Can I factor this any other way?

[moriman]

Hi,

I can't seem to find a common factor in this:

\(\displaystyle p=u^4 - 4qu^3 + 4qvu^2 - 4u^2q^2 + 2u^2v^2 - 4q^2v^2 + 8uvq^2 - 4quv^2 + 4qv^3 + v^4\)

The best I can get it down to is

\(\displaystyle 4(u-v)[(v-u) q^2-(u^2+v^2 )q]+(u^2+v^2 )^2\)

Can anyone see a tidier way of factoring this beast?

thank you

 

[tkhunny]

It might be a lot prettier in some Polar Form. Try \(\displaystyle u = r\cdot\cos(\theta)\) and \(\displaystyle v = r\cdot\sin(\theta)\) and see where it goes.

 

[moriman]

Thanks tkhunny. You wouldn't be able to give me a kick-start?
I haven't enough fingers and toes to count the number of years since I tried to convert to polar :?
It's that long, I'd even forgot about using the polar system, lol

mori

 

[tkhunny]

Hint: u^2 + v^2 = 1

That's pretty convenient.

 

[moriman]

Hint: u^2 + v^2 = 1

That's pretty convenient.

Sweet

The mists are clearing

You're a hun TK

Thank you

 

[moriman]

Would this seem like a fair result then ¿

\(\displaystyle 1-4q(u-v+q-2uvq)\)

I haven't shown the working because I'm not sure how to get Theta in TEX

Will show working for help if I'm incorrect

Many thanks

 

[moriman]

Found Theta and I think I was wrong.

Can I convert back to cartesian from polar?

I can't think of why not, but it doesn't appear to be correct.

Before I came to the forum I was working with this diophantine equation..

\(\displaystyle \dfrac{u^2+v^2}{v-u}=\dfrac{p^2+q^2}{q-p}\)

p, q, u and v > 0 and non-equal

using figures I know work

\(\displaystyle \dfrac{100^2+140^2}{140-100}=740\ \ \ \ and\ \ \ \ \dfrac{96^2+132^2}{132-96}=740\)

So, after tkhunny's kick-start, I thought of working on the above equation like this...

let \(\displaystyle u = r\cdot\cos(\theta)\) and \(\displaystyle v = r\cdot\sin(\theta)\)

so \(\displaystyle \dfrac{(r\cdot\cos(\theta))^2+(r\cdot\sin(\theta))^2}{r\cdot\sin(\theta)-r\cdot\cos(\theta)}=\dfrac{p^2+q^2}{q-p}\)

then \(\displaystyle \dfrac{1}{r\cdot\sin(\theta)-r\cdot\cos(\theta)}=\dfrac{p^2+q^2}{q-p}\)

but then I thought I could convert the remaining \(\displaystyle r\cdot\cos(\theta)\) and \(\displaystyle r\cdot\sin(\theta)\) back to cartesian, but

\(\displaystyle \dfrac{1}{v-u}=\dfrac{p^2+q^2}{q-p}\) is clearly incorrect

So on my OP, would the farthest I could go be

\(\displaystyle 1-4q(q-2r\cdot\sin(\theta)*r\cdot\cos(\theta)q-r\cdot\sin(\theta)+r\cdot\cos(\theta))\)

Thanks again for any advice

mori