How would you factor the following by grouping?

Question

1. mx+my-nx-ny
2. 5cy-10c-2y+4
3. 3ab-3ac-5b+5c
4. abm-abn-3m+3n

Answer 1

1) mx+my-nx-ny
= (mx+my) + (-nx-ny)
= m(x+y) + (-n)(x+y)
= (x+y)(m-n)

2) 5cy-10c-2y+4
= (5cy-10c) + (-2y+4)
= 5c(y-2) + (-2)(y-2)
= (y-2)(5c-2)

Can you catch on now?

Answer 2

When you distribute multiplication throughout an expression, you multiply every term by a factor. 2(a+b) = 2a+2b.
Factoring is just doing the reverse of this.
In each of your examples, the patterns are pretty clear because there are two pairs of terms that can be factored.
In the first: mx+my-nx-ny, there are two pairs of terms that can be factored - they are mx and my, the first pair, and -nx-ny, the second pair. Without breaking up your expression or changing the operations, you just rewrite it as a pair of factored expressions that make up your original expression.
mx+my becomes m(x+y). If you distributed the multiplcation in that expression, you would get your mx+my back.
The same goes for -n(x+y) which is the second half, a factored expression. if you distributed the multiplication, it would give you your orignial back: -nx-ny.
So you have m(x+y) -n(x+y) when you have factored into two groups (so it's called factored by grouping).
x+y is a common factor here. You have two terms, m and -n with the same factor. So combine them and they become one factor (m-n) and (x+y) is the other factor. This is because the m and -n are distributed to the (x+y) so all you are doing is bringing them back together. If you understand that, then you understand that there is just one (x+y) to be used as a factor.

Answer 3

1) mx+my-nx-ny
= (mx+my) + (-nx-ny)
= m(x+y) + (-n)(x+y)
= (x+y)(m-n)

2) 5cy-10c-2y+4
= (5cy-10c) + (-2y+4)
= 5c(y-2) + (-2)(y-2)
= (y-2)(5c-2)

3) 3ab - 3ac - 5b + 5c
= (3ab - 3ac) + (-5b + 5c)
= 3a(b-c) - 5(b-c)
= (3a - 5)(b - c)

4) abm - abn - 3m + 3n
= (abm - abn) + (-3m + 3n)
= ab(m-n) - 3(m-n)
= (ab - 3)(m-n)