Respuesta :

irspow
You need to factor the numerator and denominator...

(2x^2+4x-2x-4)/(2x^2-2x-2x+2)

(2x(x+2)-2(x+2))/(2x(x-1)-2(x-1))

((2x-2)(x+2))/((2x-2)(x-1))  so the (2x-2)s cancel out leaving

(x+2)/(x-1)

Answer:

[tex]\frac{x+2}{x-1}[/tex]

Step-by-step explanation:

[tex]\frac{(2x^2+2x-4)}{(2x^2-4x+2)}[/tex]

Factor top and bottom separately

[tex]2x^2+2x-4[/tex]

[tex]2x^2+4x-2x-4[/tex]

Make two groups and take out GCf

[tex]2x(x+2)-2(x+2)[/tex]

(x+2)(2x-2)

We factor the denominator

[tex]2x^2-4x+2[/tex]

[tex]2x^2-2x-2x+2[/tex]

Make two groups and take out GCf

[tex]2x(x-1)-2(x-1)[/tex]

(x-1)(2x-2)

[tex]\frac{(2x^2+2x-4)}{(2x^2-4x+2)}=\frac{(x+2)(2x-2)}{(x-1)(2x-2)}[/tex]

2x-2 cancel out

So answer is [tex]\frac{x+2}{x-1}[/tex]