How do you factor #35ab + 56a + 10kb +16k#?

1 Answer
Oct 22, 2015

Answer:

#35ab+56a+10kb+16k=(7a+2k)(5b+8)#

Explanation:

First we note that we can take out $a$ and $k$ at the left and right terms of the polynomial respectively
#35ab+56a+10kb+16k=a(35b+56)+k(10b+16)#
Now we look at common divisors of 10 and 35, these are 1 and 5. We also look at common devisors of 56 and 16, these are 2, 4, and 8. Now we make the observation that 35=75, 56=78, 10=25, 16=28. This means we can do the following:
#a(35b+56)+k(10b+16)=7a(5b+8)+2k(5b+8)=(7a+2k)(5b+8)#