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!

Adding "annihilating terms" to factor

  1. Sep 27, 2017 #1

    DS2C

    User Avatar
    Gold Member

    Not a homework problem, just something I came across in my book that makes sense but not fully.

    1. The problem statement, all variables and given/known data
    Factor: ##a^4-b^4##

    2. Relevant equations


    3. The attempt at a solution

    The book's solution:

    ##a^4-b^4##
    ##=a^4-a^3b+a^3b-a^2b^2+a^2b^2-ab^3+ab^3-b^4##
    ##=a^3\left(a-b\right)+a^2b\left(a-b\right)+ab^2\left(a-b\right)+b^3\left(a-b\right)##
    ##=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)##

    Just what is going on here? Such a tiny binomial resulting in something way larger, with what seem like arbitrary terms added in.

    Additionally, it gives another factorization of the given problem ##a^4-b^4## which is also factored into ##\left(a^2-b^2\right)\left(a^2+b^2\right)##
    You can have more than one correct solution when factoring a polynomial? In what situations would one be the "correct vs incorrect" way? Or is it synonymous with something like factoring 12 into 4*3, 6*2, 12*1 where they are all correct?
     
  2. jcsd
  3. Sep 27, 2017 #2

    fresh_42

    Staff: Mentor

    You could still perform another factorization on ##a^2-b^2##. Since it didn't say until only irreducible factors are left, it has indeed more than one solution.
     
  4. Sep 27, 2017 #3

    DS2C

    User Avatar
    Gold Member

    Man this factoring stuff goes deep. Feels like there should be an entire semester focusing on just this. Must be pretty important!
     
  5. Sep 27, 2017 #4

    fresh_42

    Staff: Mentor

    It is, and of course it's always only up to units, as one can always factor ##f(x) = c \cdot (c^{-1} \cdot f(x))## and which is of no use.
     
  6. Sep 27, 2017 #5

    DS2C

    User Avatar
    Gold Member

    Not at all sure what that means unfortunately. I'm only at factoring polynomials in the form of ##ax^2+bx+c## currently. There's obviously a lot more to learn.
     
  7. Sep 27, 2017 #6

    fresh_42

    Staff: Mentor

    Well, just apply what you did once again on: ##a^4-b^4=(a^2-b^2)(a^2+b^2)## and you get three factors, which cannot be split any further, so we call them irreducible. The terms ##a^4-b^4## and ##a^2-b^2## are reducible.
     
  8. Sep 27, 2017 #7

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Last edited: Oct 13, 2017
  9. Sep 27, 2017 #8

    DS2C

    User Avatar
    Gold Member

    Yeah thats my thread too lol. The reason for a new thread is a new aspect of the factoring. The problem posted, which was out of my book, mentions "inserting annihilating terms" into the expression to factor. This is what Im having trouble wrapping my head around. Why do this and how do you determine what "annihilating" terms to input?
     
  10. Sep 27, 2017 #9

    fresh_42

    Staff: Mentor

    It is simply another way to describe a division. Instead of calculating ##(a^4-b^4):(a-b)## the entire division is somehow written backwards. If we know the result, then we know which "annihilating terms" to be added. Otherwise it's probably a try and error. I still prefer the division. I doubt there is a system behind it.
     
    Last edited: Sep 27, 2017
  11. Sep 27, 2017 #10

    DS2C

    User Avatar
    Gold Member

    That makes more sense. Here's the text in question for reference.
     

    Attached Files:

  12. Oct 13, 2017 #11
    Factor again ##\left(a^2-b^2\right)## in ##\left(a^2-b^2\right)\left(a^2+b^2\right)##

    To see that the two factorisations are actually the same, try to re-write the above factorisation as ##\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)##. You should see what to do after the first step above.

    Normally in factorising we try to factorise each factor as far as possible, so in this case, the first part of this reply is what we normally do.
     
  13. Oct 13, 2017 #12

    fresh_42

    Staff: Mentor

    I assume you've meant ##\left(a^2-b^2\right)## in ##\left(a-b\right)\left(a+b\right)##.
     
  14. Oct 13, 2017 #13
    No. "In" not "into", as in, factor again the part ##\left(a^2-b^2\right)## in the expression ##\left(a-b\right)\left(a+b\right)##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Adding "annihilating terms" to factor
  1. Factor This (Replies: 3)

  2. Factorize (Replies: 6)

  3. Adding exponentials (Replies: 7)

  4. Adding Functions (Replies: 1)

  5. Adding cos (Replies: 5)

Loading...