Question

Factorise x^2+4y^2+4y-4xy-2x-8

Answer 1

Step-by-step explanation:

x^2+4y^2+4y-4xy-2x-8

=(x)^2+(2y)^2–2×x×2y-2x+4y-8

=(x-2y)^2–2(x-2y)-8

Let (x-2y)=p

=p^2-2p-8

=(p-4).(p+2). , putting p= x-2y

=(x-2y-4).(x-2y+2). Answer

MARK IT BRAINLIEST PLZ, IF YOU GET HELP

Answer 2

$\sf{x^2+4y^2+4y-4xy-2x-8}\\\\\sf{=x^2+(2y)^2-4xy+4y-2x-8}\\\\\sf{As\ (x-2x)^2=x^2+(2x)^2-2\times x \times 2x = x^2+(2y)^2-4xy}\\\\\sf{=(x-2y)^2+2(2y-x)-8}\\\\\sf{=(x-2y)^2-2(x-2y)-8}\\\\\sf{(Changing\ the\ sign)}\\\\\sf{Let\ us\ assume\ (x-2y)\ be\ m.}\\\\\sf{We\ get\ :-}\\\\\sf{=m^2-2m-8}\\\\\sf{= m^2-4m+2m-8}\\\\\sf{= m(m-4)+2(m-4)}\\\\\sf{= (m-4)(m+2)}\\\\\sf{Now,\ putting\ the\ value\ of\ m\ :-}\\\\\underline{\boxed{\sf{= (x-2y-4)(x-2y+2)}}}$