I'm trying to factor
$$ k^2 + 2k k_2 - 2k m\omega^2-2k_2 m\omega^2 + (m\omega^2)^2 $$
that I encountered in a coupled harmonic oscillators chapter of a book.
The answer is
$$ (k-m\omega^2)(k+2k_2 -m\omega^2) $$
My progress is only up to the following:
$$ (k^2 + 2k k_2) + (- 2k m\omega^2-2k_2 m\omega^2 + (m\omega^2)^2) \\ k(k + 2 k_2) -m\omega^2 (2k +2k_2 -m\omega^2) $$
Can someone point me to the proper factoring technique for this instance? I can't identify any of https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-strategy/a/factoring-quadratics-in-any-form as what i need :(
Cheers
Thing with factoring is that you have to 'see' which things can be factored out. Something that usually comes with a little experience. But lets begin with some basic factoring cases: $$a^2 + 2ab+b^2 = (a+b)^2$$ $$a^2 - 2ab+b^2 = (a-b)^2$$ Not that I am trying to explain you basic math as I have no clue where you stand in your mathematical experience (heck, you might be more experiences that I am). Anyway, if you can spot such a set of terms in your equation, you can factor those out: $$\left[k^2-2km\omega^2+\left(m\omega^2\right)^2\right] + 2kk_2 -2k_2m\omega^2$$ $$=\left(k-m\omega^2\right)^2 + 2kk_2 -2k_2m\omega^2$$ in the remaining terms, it can be seen $2k_2$ is multiplied to both terms, so factoring that out yields: $$=\left(k-m\omega^2\right)^2 + 2k_2\left(k -m\omega^2\right)$$ Maybe you already see where we are going to, but just to point out the last step. Both remaining terms are multiplied with $\left(k -m\omega^2\right)$, so factoring that out results in: $$=\left(k-m\omega^2\right)\left(k-m\omega^2\right) + 2k_2\left(k -m\omega^2\right)$$ $$=\left(k-m\omega^2\right)\left(\left(k-m\omega^2\right) + 2k_2\right)$$ $$=\left(k-m\omega^2\right)\left(k-m\omega^2 + 2k_2\right)$$
So a couple standard techniques have been used to derive the equation, only they didnt apply to the entire equation initially.