Question

Factorise (√2x+2y-√3z)^2​

Answer 1

${ \sf Formula \: Applied } \\ \boxed{(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca(a+b+c) } \\ \sf Solution : \\ \boxed{\begin{gathered}(\sqrt{2}x+2y-\sqrt{3}z)^2\\\\=[\sqrt{2}x+2y+(-\sqrt{3}z)]^2\\\\=(\sqrt{2}x)^2+(2y)^2+(-\sqrt{3}z)^2+2(\sqrt{2}x)(2y)+2(2y)(-\sqrt{3}z)+2(-\sqrt{3}z)(\sqrt{2}x)\\\\=2x^2+4y^2+3z^2+4\sqrt{2}xy-4\sqrt{3}yz-2\sqrt{2}\sqrt{3}zx\end{gathered} }$

Answer 2

FormulaApplied

(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca(a+b+c)

Solution:

(

2

x+2y−

3

z)

2

=[

2

x+2y+(−

3

z)]

2

=(

2

x)

2

+(2y)

2

+(−

3

z)

2

+2(

2

x)(2y)+2(2y)(−

3

z)+2(−

3

z)(

2

x)

=2x

2

+4y

2

+3z

2

+4

2

xy−4

3

yz−2

2

3

zx