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195 LearnersLast updated on October 21, 2025

When planning your sister’s birthday, if each of 40 friends gets 6 snacks, the total is 6 × 40 = 240. Writing a number or expression as a multiplication of its factors, like 6 and 40, is called the factored form.
The process of dividing a number or algebraic expression to its relevant factors is known as factoring. When you multiply these factors, it should give back the original expression.
For example:
\(x² – 7x + 12\)
We have to find two numbers whose product is 12 and sum is -7.
We have the numbers: -3 and -4. So,
\(x2 – 7x + 12 = (x - 3) (x - 4)\)
Verify the factoring by multiplying the factors together.
\((x - 3) (x - 4) = x^2 – 4x -3x + 12 = x^2 - 7x + 12\).
Since we get back the original expression, the factorization is correct.
Factoring plays a major role in algebra and is done through a series of steps. Let’s go through each steps in detail:
The largest expression in an algebraic expression is the greatest common factor(GCF), which includes both the variables and numerical coefficients; they can divide each term exactly without leaving a remainder.
For example:
Factor: \(12x^2y – 8xy^2 + 16xy\)
Solution:
To simplify an algebraic expression, we first need to simplify its GCF.
GCF of all terms = 4xy
Let’s now factor it out:
\(4xy(3x – 2y + 4)\)
Factoring by grouping works well for polynomials that have four terms. In this, the terms are grouped into two pairs, and the common factor is taken out from each group.
For example:
Factor: \(x³ + 2x² + 3x + 6\)
Solution:
Group the terms:
\((x³ + 2x²) + (3x + 6)\)
Take out the common factor from each group of terms.
\(x³(x + 2) + 3(x + 2)\)
Now, we factor out further:
\((x + 2)(x² + 3) \)
Here, \(x^2 + 3\) cannot be factorized more.
A trinomial is an algebraic expression made up of three terms. Let’s explore how to factor a quadratic expression using the AC method.
For example:
Factor: \(x² + 7x + 10\)
Solution:
Find two numbers whose product is 10 and sum is 7.
2 and 5
Rewrite the middle term using the factors:
\(x² + 2x + 5x + 10\)
We then group the terms into pairs
\((x² + 2x) + (5x + 10)\)
Factor out each group:
\(x(x + 2) + 5(x + 2)\)
Factor out further for the common binomial:
\((x + 2)(x + 5)\)
This method can be used only in cases where both terms are perfect squares separated by a minus sign.
Formula: \(a² - b² = (a + b)(a - b)\)
For example:
Factor: \(x² - 49\)
Solution:
Rewrite it as a difference of squares:
\(x² - 7² = (x + 7)(x - 7)\)
An expression with three terms is called a perfect square trinomial. A perfect square trinomial will match any one of the following forms:
\(a² + 2ab + b² = (a + b)²\)
\(a² - 2ab + b² = (a - b)²\)
Example:
Factor: \(x² + 8x + 16\)
Solution:
Identify the pattern:
\(x² + 2·4·x + 4² = (x + 4)²\)
These expressions follow the specific formulas which are given below:
\(a³ + b³ = (a + b)(a² - ab + b²)\)
\(a³ - b³ = (a - b)(a² + ab + b²)\)
For example:
Factor: \( x³ + 64\)
Solution:
Identify it as a sum of cubes:
\(x³ + 64 = x³ + 4³\)
The formula we use: \(a³ + b³ = (a + b)(a² - ab + b²)\)
Applying the formula:
\(x³ + 4³ = (x + 4)(x² - 4x + 16)\)
This method applies to factor higher-degree polynomials. Here, we substitute a repeated power with a variable.
For example:
Factor: \(x⁴ + 2x² – 15\)
Solution:
Let \(y = x²\)
So the expression becomes:
\(y² + 2y - 15\)
Now, write the expression as a multiplication of its factors:
\((y + 5)(y - 3)\)
Here, we substitute x² back for y:
\((x² + 5)(x² - 3)\)
Factored form represents an expression as a product of its factors. This method can be a little tricky for some students. We will now go through some simple tricks to help you master the concept effectively.
Factoring is a fundamental concept in mathematics. However, students often make mistakes when factoring. Here are a few common mistakes and tips to avoid them:
The factored form is a useful method for representing algebraic expressions as a product of their factors. This concept is not confined to mathematics; it has widespread practical applications in real life. Let’s now learn how it can be applied in real-world situations.
Factor: 12x + 8
\(4(3x + 2)\)
First identify the greatest common factor (GCF) of both terms:
12x and 8 → GCF = 4
Let’s now factor out the GCF:
\(12x ÷ 4 = 3x\)
\(8 ÷ 4 = 2\)
Therefore, the simplified expression is:
\(12x + 8 = 4(3x + 2)\)
Factor: x² - 25
\((x + 5)(x - 5)\)
Identify that the given expression represents a difference of squares.
x² is (x)² and 25 is (5)²
Using the formula:
\(a² − b² = (a + b)(a − b)\)
Now, apply the formula:
\(x² − 25 = (x + 5)(x − 5)\)
So, the final expression we get is:
\(x² - 25 = (x + 5)(x - 5)\).
Factor: x³ + 3x² + x + 3
\((x + 3)(x² + 1)\)
Let’s first group the terms:
\((x³ + 3x²) + (x + 3)\)
We now factor each group:
\(x²(x + 3) + 1(x + 3)\)
Factor further for the common binomial:
\((x + 3)(x² + 1)\)
Simplifying the expression:
\(x³ + 3x² + x + 3 = (x + 3)(x² + 1)\)
Factor: 3x² - 12x
\(3x(x - 4)\)
Start by finding the greatest common factor(GCF):
GCF of 3x² and \(12x = 3x\)
Now, we factor out the GCF:
\(3x² ÷ 3x = x\)
\(12x ÷ 3x = 4\)
So the simplified expression is:
\(3x² - 12x = 3x(x - 4)\).
Factor: x³ - 27
\((x - 3)(x² + 3x + 9)\)
Identify that the given expression is a difference of cubes:
\(x³ = (x)³, 27 = (3)³\)
Using the formula:
\(a³ - b³ = (a - b)(a² + ab + b²)\)
Substitute the values into the formula:
\(x³ - 27 = (x - 3)(x² + 3x + 9)\)
Factoring the expression:
\(x³ - 27 = (x - 3)(x² + 3x + 9)\).
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.






