Please factor 6x^2-28x-48?

First look for the GCF (greatest common factor). Since all the terms are even, the GCF is 2. Now you have the following:

2(3x^2 - 14x - 24)

Forget about the 2 outside the parentheses for a while. You need two numbers that, when added, give you -14 (the middle number) and when multiplied give you -72 (the product of the first and last numbers).

Several pairs of numbers multiply to be 72 (we'll deal with the negative sign momentarily): 1 & 72, 2 & 36, 3 & 24, 4 & 18, 6 & 12, and 8 & 9. Since 4 and 18 differ by 14, that is the correct choice. But, one term must be negative so the product is -72. It's either -4 & 18 or 4 & -18. Since the sum must be -14, the two numbers are 4 & -18.

Now rewrite the middle term as + 4x - 18x. You now have the following:

3x^2 + 4x - 18x - 24

The first two terms have an x in common, and the last two terms have a 6 in common. Factor these out:

x(3x + 4) - 6(3x + 4)

Notice that I factored a -6 out of the last two terms. That made the 4 change to a positive.

Since both factors have a (3x + 4), it can be factored out:

(3x + 4)(x - 6)

But don't forget about the 2!

2(3x + 4)(x - 6)

Hope this helps!

1: 6(x-3/4)(x-4) 2: (x+2)(x^3-3) I am not sure if you can go further with number two without getting really ugly numbers...

6x^2 - 28x - 48
= 2(3x^2 - 14x - 24)
= 2(3x^2 + 4x - 18x - 24)
= 2[(3x^2 + 4x) - (18x + 24)]
= 2[x(3x + 4) - 6(3x + 4)]
= 2(3x + 4)(x - 6)

2 [ 3x^2 - 14 x - 24 ]

2 ( 3 x + 4 ) ( x - 6 )

=6(x-6)(x+4/3)=2(x-6)(3x+4)

6x^2-28x-48
=2(3x^2-14x-24)
=2(x-6) (3 x+4)