Question

factorise:9x^2-12ax-y^2-z^2-2yz+4a^2

Answer 1

Let's simplify step-by-step.

9x2−12ax−y2−z2−2yz+4a2

=9x2+−12ax+−y2+−z2+−2yz+4a2

Answer:

=4a2−12ax+9x2−y2−2yz−z2

Answer 2

Answer:

Step-by-step explanation:

Given : $9x^2-12ax-y^2-z^2-2yz+4a^2$

To Find : Factorize

Solution:

$9x^2-12ax-y^2-z^2-2yz+4a^2$

$9x^2-12ax+4a^2-y^2-z^2-2yz$

$9x^2-12ax+4a^2-(y^2+z^2+2yz)$

$(3x)^2-2(3x)(2a)+(2a)^2-(y^2+z^2+2yz)$

Identity:$(a+b)^2=a^2+b^2+2ab \\(a-b)^2=a^2+b^2-2ab$

Using identity

$(3x-2a)^2-(y+z)^2$

$a^2-b^2=(a+b)(a-b)$

So, $(3x-2a+y+z)(3x-2a-y-z)$

Hence $9x^2-12ax-y^2-z^2-2yz+4a^2$ = $(3x-2a+y+z)(3x-2a-y-z)$