Question

What are the factors of


`{12c}^(2) + cd - {6d}^(2)`
F. (4c + 3d)(30 - 2d)
G. (40 - 3d)(3c + 2d)
H. (6c + d)(20 – 6d)
J. 6(2c – d)( c + d)
K. 6(2c + d)( C - d)​

Answer 1

For this case we have the following expression:

`12c ^ 2 + cd-6d ^ 2`

Rewriting the expression we have:

`12c ^ 2 + 9cd-8cd-6d ^ 2 =`

Grouping:

`(12c ^ 2 + 9cd) + (- 8cd-6d ^ 2) =`

We draw common factor from each group:

`3c (4c + 3d) -2d (4c + 3d) =\\(3c-2d) (4c + 3d)`

Answer:

`(3c-2d)(4c+3d)`