That factors to (x - 5)(x - 1)
x^2-4x/x^2-6x+8 = x(x-4)/(x-2)(x-4) = x/(x-2)
(3x2 - 6x)/3x = 3x(x-2)/3x = x-2, for x<>0
{-1,7}
(x-1)(x-5)=x^2-6x +5 Because it is negative 5, I dont see an answer.
x2 - 6x - 27 = (x-9)(x+3)
x-8
First, factorising, we get: x2 - 6x - 16 = (x - 8)(x + 2). Then, (x2 - 6x - 16) / (x - 8) = x + 2.
(6x - 1)(6x - 1)
That factors to (x - 5)(x - 1)
x^2-4x/x^2-6x+8 = x(x-4)/(x-2)(x-4) = x/(x-2)
(3x2 - 6x)/3x = 3x(x-2)/3x = x-2, for x<>0
(x^2 - 2)(x - 2)(x + 2)
(x - 3)(x^2 + 6)
It is: (3x+4)(2x-3) when factored
{-1,7}
When factored it is: (6x-1)(6x+1)