factorise c. 9x2 + 4y2 + z2 - 12xy + 4yz - 6xz
Answers
Answer:
Step-by-step explanation:
we know that,
> (a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ac
Therefore,
-9x^2= (-3x)^2
4y^2= (2y)^2
z^2 = (z)^2
So,
[(-3x)+(2y)+(z)]^2= (-3x)^2+(2y)^2+(z)^2+ 2(-3x)(2y)+2(2y)(z)+2(-3x)(z)
= +9x^2+4y^2+z^2-12xy+4yz-6xz
The factorised expression of 9x² + 4y² + z² - 12xy + 4yz - 6xz is (3x - 2y - z)²
We have to factorise ; 9x² + 4y² + z² - 12xy + 4yz - 6xz.
= 9x² + 4y² + z² - 12xy + 4yz - 6xz
= 9x² + 4y² - 12xy + z² + 4yz - 6xz
= (3x)² + (2y)² - 2(3x)(2y) + z² + 4yz - 6xz
We know, (a - b)² = a² + b² - 2ab
∴ (3x)² + (2y)² - 2(3x)(2y) = (3x - 2y)²
= (3x - 2y)² + z² + 2z(2y - 3x)
= (3x - 2y)² + z² - 2(3x - 2y)z
if (3x - 2y) = a, and z = b
then, a² + b² - 2ab = (a - b)² [ from algebraic identity]
∴ (3x - 2y)² + z² - 2(3x - 2y)z = (3x - 2y - z)²
Therefore 9x² + 4y² + z² - 12xy + 4yz - 6xz = (3x - 2y - z)².
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