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.

 

[CellCoree]

I would suggest double-checking your work and making sure you fully understand the concept of factoring polynomials before relying on someone else's verification. However, based on your explanation, your approach seems correct. You correctly grouped the terms and factored out the common factors. Your next step of changing 4 to 2^2 is also a valid method for factoring the difference of squares. Your final answer of (a+2)(t-2)(t+2) also looks correct.