Factorization in polynomials is a mathematical content that brings together techniques to write them in the form of a product between monomials or even among others polynomials. This decomposition is based on the fundamental theorem of arithmetic, which guarantees the following:
Any integer greater than 1 can be decomposed
in a product of prime numbers.
The techniques used to factorize polynomials – calls from cases in factorization – are based on the multiplication properties, especially in the distributive property. The six cases of factorization of polynomials are as follows:
1st case of factorization: common factor in evidence
Note, in the polynomial below, that there is a factor repeating itself in each of its terms.
4x + ax
to write this polynomial in the form of a product, put this factor repeating in evidence. For this, it is enough to do the inverse process of the distributive property as follows:
x (4 + a)
Note that by applying the distributive property on this factorization, we will have just the polynomial initial. See another example of the first factorization case:
4x3 + 6x2
4x3 + 6x2 = 2·2xxx + 2·3xx = 2xx (2x + 3) = 2x2(2x + 3)
For more information about this factoring case, see the text Factoring: Common factor in evidenceon here.
2nd case of factoring: grouping
It may be that, when placing factorscommon in evidence, the result is a polynomial which still has common factors. So, we must take a second step: bring common factors to the fore again.
Thus, factoring by grouping is pairfactorization by common factor.
Example:
xy + 4y + 5x + 20
at first factorization, we will highlight the common terms as follows:
y (x + 4) + 5(x + 4)
Note that the polynomial resulting has, in your terms, the common factor x + 4. putting it in evidence, we will have:
(x + 4)(y + 5)
For more information and examples about this case of factorization, see the text groupingclicking here.
3rd case of factorization: perfect square trinomial
This case is basically the opposite of productsremarkable. Note the noteworthy product below:
(x + 5)2 = x2 + 10x + 25
At perfect square trinomial factorization, we write polynomials expressed in this form as a remarkable product. See an example:
4x2 + 12xy + 9y2 = (2x + 3y)2
Note that you need to ensure that the polynomial is really a perfect square trinomial to do this procedure. Processes for this warranty can be found on here.
4th factorization case: difference of two squares
Polynomials known as two square difference have this form:
x2 - a2
Its factorization is the remarkable product known as product of sum for difference. Note the result of factoring this polynomial:
x2 - a2 = (x + a)(x - a)
For more examples and information about this case of factorization, Read the text two square difference on here.
5th factorization case: difference of two cubes
all polynomial grade 3 written in the form x3 + y3 Can be factored in the following way:
x3 + y3 = (x + y)(x2 – xy + y2)
For more examples and information about this case of factorization, Read the text two cube differenceon here.
6th case of factorization: Sum of two cubes
all polynomial grade 3 written in the form x3 - y3 Can be factored in the following way:
x3 - y3 = (x - y)(x2 + xy + y2)
For more examples and information about this case of factorization, Read the text sum of two cubeson here.
By Luiz Paulo Moreira
Graduated in Mathematics
Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/matematica/o-que-e-fatoracao-polinomios.htm