How do I factor completely the following polynomial; 396b^2-539c^2?

Question

amistre64 · Answer

with any luck its a difference of 2 squares

amistre64 · Answer

$$a^2-b^2 = (a+b)(a-b)$$

whpalmer4 · Answer

It is, but there's a constant factor to take out, too.

whpalmer4 · Answer

Here's a hint: 360 * 11 = 396

amistre64 · Answer

wheres my slide rule when i need it :)

whpalmer4 · Answer

packed away in a box somewhere, just like mine!

whpalmer4 · Answer

Uh, let's make that hint a little better: 36*11 = 396

whpalmer4 · Answer

That means we need to factor $$11(36b^2-49c^2)$$which hopefully you recognize as a difference of two squares.  Using @amistre64's formula, what are the factors?

whpalmer4 · Answer

To avoid any letter confusion, I'll rewrite the formula as
$$u^2-v^2 = (u+v)(u-v)$$$$36b^2-49c^2$$Comparing the formulas, we see that$$u^2=36b^2$$$$v^2=49c^2$$Can you solve those for $u$ and $v$?

anonymous · Answer

Hey whpalmer, although ur answer is correct, is there a different form to enter 
11(36b^2-49c^2)?

whpalmer4 · Answer

we haven't factored it all the way yet...

anonymous · Answer

oh got it....

whpalmer4 · Answer

so, if you figure out $u$ and $v$ in those little equations, the final answer will be
$$11(u-v)(u+v)$$

anonymous · Answer

Thanks for ur assistance whpalmer, I appreciate it.