Question

factorise 2(x+y)^2-9(x+y)-5

Answer 1

$\pink{\mathcal{\huge{\underline{\underline{\star\:Hey\:Dude\:\star}}}}}$

$<b><i>Solution:$

2(x + y)² - 9( x + y ) - 5

⇒2(x + y)² - 10 (x + y) + 1( x + y) - 5

⇒2(x + y)(x + y - 5 ) + 1(x + y -5 )

taking (x + y -5 ) common , 

⇒(x + y -5 )[2(x + y) + 1] 

⇒(x + y - 5)(2x + 2y +1)
_________________________
Or solving with other method..
_________________________

Let a = (x + y)

So,

2a² - 9a - 5

2a² + 10a - a - 5

2a( a + 5) - 1 (a + 5)

(a + 5) (2a - 1)

Hence, on substituting the value of a = (x + y)

(x + y + 5) [2 (x + y) - 1]

$\therefore$ (x + y + 5)(2x + 2y - 1)

Thanks for the question!

Answer 2

$\huge{\pink{\underline{\ulcorner{\star\: Answer\: \star}\urcorner}}}$

$\bold{2(x+y)^{2}-9(x+y)-5}$

$\bold{Let \:t \:= \:(x+y)}$

2t²-9t-5

2t²-10t+t-5

2t(t-5)+1(t-5)

(2t+1)(t-5)

Put the value of the in equation :

{2(x+y)+1}{(x+y)-5}

$\red{\underline{Answer\: \colon}}$

$\bold{[2x+2y+1][x+y-5]}$

Thanks