Question

factorise 4X2+y2+z2-2yz+4xz
for this we have to use the 5 identity
but I am not able to get
plsss help me guyss​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\bigstar$ Question:

Factorize $4x^{2} + y^{2} + z^{2} - 4xy - 2yz + 4xz$

$\bigstar$ Solution:

According to the question,

$4x^{2} + y^{2} + z^{2} - 4xy - 2yz + 4xz$

$\implies 2x^{2} x^{2} + y^{2} + z^{2} - 4xy - 2yz + 4xz$

$\implies (2x)^{2} + y^{2} + z^{2} - 4xy - 2yz + 4xz$

$\implies (2x)^{2} +(- y)^{2} + z^{2} - 4xy - 2yz + 4xz$

$\implies (2x)^{2} + (-y)^{2} + (z)^{2} + 2(2x)(-y)+2 (-y)(z)+2(2x)(z)$

Now, using $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac$

Where, $a=2x, \: b=-y, \: c=z$

$\implies (2x+(-y)+z)^{2}$

$\implies (2x-y+z)^{2}$

$\implies (2x-y+z)(2x-y+z)$