Question

$x^2-y^2-9z^2+6yz$ is factorized as

• A. $(x+y-3z)(x+y+3z)$
• B. $(x-y-3z)(x-y+2z)$
• C. $(x+y-3z)(x-y+3z)$
• D. $(x-2y+3z)(x-y+3z)$

Answer 1

The correct option is C $(x+y-3z)(x-y+3z)$

We have, $x^2-y^2-9z^2+6yz$

$=x^2-(y^2+9z^2-6yz)$

$=x^2-\left[\right(y{)}^2+(3z{)}^2-2(y\left)\right(3z\left)\right]$

$=x^2-(y-3z{)}^2[\because (a-b{)}^2=a^2+b^2-2ab]$

$=[x-(y-3z\left)\right][x+(y-3z\left)\right][\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$

$=(x-y+3z)(x+y-3z)$

So, $\text{C}$ is the correct option.