Factoring $3x^2 - 10x + 5$ How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).

You can try to Conplete the Square in this kind of problem. What you would want to do, is to manipulate $3x^2-10x+5$ into something like this $a^2 \pm 2ab + b^2 = (a \pm b)^2 \quad \mathbf{(1)}$.
First, factor the 3 out, like this $3x^2-10x+5 = 3 \left( x^2 - \dfrac{10}{3}x + \dfrac{5}{3} \right)$, now, look closely inside the brackets, we'll now complete the square.
$x^2$ will stand for $a^2$ in (1), so basically, we'll have $\color{red}{a = x}$ (remember this); now move onto the second term, we'll find $b$, such that, $\dfrac{10}{3}x$ must stand for $2ab$. Since $a = x$, we will have:
$\begin{align}2xb &= \dfrac{10}{3}x \\ 
\Leftrightarrow 2b &= \dfrac{10}{3} \quad \mbox{cancel } x \mbox{ from both sides} \\
\Leftrightarrow b &= \dfrac{5}{3}\end{align}$
So, back to the original expression:
The red parts, blue parts, and orange parts, although their forms do change, their values remain the same as we move down.
$\begin{align} 3x^2-10x+5 &= 3 \left( \color{red}{x^2} - \color{blue}{\dfrac{10}{3}x} + \color{orange}{\dfrac{5}{3}} \right)\\
&= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\dfrac{5}{3}} \right) \quad \mbox{make the }a^2 + 2ab \mbox{ part appear}\\
&= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2 - \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right) \quad \begin{array}{l} \mbox{make the }b^2 \mbox{ part} \\ \mbox{appear, and in other to keep}\\\mbox{the orange part the same,}\\ \mbox{we'll add and subtract simultaneously}\end{array}\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right] \quad \mbox{re-grouping}\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{25}{9} + \dfrac{5}{3}} \right]\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{10}{9}} \right]\\
&= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \dfrac{10}{9}} \right] \quad \mbox{apply } \mathbf{(1)}\\
&= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \left(\dfrac{\sqrt{10}}{3} \right)^2} \right]\\
&= 3 \left[ \left( x - \dfrac{5}{3} - \dfrac{\sqrt{10}}{3} \right) \left( x - \dfrac{5}{3} + \dfrac{\sqrt{10}}{3} \right) \right] \quad \mbox{applying difference of squares formula}\\
&= 3 \left( x - \dfrac{5+\sqrt{10}}{3}\right) \left( x + \dfrac{\sqrt{10} - 5}{3}\right)\quad \mbox{simplify it a bit}
\end{align}$
