displaystyle -\left(8y^3+16y^2+8y - 8y^2x + y^2x^2 - 8yx-2yx^2+2x^3y+x^2+2x^3-2x^4-2\sqrt{(16y^2+4yx^2+x^4)(y+2y-yx+1-x+x^2)^2}\right)$

2. Re: Is simplification possible (how to know)?

Rewrite your function as $\displaystyle g(x,y)+2\sqrt{h(x,y)}$
Let $\displaystyle f_i$ be a function of x and y which is a multinomial (does not have square roots in it). (Note that $\displaystyle f_i$ might equal zero or some other constant)
If that function does factorise then it can be written as

$\displaystyle (f_1+f_2\sqrt{f_3})(f_4+f_5\sqrt{f_3})$

$\displaystyle (f_1+f_2\sqrt{f_3})(f_4+f_5\sqrt{f_3})=f_1f_4+f_2f _3f_5+(f_2+f_5)\sqrt{f_3}$

$\displaystyle f_1f_4+f_2f_3f_5+(f_2+f_5)\sqrt{f_3}=g(x,y)+2\sqrt {h(x,y)}$

From this you can determine that
$\displaystyle f_1f_4+f_2f_3f_5=g(x,y)$

and

$\displaystyle (f_2+f_5)\sqrt{f_3}=2\sqrt{h(x,y)}$

From: $\displaystyle (f_2+f_5)\sqrt{f_3}=2\sqrt{h(x,y)}$
You can determine that
$\displaystyle f_2+f_5=2$ and $\displaystyle f_3=h(x,y)$

h(x,y) is simple enough that you can tell it cannot be factored any more (Well actually $\displaystyle (16y^2+4yx^2+x^4)$ can be factorised but it involves imaginary numbers so ignore that)

Since $\displaystyle f_1f_4+f_2f_3f_5=g(x,y)$
You can rewrite that as $\displaystyle \frac{f_1f_4}{f_3}+f_2f_5=\frac{g(x,y)}{f_3}$

You know g(x,y) and $\displaystyle f_3$ so you can do the long division. When you do the long division you will get a remainder. The remainder is equal to $\displaystyle f_1f_4$

That's as much progress as I can make towards getting the factors. Maybe you can work out the next step. This is very involved for pre-university maths.

3. Re: Is simplification possible (how to know)?

First of all, thanks for your very helpful feedback. It gives me quite some things work with.

I missed a square in the expression, it should read

$\displaystyle -\left(8y^3+16y^2+8y - 8y^2x + y^2x^2 - 8yx-2yx^2+2x^3y+x^2+2x^3-2x^4-2\sqrt{(16y^2+4yx^2+x^4)(y^2+2y-yx+1-x+x^2)^2}\right)$

$\displaystyle (f_1+f_2\sqrt{f_3})(f_4+f_5\sqrt{f_3})=f_1f_4+f_2f _3f_5+(f_2+f_5)\sqrt{f_3}$
$\displaystyle (f_1+f_2\sqrt{f_3})(f_4+f_5\sqrt{f_3})=f_1f_4+f_2f _3f_5+(f_2f_4+f_1f_5)\sqrt{f_3}$
$\displaystyle (f_1-\sqrt{f_2})^2 = f_1^2+f_2-2\sqrt{f_2}f_1$