How to factor $a^2+3b^2+4ab+2ac+6bc-4b+4c-4$? Can someone please explain in the simplest language they can manage strategies (best if one that applies to most polynomial factorization) to complete the formula above? And how in general to improve factorizing? I'm only fourteen with limited knowledge (duh) so I might not understand what you're saying but say it anyways cause then at least I'll know where to look/ get the gist of it. Thanks in advance!
Usually it amounts to noticing differences of squares and such consider this
$$a^2+4ab+3b^2=(a+2b)^2-b^2\\(a+2b)^2-b^2+2ac+6bc-4b+4c-4\\-b^2-4b-4=-(b+2)^2\\(a+2b)^2-(b+2)^2+2ac+6bc+4c\\2ac+6bc+4c=2c(a+3b+2)\\(a+2b)^2-(b+2)^2+2c(a+3b+2)=\\(a+2b+b+2)(a+2b-b-2)+2c(a+3b+2)=\\(a+3b+2)(a+b-2)+2c(a+3b+2)=\\(a+3b+2)(a+b-2+2c)$$
Though this amounts to trial and error,usually completing to square and noticing well known identities.
To factor it, I will use the &quot;Crossover Method&quot;, I don't know if it exists, but it's the way I call it.
Well, let's go!
\begin{align}
a^2+3b^2+4ab+2ac+6bc-4b+4c-4\\
= a^2+4ab+2ac+3b^2+6bc-4b+4c-4\\
= a^2+2(2b+c)a+(3b^2+6bc-4b+4c-4)\\
= a^2+2(2b+c)a+[3b^2+(6c-4)b+4(c-1)]\\
= a^2+2(2b+c)a+[3b^2+(6c-4)b+2.2(c-1)]\\
= a^2+2(2b+c)a+[b+2(c-1)] (3b+2)\\
= a^2+2(2b+c)a+(b+2c-2)(3b+2)\\
= [a+(b+2c-2)][a+(3b+2)]\\
= (a+b+2c-2)(a+3b+2)
\end{align}
So, that's it, I hope I have helped. Thank you!