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B What are the different methods of factoring polynomials?

  1. Jul 1, 2017 #1
    Methods of factoring .

    Method of common factors
    Factorization by regrouping terms
    Factorization using identities
    Factors of the form ( x + a) ( x + b)
    Factor by Splitting

    factoring_polynomials.png

    factoing_polynomials_2.png
    factoring_polynomials_3.png

    Is this all the factoring methods out there ? Or are there more ?

    I am also looking for a book with lots of practice problems of factoring polynomials .

    Please help
     
  2. jcsd
  3. Jul 1, 2017 #2
    If you want to factorize school examples that are designed to have nice results, you can use whatever tricks come to your mind.
    If you want to do it in practice, you find the roots first, either with a formula (probably only with quadratic polynomials) or using one of several iterative methods (I like Laguerre's method but there are dozens).
    Once you have the roots, say a, b, c, d then the factorization is ##k(x-a)(x-b)(x-c)(x-d)##. Obviously the number of factors depends on the degree of the polynomial.

    Edit: forgot to put in the constant multiplier ##k## that is the same as the coefficient of the highest degree, e.g. 5 in ##5x^3-2x+1##.
     
  4. Jul 1, 2017 #3
    Thanks for the reply ,
    One reason why i like this forum is that you always get a fast reply no matter how stupid your questions are .

    :-)
     
  5. Jul 1, 2017 #4

    I like Serena

    User Avatar
    Homework Helper

    You may like the Rational root theorem.

    If we can split a polynomial with integer coefficients, like ##3x^3 - 5x^2 + 5x - 2##, in the form ##a(x-b)(x^2+...)## where ##b## is an integer or a rational number, then ##b## has to be plus or minus a divider of ##2## divided by a divider of ##3##.
    That is, b is one of ##\pm \frac 11, \pm\frac 21, \pm\frac 13, \pm \frac 23##.
    And indeed, if we substitute ##x=\frac 23##, we find that it's a root.
    So we can split it as ##3(x- \frac 23)(x^2 + ..)## or more cleanly as ##(3x- 2)(x^2 + ..)##.
     
  6. Jul 1, 2017 #5
    Thanks for the reply ,
    Currently i am only following the materials available from these resources .

    http://ncert.nic.in/textbook/textbook.htm
    https://2012books.lardbucket.org/books/beginning-algebra/
    http://ins.sjtu.edu.cn/people/mtang/textbook.pdf

    Most of those resources don't have factorization of Cubic polynomials in it , Don't know why though .
    I guess those methods are more advanced than the usual methods .

    I am only getting comfortable with explanations like these ,

    A whole number has "pairs of factors" in it which when multiplied together gives you the whole number
    I like that explanation because it says "pairs of factors" , you can then chose the "pairs of factors" that goes to the middle part of the polynomial when factoring it .

    I am only trying to find more examples of "Factor by Splitting"

    :-)
     
  7. Jul 1, 2017 #6

    Mark44

    Staff: Mentor

    Because factoring a cubic polynomial is generally very complicated. There are a few cubics that have nice factorizations, such as ##(x^3 + a^3) = (x + a)(x^2 - ax + a^2)## and ##(x^3 - a^3) = (x - a)(x^2 + ax + a^2)##.
     
  8. Jul 2, 2017 #7
    Thanks for the explanations :-)
     
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