Im friggin stumped plz help!!! how do I factor 14x^2 - 49x -28???

Question

How do I factor this by grouping??? theres too many factors for -392 I dont know which ones to use. uhhgg

Answer 1

Divide it all by 7 first to get it down to size.

7(2x^2-7x-4)
7(2x+1)(x-4)

Answer 2

You have a negative term at the end, so your factors are going to have one plus and one minus. They're going to take the form of:
(ax + b)(cx - d)

We need to have ac = 14, bc - ad = -49, and bd = 28. The integer factors of 14 are (1,14) and (2,7). The integer factors of 28 are (1,28), (2,14), and (4,7). We need to find a combination of these that adds up to -49. Start with (1,14) and go through the others:
1*1 - 14*28 doesn't give is -49 or 49. Neither does 1*14 - 1*28.
1*2 - 14*14 doesn't work, and neither does 1*14 - 14*2.
1*4 - 14*7 doesn't work. But notice that 1*7 - 14*4 = -49. So we've found our factors:
(14x + 7)(x - 4)

Notice that you can factor a 7 out of the first term and put it into the second term:
7(2x + 1)(x - 4)
(2x + 1)(7x - 28)

So this gives us another pair of factors. If we started off checking (2,7) for the factors of 14, we might have gotten this as a result instead.

Answer 3

I find completing the square a good fallback
= 14( x^2 - 7x/2 -2)
= 14[( x - 7/4)^2 -2-49/16 ]
= 14[(x-7/4)^2 -81/16 ]
= 14( x-7/4+9/4)(x-7/4-9/4)
= 14( x+1/2)(x-4)
= 7(2x+1)(x-4)

so I agree with bookworm who spotted the groupings better.

Answer 4

Look at the coefficients:
all are divisible by 7:
7(2x^2 - 7x - 4) = 7(2x+1)(x-4)
Do not give up; just keep trying and it will become obvious.

Answer 5

just factor out a 7. 7(2x^2-7x-4). 7(2x+1)(x-4). if that is equal to zero then x=-1/2, 4.

Answer 6

Hear's what I got:
(2x+1)(7x-28)

Good Luck!

Answer 7

14x^2-49x-28
7(2x^2-7x-4)
7(2x+1)(x-4)