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).

$3x^2 - 10x + 5$ 
Testing the polynomial using the discriminant (the $\;\sqrt{b^2 - 4ac}\;$ portion of the quadratic formula,), 
with $$b = -10, \,a=3,\,c = 5$$ gives you $$\sqrt{100 - 4\cdot 3 \cdot 5} = \sqrt{100 - 60} = \sqrt{40} = \sqrt{4\cdot 10} = \sqrt{4}\cdot \sqrt{10} = 2\sqrt{10},$$ which tells you that there aren't any rational, let alone integer, roots. 
So we need to find the roots using the entire quadratic formula: For $ax^2 + bx + c$, we get two roots: $$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a},\;\; x_2 =\frac{-b - \sqrt{b^2 - 4ac}}{2a}$$ and we already know that $\sqrt{b^2 - 4ac} = 2\sqrt{10}$ for our polynomial:
$$x=\dfrac {10 \pm 2\sqrt{10}}{6} =\dfrac 16 (2\cdot 5 \pm 2\sqrt{10})= 2\cdot\frac 16(5 \pm \sqrt{10}) = \frac13 (5 \pm \sqrt{10})$$ 
In general, $ax^2+bx+c=a(x-x_1)(x-x_2)$:
So, to express your polynomial in factored form you need $$3x^2 - 10x + 5\quad = \quad 3\left(x-\frac 13(5 + \sqrt{10})\right)\left(x-\frac 13(5 - \sqrt{10})\right)$$
