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

Mentor
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.

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