math help

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• Jan 13th 2007, 12:31 PM
fancyface
math help
factor completely

16x2-2x-3
• Jan 13th 2007, 02:18 PM
ThePerfectHacker
Quote:

Originally Posted by fancyface
factor completely

16x2-2x-3

$(8x + 3)(2x - 1)$
I cannot really give a good explanation to how I factored. It was more of a guess.
I just saw that 3 was prime and only has 2 factors: 3 and 1.
Thus, they must appear in the end of the factor.
• Jan 13th 2007, 04:04 PM
topsquark
Quote:

Originally Posted by fancyface
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:
$16x^2-2x-3 = 16x^2 + 6x - 8x - 3$

$= (16x^2 + 6x) - (8x + 3)$

$= 2x \cdot (8x + 3) - 1 \cdot (8x + 3)$

$= (2x - 1)(8x + 3)$

If such a pair of factors cannot be found, the quadratic does not factor over the rational numbers.

-Dan