# Is simplification possible (how to know)?

• Sep 16th 2013, 12:10 PM
niaren
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.

$-\left(8y^3+16y^2+8y - 8y^2x + y^2x^2 - 8yx-2yx^2+2x^3y+x^2+2x^3-2x^4-2\sqrt{(16y^2+4yx^2+x^4)(y+2y-yx+1-x+x^2)^2}\right)$
• Sep 16th 2013, 03:10 PM
Shakarri
Re: Is simplification possible (how to know)?
Rewrite your function as $g(x,y)+2\sqrt{h(x,y)}$
Let $f_i$ be a function of x and y which is a multinomial (does not have square roots in it). (Note that $f_i$ might equal zero or some other constant)
If that function does factorise then it can be written as

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

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

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

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

and

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

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

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

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

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

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.
• Sep 16th 2013, 11:35 PM
niaren
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

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

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

I was hoping that I could do the following factorization
$(f_1-\sqrt{f_2})^2 = f_1^2+f_2-2\sqrt{f_2}f_1$
but that seems like not possible.