factor completely
16x2-2x-3
Multiply the coefficient of the leading term by the constant term:
16 x -3 = -48
Now find two factors of -48 that add to the coefficient of the linear term: -2
Factors......Sum
1, -48........-47
2, -24........-22
3, -16........-13
etc.
I get that 6 x -8 = -48 and 6 + -8 = -2
So:
$\displaystyle 16x^2-2x-3 = 16x^2 + 6x - 8x - 3$
$\displaystyle = (16x^2 + 6x) - (8x + 3)$
$\displaystyle = 2x \cdot (8x + 3) - 1 \cdot (8x + 3)$
$\displaystyle = (2x - 1)(8x + 3)$
If such a pair of factors cannot be found, the quadratic does not factor over the rational numbers.
-Dan