- #1

- 402

- 14

$$ at^2-4a + 2t^2-8$$

I first grouped the values: [itex] (at^2-4a) + (2t^2-8) [/itex]

I then factorised these equations into: [itex] a(t^2-4a) + 2(t^2-4) [/itex]

I then regrouped: [itex] (a+2) (t^2-4) [/itex]

This is the part I am not sure is right, I the thought to factor more I could change the [itex] 4 [/itex] to [itex] 2^2 [/itex] which would give [itex] (t^2-2^2) [/itex] which gave me the difference of two squares; right?

I then got the final answer of: [itex] (a+2)(t-2)(t+2) [/itex].

I would just like to know if this is the right way of doing a equation like this, if not could someone show where I have gone wrong.