Quick check of 4 term polynomial factorised

  • #1
402
14

Main Question or Discussion Point

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: [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.
 

Answers and Replies

  • #2
441
0
Looks good to me. You can always multiply out your final answer and see if it comes out to your original equation.
 
  • #3
33,646
5,315
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: [itex] (at^2-4a) + (2t^2-8) [/itex]

I then factorised these equations into: [itex] a(t^2-4a) + 2(t^2-4) [/itex]
You have an error (maybe a typo) above. The first term should be a(t2 - 4).
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.
 
  • #4
402
14
Yeah it was a typo, my bad.
 

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