Probably simple factoring problem I came across this in a friend's 12th grade math homework and couldn't solve it.  I want to factor the following trinomial:
$$3x^2  -8x + 1.$$  
How to solve this is far from immediately clear to me, but it is surely very easy.  How is it done?

There's an approach called (by some) the "$ac$ method". Suppose $\alpha,\beta,\gamma,\delta$ are some constants, and consider the expansion $$(\alpha x+\beta)(\gamma x+\delta)=\alpha\gamma x^2+(\alpha\delta+\beta\gamma)x+\beta\delta=ax^2+bx+c.$$ Note that $\alpha\delta$ and $\beta\gamma$ are factors of $ac=\alpha\beta\gamma\delta$ and that $b=\alpha\delta+\beta\gamma$. In fact, they are paired factors of $ac$, in that $\alpha\delta\cdot\beta\gamma=ac$. The idea of the $ac$ method of factoring is to find such paired factors of $ac$ whose sum is $b$.
To see how this can be useful, consider the example $3x^2+13x-10$. Here, $ac=-30$ and $b=13$. We need to find two numbers that add up to $13$ and multiply to get $-30$. The only pair that works is $15$ and $-2$. Thus, we can rewrite $13$ as $15-2$ or $-2+15$. How is this useful? Well, $$3x^2+13x-10=3x^2+15x-2x-10=3x(x+5)-2(x+5)=(3x-2)(x+5)$$ and $$3x^2+13x-10=3x^2-2x+15x-10=x(3x-2)+5(3x-2)=(x+5)(3x-2),$$ so either way, we obtain our factorization without too much difficulty.
In this case, unfortunately, we have $ac=3$ and $b=-8$, so there is no obvious pair of factors! Such a pair does exist, namely $-4+\sqrt{13}$ and $-4-\sqrt{13}$, but even knowing the pair may not make the factorization simple in all cases (in this case it is, because $c=1$).
In general, if trying to factor something intractible to the $ac$ method, I recommend completing the square, as Marvis has demonstrated so well.
