Question

what are the zeroes of this polynomial
$p(x) = {x}^{2} - x - 12$

Answer 1

= x^2 - x - 12
= x^2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x+3) (x-4)
x = (-3)
x = 4