Question

Factorize the following quadratic polynomial by using the method of completing the square:
$z^2-4z-12$

Answer 1

Given: $z^2-4z-12$
Since, the coefficient of $z^2$ is unity.
$\therefore \text{Adding and subtracting}{(\frac{1}{2}\times \text{coefficient of z})}^2$
i.e., $(-2{)}^2=(2{)}^2$
$\Rightarrow z^2-4z-12$
$=z^2-4z+2^2-2^2-12$
$=(z^2-4z+2^2)-4-12$
$=(z-2{)}^2-16$
$=(z-2{)}^2-4^2$
= (z - 2 + 4)(z - 2 -4)
=(z + 2)(z - 6)
Hence, $z^2-4z-12=(z+2)(z-6)$