Quick check of 4 term polynomial factorised

[Taylor_1989]

Could someone quickly go over my working, as I am not 100% sure I have done it the right way. I will show and explain my working step by step.

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

I first grouped the values: $(at^2-4a) + (2t^2-8)$

I then factorised these equations into: $a(t^2-4a) + 2(t^2-4)$

I then regrouped: $(a+2) (t^2-4)$

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

I then got the final answer of: $(a+2)(t-2)(t+2)$.

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.

 

[Diffy]

Looks good to me. You can always multiply out your final answer and see if it comes out to your original equation.

 

[Mark44]

Could someone quickly go over my working, as I am not 100% sure I have done it the right way. I will show and explain my working step by step.

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

I first grouped the values: $(at^2-4a) + (2t^2-8)$

I then factorised these equations into: $a(t^2-4a) + 2(t^2-4)$

You have an error (maybe a typo) above. The first term should be a(t² - 4).

I then regrouped: $(a+2) (t^2-4)$

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

I then got the final answer of: $(a+2)(t-2)(t+2)$.

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.

 

[Taylor_1989]

Yeah it was a typo, my bad.