SOLUTION: Can you explain and factorise 5(3x+2y)-9kx-6ky, 2p(x-5y)-(x-5y), (x+1)(x-2)+(2-x)(3-x), xy+3y-2x-6, 12ah-2bh-30ak+5bk

5(3x+2y)-9kx-6ky, 

Factorise last two terms: -9kx-6ky as -3k(3x+2y)

5(3x+2y)-3k(3x+2y)

Let (3x+2y) = U

5U-3kU

U(5-3k)

Replace U by (3x+2y)

(3x+2y)(5-3k)

--------------------------

2p(x-5y)-(x-5y)

Let U = (x-5y)

2pU-U

U(2p-1)

Replace U by (x-5y)

(x-5y)(2p-1)

--------------------------

(x+1)(x-2)+(2-x)(3-x)

2-x is not in standard order
Write it as -x+2
Take out common factor -1
-1(x-2)

3-x is not in standard order
Write it as -x+3
Take out common factor -1
-1(x-3)

(x+1)(x-2)+[(-1)(x-2)(-1)(x-3)]

(x+1)(x-2)+[(-1)(-1)(x-2)(x-3)]

(x+1)(x-2)+[(+1)(x-2)(x-3)]

(x+1)(x-2)+(x-2)(x-3)

Let U = (x+1)
Let V = (x-2)
Let W = (x-3)

UV + VW

V(U+W)

Use brackets to hold parentheses:

V[U+W]

Replace U, V and W by what we let 
them equal to:

(x-2)[(x+1)+(x-3)]

(x-2)[x+1+x-3]

(x-2)[2x-2]

(x-2)[2(x-1)]

2(x-2)(x-1)

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 xy+3y-2x-6

Factorize first two terms: xy+3y as y(x+3)
Factorize last two terms: -2x-6 as -2(x+3)

y(x+3)-2(x+3)

Let U = (x+3)

yU-2U

U(y-2)

replace U by (x+3)

(x+3)(y-2)

===================

12ah-2bh-30ak+5bk

Factorise first two terms: 12ah-2bh as 2h(6a-b)
Factorise last two terms: -30ak+5bk as -5k(6a-b)

2h(6a-b)-5k(6a-b)

Let U = (6a-b)

2hU-5kU

U(2h-5k)

Replace U by (6a-b)

(6a-b)(2h-5k)

Edwin