Question

If 2 is one of the zeros of $x^{3}-3x^{2} -4x-12$ then find the remaining
two zeroes.

Answer 1

Step-by-step explanation:

$x = 2 \\ x - 2 = 0 \\ {x}^{3} - 3 {x}^{2} - 4x + 12 = 0 \\ {x}^{3} - 2 {x}^{2} - {x}^{2} + 2x - 6x + 12 = 0 \\ {x}^{2} (x - 2) - x(x - 2) - 6(x - 2) = 0 \\ (x - 2)( {x}^{2} - x - 6) = 0 \\ {x}^{2} - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ (x + 2)(x - 3) = 0 \\ x = 3 , - 2$other two roots are 3,-2