-10
(x - 12)(x + 12)
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)
x2 + 6x + 12 = 0 x2 + 6x + 9 = -3 (x + 3)2 = -3 x + 3 = ± √-3 x = -3 ± i√3
(x + 6)(x - 2)
x2 + x - 12 = (x + 4)(x - 3)
x2 + x - 12 = (x + 4)(x - 3)
x2 - 10x - 24= x2 - 12x + 2x - 24= x(x - 12) + 2(x - 12)= (x - 12)(x + 2)
14
If: x2+x = 12 Then: x2+x-12 = 0 And using the quadratic formula: x = -4 or x = 3
(x2-x-12)/(x-4) = (x+3)
x2 -11x-12= (x-12)(x+1)
x2 - 8x - 48 = x2 - 12x + 4x - 48 = x(x-12)+4(x-12) = (x-12)(x+4) So, factors are (x-12) & (x+4).
14
-10
No. x2 -4x - 12 = (x + 2)(x - 6)
x2-13x+12 = (x-1)(x-12) when factored