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?
 A: Hint: Use the quadratic formula.
A: We have the following theorem (Viéte formulae) for the quadratic case:
If $\,\alpha\,,\,\beta\,$ are roots (i.e., solutions) of the quadratic equation $\,ax^2+bx+c=0\,$ , then we have that$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
This is the reason why Eugene hinted you at using the quadratic formula
A: A standard way of factorizing, when it is hard to guess the factors, is by completing the square.
\begin{align}
3x^2 - 8x + 1 & = 3 \left(x^2 - \dfrac83x + \dfrac13 \right)\\
& (\text{Pull out the coefficient of $x^2$})\\
& = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \dfrac13 \right)\\
& (\text{Multiply and divide by $2$ the coefficient of $x$})\\
& = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \left(\dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\
& (\text{Add and subtract the square of half the coefficient of $x$})\\
& = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\
& (\text{Complete the square})\\
& = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac13 \right)\\
& = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac39 \right)\\
& = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{13}9\right)\\
& = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac{\sqrt{13}}3 \right)^2\right)\\
& = 3 \left(x - \dfrac43 + \dfrac{\sqrt{13}}3\right) \left(x - \dfrac43 - \dfrac{\sqrt{13}}3\right)\\
& (\text{Use $a^2 - b^2 = (a+b)(a-b)$ to factorize})
\end{align}
The same idea works in general.
\begin{align}
ax^2 + bx + c & = a \left( x^2 + \dfrac{b}ax + \dfrac{c}a\right)\\
& = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \dfrac{c}a\right)\\
& = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \left( \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\
& = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\
& = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} + \dfrac{c}a\right)\\
& = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{b^2-4ac}{4a^2} \right)\right)\\
& = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{\sqrt{b^2-4ac}}{2a} \right)^2\right)\\
& = a \left( x + \dfrac{b}{2a} + \dfrac{\sqrt{b^2-4ac}}{2a}\right) \left( x + \dfrac{b}{2a} - \dfrac{\sqrt{b^2-4ac}}{2a}\right)\\
\end{align}
A: 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.
