# 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!

• Do you know that this can actually be factored? – Michael Burr Apr 29 '17 at 14:16

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.

Let's begin by looking at the given polynomial: $$a^2+3b^2+4ab+2ac+6bc-4b+4c-4.$$ This polynomial has only quadratic terms (the highest powers are products of two variables), so we expect (hope) that it could factor into two linear terms: $$(\lambda_1a+\lambda_2b+\lambda_3c+\lambda_4)(\mu_1a+\mu_2b+\mu_3c+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Observe that the RHS doesn't have a $c^2$ term. Since the $c^2$ term on the LHS has a coefficient of $\lambda_3\mu_3$, one of these two is zero, so let's choose $\mu_3=0$. This makes the equation $$(\lambda_1a+\lambda_2b+\lambda_3c+\lambda_4)(\mu_1a+\mu_2b+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Next, observe that the coefficient of $a^2$ on the RHS is $1$ and the coefficient of $a^2$ on the LHS is $\lambda_1\mu_1$. Since this product must be $1$, let's assume that both $\lambda_1=1$ and $\mu_1=1$. Now, we have the equation $$(a+\lambda_2b+\lambda_3c+\lambda_4)(a+\mu_2b+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Now, let's look at $ac$. The coefficient of $ac$ is $2$ on the RHS, and it only appears with coefficient $\lambda_3$ on the LHS. Therefore, $\lambda_3=2$. This gives the equation $$(a+\lambda_2b+2c+\lambda_4)(a+\mu_2b+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Next, we look at $bc$. On the RHS, the coefficient is $6$, while on the LHS, the coefficient is $2\mu_2$. Therefore, $\mu_2=3$ and we have $$(a+\lambda_2b+2c+\lambda_4)(a+3b+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

We now consider $b^2$. The coefficient of this is $3$ on the RHS, while the coefficient on the LHS is $3\lambda_2$. Hence $\lambda_2=1$ and we have $$(a+b+2c+\lambda_4)(a+3b+\mu_4)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Now, look at the $c$ term, on the RHS its coefficient is $4$, while on the LHS, its coefficient is $2\mu_4$. Hence, $\mu_4=2$. Now we have $$(a+b+2c+\lambda_4)(a+3b+2)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Finally, by looking at the constant term, we see that the constant term on the RHS is $-4$ while the constant term on the LHS is $2\lambda_4$ so $\lambda_4=-2$. Now, we have the complete factorization $$(a+b+2c-2)(a+3b+2)=a^2+3b^2+4ab+2ac+6bc-4b+4c-4$$

Maybe the most basic factorization strategy is to write down the "form" of the factorization, multiply it out and see what that says about the coefficients in the form. In your example, you expect the factorization to look something like

$$(xa+yb+zc+w)(ra+sb+tc+u).$$

You can multiply that out, but it's better to think a bit first. There's no $c^2$ in the polynomial, so you can take $z=0$. The coefficient on $a^2$ is $1$, so probably $x=r=1$. The coefficient on $b^2$ is $3$, so one of $y$ and $s$ is $3$ and the other is $1$. So we're looking at something like

$$(a+3b+w)(a+b+tc+u)$$

or

$$(a+b+w)(a+3b+tc+u)$$

We have in the first case, $3tcb = 6cb$ and in the second case $tcb=6cb.$ $t=6$ isn't going to work, so probably the first case is the right one, with $t=2.$ Using that $wu=-4,$ there are not many possibilities to try. In fact it's not hard to deduce that $w=2$ and you're done.