Factorise $32 (x + y)^{2} - 2 x - 2 y$

(x + y) (x + y + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (x + y) (16 x + 16 y - 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(x + y) (16 x + 16 y - 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 (x + y) (16 x + 16 y - 1)$
$32 (x + y)^{2} - 2 x - 2 y = 32 (x + y)^{2} - 2 (x + y) = 2 (x + y) {16 (x + y) - 1} = 2 (x + y) (16 x + 16 y - 1)$

Suggest Corrections

Similar questions

2 (x + y)^{2} - 6 (x + y)

x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1, x^{3} - 1, x^{3} + x^{2} + x

x^{2} y + y x + y

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

(x - y)^{2} a^{2} + 2 (x - y) (x + y) a b + b^{2} (x + y)^{2}

x^{3} - 2 x^{2} - x + 2

x^{3} - 3 x^{2} - 9 x - 5

x^{3} + 13 x^{2} + 32 x + 20

2 y^{3} + y^{2} - 2 y - 1

Join BYJU'S Learning Program