High School What are the different methods of factoring polynomials?

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The discussion covers various methods of factoring polynomials, including common factors, regrouping, identities, and splitting. Participants express interest in additional factoring techniques and seek resources for practice problems, particularly for cubic polynomials. The Rational Root Theorem is highlighted as a useful tool for finding integer or rational roots. There is a consensus that while some cubic polynomials can be factored neatly, many require more advanced methods. Overall, the conversation emphasizes the complexity of polynomial factorization and the need for more educational resources.
awholenumber
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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

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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
 
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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##.
 
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Likes awholenumber
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 .

:-)
 
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Likes Brian blake science
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 + ..)##.
 
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Likes Greg Bernhardt and awholenumber
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"

:-)
 
awholenumber said:
Most of those resources don't have factorization of Cubic polynomials in it
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)##.
 
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Likes awholenumber
Thanks for the explanations :-)
 

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