Factorise completely

1. Factor out the common factor from both terms.

$2p^4-32s^4=2(p^4-16s^4)$

2. Recognise the remaining expression as a difference of squares.

$p^4-16s^4=(p^2)^2-(4s^2)^2$
$=(p^2+4s^2)(p^2-4s^2)$

3. Factorise the remaining difference of squares completely.

$p^2-4s^2=p^2-(2s)^2=(p+2s)(p-2s)$
Therefore,
$2p^4-32s^4=2(p^2+4s^2)(p+2s)(p-2s).$

1. Group the four terms into two pairs with common factors.

$12cd-9cx+6xy-8dy=(12cd-9cx)+(6xy-8dy)$

2. Factorise each pair to obtain a common binomial factor.

$12cd-9cx+6xy-8dy=3c(4d-3x)-2y(4d-3x)$

3. Extract the common binomial factor completely.

$3c(4d-3x)-2y(4d-3x)=(4d-3x)(3c-2y)$