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Last updated on October 16, 2025
Expressions whose product gives the original polynomial are called its factors. Using techniques such as common factoring, factoring by grouping, or applying special identities like difference of squares, a polynomial can be factored into linear or quadratic expressions.
The opposite of multiplying factors of polynomials is factoring the polynomial. A polynomial of degree “n” in variable x is an expression of the form axn+bxn-1+kcxn-2+ ... +kx+i, where each variable has a constant associated with it as its coefficient. Therefore, a polynomial is an expression made up of one or more terms, where each term has a constant multiplied by a variable to a whole-number power, and the terms are separated by addition or subtraction.
What is Factorization of a Polynomial?
Factorization is the process of determining the numbers or expressions whose product is the original value.
Likewise, when it comes to polynomials, the factors are the polynomials that are multiplied to create the original polynomial. The factors of x2+5x+6 are, for example, (x + 2) (x + 3). The original polynomial is produced when multiplying them. We can also determine the polynomial’s zeros after factorization. The zeros are x = -2 and x = -3.
Polynomials can be factorized using six techniques.
How to Solve Polynomials?
Polynomials can be factorized in several ways. Let’s talk about these techniques.
Greatest Common Factor
To factorize a polynomial, first find the greatest common factor of the polynomial. This procedure is simply the distributive law applied in reverse, like
p(q + r)=pq + pr
However, factorization is simply the opposite process;
pq+pr=p(q+r)
Where p is the greatest common factor.
Factoring Polynomials by Grouping
Another name for this method is factoring by pairs. Here, the zeros are found by distributing or grouping the given polynomial in pairs. Let us look at an example.
Factorize x2-15x+50
Now, find the two numbers which, when added, give -15 and when multiplied, give 50.
-5 and -10 are the two numbers such that,
(-5)+(-10)=-15
(-5) × (-10)=50
Now, the polynomial is as follows,
x2-5x-10x+50
x(x-5)-10(x-5)
Let us take x-5 as a common factor, and then we get:
(x-5)(x-10)
Therefore, the factors are (x-5) and (x-10).
Factoring Polynomial with Four Terms
Let us now see how polynomials with four terms are factorized. For example, the polynomial is x3+x2-x-1
Now, separate the given polynomial into two.
(x3+x2)+(-x-1)
Now find the highest common factor and take that factor out of the bracket.
Here x2 is the greatest common factor, and from the second part, we can factor out -1. Thus,
x2(x+1)-1(x+1)
x2(x+1)-1(x+1)=(x2-1)(x+1)
=(x-1)(x+1)(x+1)
=(x-1)(x+1)2
Now, group the terms as factors.
(x2-1)(x+1)
Thus, the factorization of x3+x2-x-1 gives (x2-1)(x+1)
This shows how factors help in solving problems in real-world problems. Let us see some applications of the factors of polynomials.
Students often make mistakes, such as overlooking common factors and using incorrect factoring methods. Let us look at those mistakes and how to quickly correct them.
Factorize x2+5x+6
(x+2)(x+3)
Since the numbers two and three multiply to six and add to five. The polynomial can be factored as (x+2)(x+3)
Factor the polynomial x2-36
(x-6)(x+6)
This is an example of the difference between squares. 36 and x2 are perfect squares, and the formula a2-b2=(a-b)(a+b) is used in this case. Thus,x2-36=(x-6)(x+6).
Factorize 3x2+12x
3x(x+4)
First, find that 3x is the greatest common factor (GCF) between the two terms. The result of factoring that out is 3x2+12x=3x(x+4). The expression becomes simpler and facilitates additional analysis when the GCF is factored out.
Factorize x2+4x+4
(x+2)2
The trinomial x2+4x+4 is a perfect square. The square root of x2 is x, and the square root 4 is 2. Since 2 × x × 2=4x, it fits the pattern a2+2ab+b2=(a+b)2. Therefore, x2+4x+4=(x+2)2
Factorize x3-3x2-x+3
(x-1)(x2-2x-3)=(x-1)(x-3)(x+1)
To factor the polynomial x3-3x2-x+3, group the terms as (x3-3x2)+(-x+3). Factor every group: x2(x-3)-1(x-3). Factor out the common binomial (x-3) to obtain (x2-1)(x-3). Then factor x2-1 as a difference of squares: (x-1)(x+1). As a result, (x-1)(x+1)(x-3) is the final factorized form.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.