factor completely

16x2-2x-3

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- Jan 13th 2007, 12:31 PMfancyfacemath help
factor completely

16x2-2x-3 - Jan 13th 2007, 02:18 PMThePerfectHacker
- Jan 13th 2007, 04:04 PMtopsquark
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