Factorise the following:

e) $(x + 4)^{2} - 5 x y - 20 y - 6 y^{2}$

Open in App

Solution

Given: $(x + 4)^{2} - 5 x y - 20 y - 6 y^{2}$

$= (x + 4)^{2} - 5 y (x + 4) - 6 y^{2}$

By splitting the middle term, we get,

$= (x + 4)^{2} - 6 y (x + 4) + y (x + 4) - 6 y^{2}$

$= ((x + 4) (x + 4 - 6 y)) + y (x + 4 - 6 y)$

$= (x + 4 - 6 y) (x + 4 + y)$

$= (x - 6 y + 4) (x + y + 4)$

Hence, the answer is $(x - 6 y + 4) (x + y + 4)$ .

Suggest Corrections

Similar questions

Find $\frac{x}{y}$ ; when $x^{2} + 6 y^{2} = 5 x y$ .

$(x + 4)^{2} - 5 x y - 20 y - 6 y^{2} =$

[2 M a r k s]

2 x^{2} + (\frac{x}{6}) - 1

(2 p + q)^{2} - 10 p - 5 q - 6

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program