Question

The factorization of z² -4z -12 is
I)(z+6)(z+2)
ii)(z-6)(z-2)
iii)(z-6)(z+2)

Answer 1

Step-by-step explanation:

${z}^{2} - 4z - 12$

$= {z}^{2} - (6 - 2)z - 12$

$= {z}^{2} - 6z + 2z - 12$

$= z(z - 6) + 2(z - 6)$

$= (z - 6)(z + 2)$