This is not a difficult factorisation, as it is merely a normal factorisation with "a" stuck on the end of each number. 25a(2) - 20ab + 4b(2) (2b-5a)(2b-5a) This is: 2bx2b=4b(2) 2bx-5a=-10ab (twice) -5ax-5a=25a
(b-c)(a+b)-ac
(x - 5)(x + 12)
(2x - 1)(5x + 8)
(x-1)(x-4)
(x + 2)(3x - 1)(3x + 1)
There is no rational factorisation.
There are no rational factors.
(a - 2)(a^2 + 6)
(b-c)(a+b)-ac
(3y - 5)(y + 5)
(x + 3)(3x - 2)
(7a - 1)(6a + 1)
it is (x-y)(x-y) :)
(3r - 2)(3r - 2) or (3r - 2) squared.
6(b - ac + b2 - bc)
6(ab - ac + b2 - bc)
Factor out -2: -2(x squared +11x-20). This is fully factored.