# How do you find all the zeros of x^2 - 8x - 48 = 0?

The important thing is to be able to identify factors of $48$ that add / subtract to give $- 8$.
The important thing is to be able to identify factors of $48$ that add / subtract to give $- 8$.
$48$ could be $6 \cdot 8$ which doesn't work, or it could be $4 \cdot 12$ which does work - $4 - 12 = - 8$
Then ${x}^{2} - 8 x - 48 = \left(x - 12\right) \left(x + 4\right)$