Question

2a^2 - 4a + 2 / 3a^2 - 3

Answer 1

The problem is \[\Large \frac{2a^2 - 4a + 2}{3a^2-3}\] right?

Answer 2

yes

Answer 3

alright, thanks

Answer 4

We can factor 2a^2 - 4a + 2 to get


2a^2 - 4a + 2

2(a^2 - 2a + 1)

2(a-1)(a-1)

Answer 5

Then we can factor 3a^2-3 to get

3a^2-3

3(a^2-1)

3(a-1)(a+1)

Answer 6

So

\[\Large \frac{2a^2 - 4a + 2}{3a^2-3}\]

turns into

\[\Large \frac{2(a-1)(a-1)}{3(a-1)(a+1)}\]

Answer 7

Do you see where to go from here?

Answer 8

yes thanks

Answer 9

you're welcome