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110 LearnersLast updated on October 30, 2025

Factoring polynomials means breaking them into smaller expressions that multiply together to form the original polynomial. We can use different methods, like finding the greatest common factor (GCF) or dividing polynomials. This article will help us learn how to factor polynomials.
A polynomial is an algebraic expression made up of terms that include constants and variables raised to non-negative integer exponents, combined using addition, subtraction, or multiplication.
They do not include square roots, negative powers, or variables in denominators, but include operations like addition, subtraction, and multiplication.
Factoring a polynomial requires breaking it down into smaller factors that, when multiplied, give the original expression. This method helps solve equations and better understand the behavior of the polynomial.
A polynomial is typically represented as axn + bxn-1 +... + px + q and can be factored using grouping, substitution, or identities.
The highest power of x, called the degree, helps determine how many solutions (or zeros) the polynomial can have. Factoring changes a complicated polynomial into smaller pieces, like expressions with x or x², that are easier to work with.
For instance, x2 + x(a + b) + ab becomes (x + a)(x +b)
To factorize a polynomial, follow these steps:
Step 1: Look at all the terms and take out the greatest common factor (GCF), if there is one.
Step 2: Use a factoring method that fits the expression, such as grouping, using algebraic identities, or trying substitution.
Step 3: Rewrite the polynomial as a product of the smaller expressions.
The six different types of methods used for factorizing a polynomial are listed below:
In this section, we will focus on three main methods used to solve polynomials.
Greatest Common Factor
In this method, we take out the largest number (and variable, if any) that all the terms have in common. It’s like reversing the distributive property.
Distributive property:
p(q + r) = pq + pr
Factored form:
pq + pr = p(q + r), where p is the greatest common factor.
Example: Factor the polynomial 12x + 18
Step 1: Find the GCF
Break each term into its factors:
\(12x = 3 × 4 × x\) or \(6 × 2 × x\)
\(18 = 3 × 6\)
So, the GCF is 6.
Step 2: Factor it out
\(12x + 18 = 6(2x + 3)\)
Step 3: Check by distributing
\(6(2x + 3) = 6 \times 2x + 6 \times 3 = 12x + 18\)
Factoring polynomials by grouping
This method works well while trying to factor trinomials (three-term expressions) that can’t be factored just by taking out a common factor. The idea is to split the middle term into two parts so we can group the terms and factor in pairs.
Example: Factor \(x^2 + 11x + 24 = (x + 3)(x + 8)\)
Step 1: Find two numbers
Look for two numbers that:
Add up to 11 (the middle term), and
Multiply by 24 (the last term)
The numbers 3 and 8 work because:
\(3 + 8 = 11\)
\(3 × 8 = 24\)
Step 2: Rewrite the middle term using these two numbers
\(x^2 + 11x + 24 = x^2 + 3x + 8x + 24\)
Step 3: group the terms and factor
\((x^2 + 3x) + (8x + 24)\)
\(x(x + 3) + 8(x + 3) \)
Take the common binomial factor,
\((x + 3)(x + 8)\)
So, \((x + 3)(x + 8)\) are the factors of \(x^2 + 11x + 24\)
Factoring using Identities
Algebraic identities can also help in factorizing polynomials. Commonly used identities for this purpose are:
\((a + b)^2 = a^2 + 2ab + b^2\)
\((a - b)^2 = a^2 - 2ab + b^2\)
\(a^2 - b^2 = (a + b)(a - b)\)
Let’s take examples for each identity listed above.
Example 1: Factorize \(x^2 + 6x + 9\), using (a + b)2
Step 1 - Compare with the identity
\(x^2 + 6x + 9 = x^2 + 2 \times 3 \times x + 3^2\)
We can see the pattern a2 + 2ab + b2, where a = x and b = 3
Step 2 - Apply the identity
\(x^2 + 6x + 9 = (x + 3)^2\)
So, (x + 3)2 is the answer.
Example 2: Factor \(x^2 - 10x + 25\) using identity (a - b)2
Step 1 - Match with the identity
\(x^2 - 10x + 25 = x^2 - 2 \times 5 \times x + 5^2\)
Here, a = x and b = 5
Step 2 - Apply the identity
\(x^2 - 10x + 25 = (x - 5)^2\)
(x - 5)2 is the answer.
Example 3: Factor x2 - 49 using identity a2 - b2
Step 1 - It needs to be the difference between two perfect squares
\(x^2 - 49 = x^2 - 7^2\)
Step 2 - Apply Identity a2 - b2 = (a + b)(a - b)
So, \(x^2 - 49 = (x + 7)(x - 7)\)
Factoring polynomials can seem tricky at first, but with the right techniques and a little practice, it becomes simple. These tips and tricks make the process faster and more accurate.
The steps involved in factoring polynomials can sometimes be confusing. This might cause mistakes; however, many of these mistakes can be avoided with the right knowledge and practice. Let’s see some common mistakes students make and how to avoid them.
Factoring polynomials is an important algebraic concept that helps design structures and predict profits. It has a wide range of real-life applications, and some of them are mentioned below:
Factor the polynomial 6x3 + 12x2
6x2(x + 2)
The GCF of 6x3 and 12x2 is 6x2
After factoring, 6x3 + 12x2 = 6x2(x + 2)
Factor the trinomial x2 + 7x + 12
(x + 3)(x + 4)
Since we are factoring a trinomial, we will split the middle term.
The two numbers are 3 and 4 because\( 3 + 4 = 7\) and \(3 × 4 = 12.\)
So, \(x^2 + 7x + 12 = (x + 3)(x + 4)\)
Factor the polynomial x2 + 5x + 3x + 15
(x + 5)(x + 3)
Group the terms in like pairs,
\((x^2 + 5x) + (3x + 15)\)
Then, factor each group,
\(x(x + 5) + 3(x + 5)\)
And now, factor out the common binomial,
\((x + 5)(x + 3)\)
Factor the expression 9x2 - 16
(3x + 4)(3x - 4)
We use the identity \(a^2 - b^2 = (a + b)(a - b)\)
\((3x)^2 - 4^2 = (3x + 4)(3x - 4)\)
Factor the expression x2 - 10x + 25
(x - 5)2
We can use the identity \(a^2 - 2ab + b^2 = (a - b)^2\)
Here, x2 = a2 so, a = x
-10x = - 2ab, then b = 5
25 = b2, so b = 5
We see that all conditions are matching, so
\(x^2 - 10x + 25 = (x - 5)^2\)
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.






