x2 - 4x - 32 = (x - 8)(x + 4)
6(x2 -4x - 12) 6(x-6)(x+2)
4x3 + 16x2 - 180x = 4x(x2 + 4x - 45) = 4x(x + 9)(x - 5)
f(x) = x2-4x-5 = x2-5x+x-5 = x(x-5)+1(x-5) = (x+1)(x-5) Factors of f(x) are (x+1) and (x-5).
If that's -5 the answer is (x - 5)(x + 1)
No. x2 -4x - 12 = (x + 2)(x - 6)
x2-4x-12 = (x-6)(x+2) when factored
x2 + 4x - 96 = (x - 8)(x + 12).
x2 - 4x = 8 x2 - 4x + 4 = 8 + 4 (x - 2)2 = 12 x - 2 = +&- sq. root of 12 x = 2 +&- 2(sq. root of 3)
4x-x2 = 2
x2+4x-12 = (x-2)(x+6) when factored
If you are looking to factor this expression, then it does not have a real solution.
14
6
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
(x + 6)(x - 2)
(x + 12)(x - 8)