Question

3x^2-x-4, factor completely. <- HELP

Answer 1

Please help.

Answer 2

ok... here is a method that works every time for a quadratic

\[ax^2 + bx + c\]

in your question a = 3 and c = -4

what is 3 x -4 = ?

the find the factors that add to b, -1

Answer 3

So I would have to multiply 3 time 4 which gets me 12. The factors would be 1, 2, 6, and 12. None that I see add to -1. :\ @campbell_st

Answer 4

Can you tell me the factors of -4?

Answer 5

well 3 * -4 = -12

so find the factors that add to -1, the larger factor is negative...

Answer 6

then its simply a case of

\[\frac{(ax + factor~1)(ax + factor~2)}{a}\]

Answer 7

and you missed 2 factors of 12... 3 and 4

so if you use -4 and 3 they add to -1 and multiply to -12

so you'll get

\[\frac{(3x + 3)(3x -4)}{3}\]

can you see a common factor in the 1st binomial that cancels with the denominator?

Answer 8

just remove the common factor for the answer

Answer 9

Try to set up like this:
(3x )(x )
Because of the -4, one factor has negative sign and other must have positive sign.
So it can be like this
(3x-#)(x+#) or (3x+#)(x-#)