Factorise 56ad + 16bd +21ac + 6bc?

Hi Adrianna,

First, we will try to find two pair of two terms which do not overlap each other.
Notice that c and d do not overlap each other, as c is only found in 21ac and 6bc, whereas d is only found in 56ad and 16bd.
We will take out c and d from both pairs of terms.

56ad + 16bd +21ac + 6bc
= d(56a + 16b) + c(21a + 6b)

Notice that there are some common integers between 56a and 16b, as well as 21a and 6b. We will take them out as well.

d(56a + 16b) + c(21a + 6b)
= 8d(7a + 2b) + 3c(7a + 2b)

Now, (7a + 2b) is present on both sides, so we can factorise the entire thing directly to get

8d(7a + 2b) + 3c(7a + 2b)
= (8a + 3c)(7a + 2b)

Cheers.

56ad + 16bd +21ac + 6bc
=8d(7a+2b)+3c(7a+2b)
=(7a+2b)(8d+3c) answer//

56ad + 16bd +21ac + 6bc = 8d(7a + 2b) 3c(7a + 2b) = (8d + 3c) (7a + 2b)

answer >>> (8d + 3c) (7a + 2b)

56ad + 16bd + 21ac + 6bc
= 8d(7a + 2b) + 3c(7a + 2b)
= (7a + 2b)(8d + 3c)