Question

Factorize: $x^2+4y^2-9z^2-4xy$

• A. $\left[\right(x-4y-3z\left)\right(x-4y+3z\left)\right]$
• B. $\left[\right(x-2y-3z\left)\right(x-2y+3z\left)\right]$
• C. $\left[\right(x+2y-3z\left)\right(x+2y+3z\left)\right]$
• D. none

Answer 1

Answer: (B)

$\left[\right(x-2y-3z\left)\right(x-2y+3z\left)\right]$

The given expression $x^2+4y^2-9z^2-4xy$ can be factorized as shown below:

$x^2+4y^2-9z^2-4xy=(x^2+4y^2-4xy)-9z^2=[x^2+(2y{)}^2-(2\times x\times 2y)]-9z^2=(x-2y{)}^2-9z^2(\because (a-b{)}^2=a^2+b^2-2ab)=(x-2y{)}^2-(3z{)}^2=(x-2y-3z\left)\right(x-2y+3z\left)\right(\because a^2-b^2=(a-b)(a+b))$

Hence, $x^2+4y^2-9z^2-4xy=(x-2y-3z)(x-2y+3z)$.