Math, asked by panditadhruv7, 5 days ago

(ii) 2x2 + y2 + 8z2 - 2/2 xy + 42 yz - 8xz​

Answers

Answered by ll5119507
0

Answer:

Answer: The answer is (-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z).(−

2

x+y+2

2

z)(−

2

x+y+2

2

z). )

Step-by-step explanation: We are given to factorise the following expression:

E=2x^2+y^2+8z^2-2\sqrt2xy+4\sqrt 2yz-8xz.E=2x

2

+y

2

+8z

2

−2

2

xy+4

2

yz−8xz.

We will be using the following factorisation formula:

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca.(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca.

We have

\begin{gathered}E\\\\=2x^2+y^2+8z^2-2\sqrt2xy+4\sqrt 2yz-8xz\\\\=(-\sqrt2x)^2+y^2+(2\sqrt2z)^2+2(-\sqrt2x)y+2y(2\sqrt2z)+2(2\sqrt2z)(-\sqrt2x)\\\\=(-\sqrt2x+y+2\sqrt2z)^2\\\\=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z).\end{gathered}

E

=2x

2

+y

2

+8z

2

−2

2

xy+4

2

yz−8xz

=(−

2

x)

2

+y

2

+(2

2

z)

2

+2(−

2

x)y+2y(2

2

z)+2(2

2

z)(−

2

x)

=(−

2

x+y+2

2

z)

2

=(−

2

x+y+2

2

z)(−

2

x+y+2

2

z).

Thus, the answer is (-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z).(−

2

x+y+2

2

z)(−

2

x+y+2

2

z)

Similar questions