1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quick check of 4 term polynomial factorised

  1. Sep 26, 2012 #1
    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.
     
  2. jcsd
  3. Sep 26, 2012 #2
    Looks good to me. You can always multiply out your final answer and see if it comes out to your original equation.
     
  4. Sep 26, 2012 #3

    Mark44

    Staff: Mentor

    You have an error (maybe a typo) above. The first term should be a(t2 - 4).
     
  5. Sep 27, 2012 #4
    Yeah it was a typo, my bad.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quick check of 4 term polynomial factorised
Loading...