Question

Factorize : (x-2y)^3 – 5x^2 – 20y^2 + 6x – 12y + 20xy.​

Answer 1

Step-by-step explanation:

(x-2y)^3 - 5x^2 + 20xy - 20y^2 + 6x - 12y

= (x-2y)^3 - 5 (x^2 - 4xy + 4y^2) + 6 (x-2y)

= (x-2y)^3 - 5 (x^2 - 2×2×xy + (2y)^2) + 6 (x-2y)

= (x-2y)^3 -5 (x-2y)^2 + 6 (x-2y)

Taking (x-2y) common you will get

=(x-2y)[(x-2y)^2 - 5 (x-2y) +6]

=(x-2y) [(x-2y)^2 -3 (x-2y) -2 (x-2y) +6]

=(x-2y) [(x-2y)(x-2y-3) -2 (x-2y-3)]

=(x-2y) (x-2y-2) (x-2y-3) ans.