x^2 - 4x -12 = 0
(x + 2) (x - 6) = 0
x + 2 = 0 or x - 6 = 0
x + 2 - 2 = 0 -2 or x - 6 + 6 = 0 + 6
x = -2 or x = 6
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).
Y = X2 - 4X + 12 set to 0 X2 - 4X + 12 = 0 subtract 12 from each side X2 - 4X = - 12 now, take the linear term ( - 4 ), halve it, square it and add it to both sides X2 - 4X + 4 = - 12 + 4 factor the left side, gather terms on the right (X - 2)2 = - 8 add 8 to each side (X - 2)2 + 8 = 0 ------------------------ (2, 8) ------------the vertex of this function
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.
6
14
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 + 12)(x - 8)