Question

3r(2p-5)-p(2p-5) factorise it

Answer 1

3r(2p - 5) - p(2p - 5)

let a = 2p - 5

3r(2p - 5) - p(2p - 5) = 3ra - pa = a(3r - p)

let's substitute back the value of a.

a(3r - p) = (2p - 5)(3r - p)

Therefore: 3r(2p - 5) - p(2p - 5) = (2p - 5)(3r - p)

Answer 2

3r(2p - 5) - p(2p - 5)
The common factor is (2p - 5), therefore we can write:
(2p - 5)(3r - p)