Factorize completely :

$x^{2} - 16 y^{2} - 2 x - 8 y$

$(x + 4 y) (x - 4 y - 2)$

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

$(x + 4 y) (x + 4 y - 2)$

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

$(x - 4 y) (x + 4 y - 2)$

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

$(x + 4 y) (x - 4 y + 2)$

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

Open in App

Solution

The correct option is A
$(x + 4 y) (x - 4 y - 2)$

$\Rightarrow [(x^{2} - 16 y^{2}) - 2 x - 8 y]$

$= [(x - 4 y) (x + 4 y) - 2 x - 8 y]$

$= [(x - 4 y) (x + 4 y) - 2 (x + 4 y)]$

$= (x + 4 y) [(x - 4 y) - 2]$

$= (x + 4 y) (x - 4 y - 2)$

Suggest Corrections

Similar questions

x^{2} + x y + 8 x + 8 y

x^{2} + 2 y^{2} - 2 x + 12 y + 10 = 0

x^{2} + 4 y^{2} - 4 x + 24 y + 31 = 0

4 x^{2} + y^{2} - 8 x + 2 y + 1 = 0

3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0

4 x^{2} + 16 y^{2} - 24 x - 32 y - 12 = 0

x^{2} + 4 y^{2} - 2 x = 0

x^{2} - 1 - 2 x - x^{2}

Circles $x^{2} + y^{2} - 2 x - 4 y = 0 a n d x^{2} + y^{2} - 8 y - 4 = 0$

Join BYJU'S Learning Program