Factorization Notes

This article is translated from my website.

Definition#

The process of transforming a polynomial into the product of several polynomials within a certain range is called factorization.

Methods of Factorization#

Common Factor Method#

The common factor method is the most basic method of factorization, which involves extracting the common factors from each term in the polynomial. Before using any other method, the common factor method should be used to remove redundant parts to avoid affecting the subsequent problem-solving.

Formula Method#

When using other methods, it is important to decompose the higher-order factors completely and consider whether the formula method or other methods can be used for further factorization.

The method of decomposing a polynomial into factors using the reverse multiplication formula is called the formula method. Common formulas include:

Difference of squares:
$a^2-b^2=(a+b)(a-b)$
Perfect square:
$a^2 \pm 2ab+b^2=(a \pm b)^2$
Sum and difference of cubes:
$a³+b³=(a+b)(a²-ab+b²)$
$a³-b³＝(a-b)(a²+ab+b²)$
Trinomial square:
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Perfect cube:
$(a+b)³＝a³+3a²b+3ab²+b³$
$(a-b)³＝a³-3a²b+3ab²-b³$
Euler's formula and its corollary:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
If $a+b+c=0$ , then $a^3+b^3+c^3=3abc$
If $a^3+b^3+c^3=3abc$ , then $a+b+c=0$ or $a=b=c$
When encountering Euler's formula in factorization, the corollary is usually used together.

This method does not have any good examples, just apply the formulas.

Cross Multiplication Method#

Definition: The method of decomposing a quadratic or cubic polynomial into factors using cross multiplication is called the cross multiplication method.
Mnemonic: Split the first and last terms, multiply diagonally, and sum to get the middle term.

The cross multiplication method is to use the cross method to decompose expressions such as abx²+(an+bm)x+mn and abx²+(an+bm)+mny² into (ax+m)(bx+n) and (ax+my)(bx+ny) respectively.

After reading this example, you should know that cross multiplication is to split the constant term or quadratic term at the end (considering the smaller difference) so that they can add up to the coefficient of the middle term.

If the coefficient of the x² term is not one, then we can split the x² term into the product of two terms with x, such as 4x²=2x·2x=x·4x, which requires more consideration. If one method doesn't work, don't doubt yourself, it may be that this method cannot solve the problem.

Grouping Method#

Definition: The method of factorization using grouping is called the grouping method.
For example: ax+ay+bx+by=(ax+ay)+(bx+by)=a(x+y)+b(x+y)=(a+b)(x+y)
There is no unique grouping method: ax+ay+bx+by=(ax+bx)+(ay+by)=x(a+b)+y(a+b)=(x+y)(a+b)

Decomposition steps:

Group the polynomial based on letters, coefficients, and exponents, and factorize each group separately for continuous development.
After grouping, further factorize using other methods.

Example 1: Factorize: 4xy+2x+2y+1
Original expression=(4xy+2x)+(2y+1)
=2x(2y+1)+(2y+1)
=(2x+1)(2y+1)

Example 2: Factorize: ac²+bd²-ad²-bc²
Approach: Group the terms with c² together and the terms with d² together, and then factorize each group separately.
Original expression=(ac²-bc²)+(bd²-ad²)
=c²(a-b)-d²(a-b)
=(a-b)(c²-d²) //Note whether it is completely factorized
=(a-b)(c+d)(c-d)

Example 3: Factorize: x⁵+x⁴+x³+x²+x+1
Approach: Group the terms based on their exponents, three terms per group, and then factorize each group separately.
Original expression=(x⁵+x⁴+x³)+(x²+x+1)
=x³(x²+x+1)+(x²+x+1)
=(x+1)(x²-x+1)(x²+x+1)

Principal Element Method#

Definition#

The principal element method is used to decompose algebraic expressions with multiple variables. In this method, one of the variables is selected as the principal element (element refers to the unknown), and it is treated as a constant. The algebraic expression is then rearranged into a polynomial in descending or ascending order of the principal element, and other methods are used to further factorize it.

Factorization Steps#

Choose a principal element, here we use x, and rearrange the expression into the form of ax²+bx+c.
Use other methods, such as cross multiplication, to further factorize the rearranged expression.

Example#

Example 1: Factorize: x²-mx²+mx-3x+2
Approach: In this case, if the principal element m is used, there will be no quadratic term left. It is simpler than using the principal element x.
Original expression=(-x²+x)m+x²-3x+2
=-x(x-1)m+(x-1)(x-2)
=(x-1)(x-mx-2)

Example 2: Factorize: x²-a(3x-2a+b)-b²
Principal element a
Original expression=x²-3xa+2a²-ba-b²//It is recommended to put the principal element at the end of each term, which is clearer
=2a²-(3x+b)a+x²-b²
=2a²-(3x+b)a+(x+b)(x-b)
=(x-b)(2a-x+b)

Split and Fill Method#

Concept#

Splitting means splitting an algebraic expression into several terms of algebraic sum, such as $a^2=2a^2-a^2$
Filling means filling in two opposite terms in an algebraic expression, which is called filling. For example, $a^2+1=a^2+2a+1-2a$

When factorizing a given polynomial, if it is difficult to directly factorize by grouping, the split and fill method can often be used to create conditions for extracting common factors or applying the formula method, making it easier to factorize some terms and facilitate factorization using the grouping method.

Example#

Example 1: Factorize: $a^3-4a+3$
Solution:
Method 1 (Filling):
Original expression $=a^3-a^2+a^2-4a+3$
$=a^2(a-1)+(a-1)(a-3)$
$=(a-1)(a^2+a-3)$
Method 2 (Splitting):
Original expression $=a^3-a-3a+3$
$=a(a+1)(a-1)-3(a-1)$
$=(a-1)(a^2+a-3)$

Example 2: Factorize: $a^4+2a^3b+3a^2b^2+2ab^3+b^4$
Solution: Original expression $=(a^4+a^3b+a^2b^2)+(a^3b+a^2b^2+ab^3)+(a^2b^2+ab^3+b^4)$
$=a^2(a^2+ab+b^2)+ab(a^2+ab+b^2)+b^2(a^2+ab^2)$
$=(a^2+ab+b^2)^2$

Substitution Method#

Substitution refers to using other variables (elements) to represent expressions when factorizing expressions with the same or related terms. This often makes the original expression clearer, reduces the number of terms in the polynomial, and simplifies complex problems.

Example: Factorize: $(x^2+x+1)(x^2+x+2)-12$
Solution: Let $t=x^2+x$ ,
Then the original expression $=(t+1)(t+2)-12$
$=t²+3t+2-12$
$=t²+3t-10$
$=(t-2)(t+5)$
$=(x²+x-2)(x²+x+5)$
$=(x-1)(x+2)(x²+x+5)$

Trial and Error Method#

Introduction to the trial and error method, you need to know the following concepts: algebraic expressions about x, remainder theorem, and factor theorem.

Algebraic Expressions about x#

Algebraic expressions about x can be represented as f(x) or g(x), etc.
Here, f represents a series of operations on the letter inside the parentheses, and f(3) represents the value of the algebraic expression f(x) when x=3.

Remainder Theorem and Factor Theorem#

The quotient of dividing polynomial f(x) by polynomial g(x) is q(x), and the remainder is r(x).
[latex]f(x) \div g(x)=q(x) \cdots r(x)[/latex], which can be written as f(x)=g(x)q(x)=r(x).
When g(x) is a linear polynomial (ax+b), the remainder r(x) can only be a constant, and in this case, the remainder is called the remainder, denoted as r, so it can be written as f(x)=(ax+b)·q(x)+r

Remainder theorem: The remainder of dividing polynomial f(x) by (ax+b) is equal to [latex]f(-\frac{b}{a})[/latex].
Factor theorem: If the polynomial f(x) has a factor (ax+b), then [latex]f(-\frac{b}{a})=0[/latex], and vice versa.

From this, we can see that the factor theorem is a special case of the remainder theorem when r=0.

Definition#

By using the factor theorem and long division, we can obtain another method of factorization, called the trial and error method.

(1) For a polynomial with integer coefficients $f(x)=a_0x^{n}+a_1x^{n-1}+…+a_{n-1}x+a_n$ :
Property 1: If the leading coefficient $a_0=1$ and it has a factor (x-q) (q is an integer), then q must be a factor of the constant term $a_n$
Property 2: If the leading coefficient $a_0\ne1$ and it has a factor (px-q) (p and q are coprime), then p must be a factor of the leading coefficient $a_0$ , and q must be a factor of the constant term $a_n$
(2) Trial and Error: When factorizing an integer coefficient polynomial $f(x)$ , according to the factor theorem, if a root $x=q$ or $x=\frac{q}{p}$ of the equation $f(x)=0$ can be found, the remaining part can be obtained by long division. However, it is important to note whether it is completely factorized.

However, it should be noted that even if rational roots cannot be found, it does not mean that the original expression cannot be factorized within the range of rational numbers. It can only indicate that there is no linear factor.

Example

Example 1: Factorize: $x⁵+5x⁴-5x³-25x²+4x+20$
Solution: Trial and error: When $x=±1$ , the original expression is 0. Therefore, the original expression must have the factor (x+1)(x-1).
Using long division,
Original expression $=(x+1)(x-1)(x³+5x²-4x-20)$
$=(x+1)(x-1)[x²(x+5)-4(x+5)]$
$=(x+1)(x-1)(x+5)(x²-4)$
$=(x+1)(x-1)(x+2)(x-2)(x+5)$

Example 2: Factorize: $6x^4+5x^3+3x^2-3x-2$
Trial and error: When $x=-\frac{1}{2},x=\frac{2}{3}$ , the original expression is 0. Therefore, the original expression must have the factor (2x+1)(3x-2).
Using long division, the original expression is $(2x+1)(3x-2)(x^2+x+1)$

Example 3: Factorize: $3x^3-5x^2y+x^2y+x^3$
Trial and error: When $x=y$ , the original expression is 0. Therefore, the original expression must have the factor (x-y).
Using long division,
Original expression $(x-y)(3x^2-2xy-y^2)$
$(x-y)^2(3x+y)$

Cyclic Symmetry Method#

Concept#

A symmetric expression refers to a polynomial with several elements, and the original polynomial remains unchanged when any two elements are exchanged.
A cyclic expression refers to a polynomial in which the original polynomial remains unchanged when the elements are cyclically permuted in any order.
A homogeneous expression refers to a polynomial in which all terms have the same degree n. Such a polynomial is called an nth degree homogeneous polynomial.

Cyclic and symmetric expressions have the following properties:

Symmetric expressions are always cyclic expressions, but cyclic expressions are not necessarily symmetric expressions.
The sum, difference, product, and quotient (when the divisor is not zero) of cyclic or symmetric expressions about the same letter are still cyclic or symmetric expressions.
If a cyclic or symmetric expression contains a certain form of expression, there must be expressions of the same type.
For example, a cyclic symmetric expression about x, if it contains an ax² term, there must be ay² and az² terms; if it contains a bxy term, there must be a byz and a bzx term.

Solution Steps

Trial and Error

Trial and error means making the original expression equal to 0, as mentioned in the "Trial and Error Method" above.
Trial and error has a range, as shown in the table below:

Trial	Factor
x=0	xyz
x=y	(x-y)(y-z)(z-x)
x=-y	(x+y)(y+z)(z+x)
x=y+z	(x-y-z)(y-z-x)(z-x-y)
x=-(y+z)	(x+y+z)

Compare Degrees

Subtract the degrees of the original expression from the degrees of the factors that must exist, and then multiply the factors obtained by the factors corresponding to the difference in degrees.
The factors corresponding to the difference in degrees are shown in the table below:

Degree	Factor
First degree	a(x+y+z)
Second degree	a(x²+y²+z²)+b(xy+yz+zx)
Third degree

Compare Coefficients

By using methods such as assigning values and comparing coefficients, calculate the coefficients of the factors to complete the factorization.

Example

Example 1: Factorize: $(a+b+c)^4-(a+b)^4-(b+c)^4-(c+a)^4+a^4+b^4+c^4$
Solution: Trial and error: When a=0, the original expression is 0.
According to the factor theorem and the properties of cyclic expressions, the original expression must have the factor abc.
Compare degrees, assume the original expression is kabc(a+b+c)
Let a=b=1,c=-1, we get 1-16-0+1+1+1=-k
Solve k=12
∴The original expression is 12abc(a+b+c)

Example 2: Factorize: $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc$
Solution: Trial and error: When a=-b, the original expression is 0.
According to the factor theorem and the properties of cyclic expressions, the original expression must have the factor (a+b)(b+c)(c+a)
Compare degrees, assume the original expression is k(a+b)(b+c)(c+a)
Let a=0,b=1,c=2, we get 4+2=6k
∴k=1
∴The original expression is (a+b)(b+c)(c+a)

Method of Undetermined Coefficients#

This method is not very important, and the process is complicated. Factorization is just a small step in solving practical problems. If the method of undetermined coefficients is used, the problem will become more complex, so it is omitted.