(2x^2-8x-42)/[(x^2-9)/(x^2-3x)]
=2(x^2-4x-21)/[(x+3)(x-3)/(x(x-3))]
=2(x-7)(x+3)/[(x+3)/x]
When we canceled out "(x-3)," it set up a domain restriction: x≠3. This is correct because we are not allowed to divide by 0 at any point, ever! Moving on...
=2(x-7)(x+3)(x)/(x+3)
=2(x-7)(x)
Now, x≠-3 for the same reason stated above. Moving on...
=2x(x-7)
=2x^2-14x
3x2-2x-8 = (3x+4)(x-2) when factored
3x(2x + 1)
(x - 1)(3x + 5)
(x - 2)(2x + 7)
6x3 + 9x2 - 2x2 - 3x Before any grouping, take out the common factor, x, to give x*(6x2 + 9x - 2x - 3) = x*[3x*(2x + 3) - 1*(2x + 3)] = x*(2x + 3)*(3x - 1)
3x squared - x squared = 2x squared
2x squared - 3x
x(3x - 2)
2x2 x 3x2 = 6 x4 (2x)2 x (3x)2 = 36 x4
5
6x2 + 10x = 2x*(3x + 5)
3x2-2x-8 = (3x+4)(x-2) when factored
2 squared X 3 squared = 5 squared
2x+x is 3x
(6x^3-9x^2) / (2x-3) = 3x^2((2x-3) / (2x-3) = 3 x ^2 (3 x squared)
3x(2x + 1)
(x + 1)(2x - 5)