Factorization was first considered for integers by an ancient Greek mathematician. He proved the Main Theorem of Arithmetic that every positive integer can be factored into a product of prime numbers, and integers greater than 1 cannot be further factored. Moreover, this factorization is unambiguous down to the order of the factors. Integer factorization is a type of inverse function of multiplication. Polynomial factorization has also been studied for centuries. In particular, univariate polynomials with complex coefficients allow a unique factorization (up to a degree) into linear polynomials, which is a version of the Main Theorem of Algebra. In this case, the factorization can be performed using a root-finding algorithm. In this article, we will see various solved examples and practice questions.

What is Factorization?

In mathematics, factoring is breaking a number down into smaller numbers that can be multiplied together to get the original number. Dividing a number into factors or divisors is called factoring. For example, factoring the number 12 gives 3 times 4.

The six methods are as follows:

• Greatest Common Factor (GCF)
• Grouping Method
• Sum or Difference in Two Cubes
• Difference in Two Squares Method
• General Trinomials
• Trinomial Method

Factoring in mathematics is the process of expressing a mathematical expression as a product of simpler expressions or factors. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12. The number 12 can be expressed as a product of its factors.

For example, 12 = 1 × 12, 2 × 6, or 4 × 3.

Important Formulas

• a² – b² = (a – b)(a + b)
• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• a³ – b³ = (a – b)(a² + ab + b²)
• a³ + b³ = (a + b)(a² – ab + b²)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ca

Practice Questions on Factorization

Q1: Factor: x² + 5x + 6

Solution:

=x² + 5x + 6

= x² + (3 + 2)x + 6

= x² + 3x + 2x + 6

= x(x + 3) + 2(x + 3)

= (x + 2)(x + 3)

Q1: Factor completely: 12x² – 27

Solution:

Steps:

First, factor out the greatest common factor: 3(4x^2 – 9)

The remaining expression is a difference of squares: a^2 – b^2 = (a+b)(a-b)

Here, a = 2x and b = 3

So we get: 3(2x – 3)(2x + 3)

3).Factor: 2x² + 7x – 15

Solution: (2x – 3)(x + 5)

Steps:

We’re looking for two numbers that multiply to give -30 and add up to 7

These numbers are -3 and 10

So we can factor it as (2x – 3)(x + 5)

4).Factor the difference of squares: 49a^2 – 64b^2

Solution: (7a + 8b)(7a – 8b)

Steps:

This is a difference of squares: a^2 – b^2 = (a+b)(a-b)

Here, a = 7a and b = 8b

So we get: (7a + 8b)(7a – 8b)

5).Greatest Common Factor (GCF)

Problem: Factor 12x²y + 18xy² – 24x²

Solution:

Step 1: Identify the GCF of all terms

GCF = 6xy

Step 2: Factor out the GCF

12x²y + 18xy² – 24x² = 6xy(2x + 3y – 4x)

Step 3: Simplify

6xy(2x + 3y – 4x) = 6xy(-2x + 3y)

Therefore, 12x²y + 18xy² – 24x² = 6xy(-2x + 3y)

6).Complex Factoring

Problem: Factor x⁴ – 16y⁴

Solution:

Step 1: Recognize this as a difference of squares

x⁴ – 16y⁴ = (x²)² – (4y²)²

Step 2: Apply the difference of squares formula

(x² + 4y²)(x² – 4y²)

Step 3: The second factor is also a difference of squares

(x² + 4y²)(x + 2y)(x – 2y)

Therefore, x⁴ – 16y⁴ = (x² + 4y²)(x + 2y)(x – 2y)

7).Factoring with Fractional Coefficients

Problem: Factor 3/2x² – 5/4x – 3/8

Solution:

Step 1: Multiply all terms by the LCD (8) to eliminate fractions

8(3/2x² – 5/4x – 3/8) = 12x² – 10x – 3

Step 2: Factor this trinomial

Find two numbers that multiply to give ac (12×-3=-36) and add up to b (-10)

These numbers are -12 and 2

Step 3: Rewrite the middle term: 12x² – 12x + 2x – 3

Step 4: Group terms: (12x² – 12x) + (2x – 3)

Step 5: Factor out common factors: 12x(x – 1) + 1(2x – 3)

Step 6: Factor out the greatest common binomial:

(6x + 1)(2x – 3)

Step 7: Divide by 8 to return to the original terms

(3/4x + 1/8)(2x – 3)

Therefore, 3/2x² – 5/4x – 3/8 = (3/4x + 1/8)(2x – 3)

8).Given that one of the factors of (5×2 + 70x – 160) is (x – 2), find the other factor.

Solution

We’re given that (x – 2) is one factor of (5x² + 70x – 160).

Let’s call the other factor (ax + b), where a and b are constants we need to find.

So, we can write:

5x² + 70x – 160 = (x – 2)(ax + b)

Let’s expand the right side:

(x – 2)(ax + b) = ax² + bx – 2ax – 2b

= ax² + (b – 2a)x – 2b

Comparing coefficients with the original expression:

a = 5

b – 2a = 70

-2b = -160

From the first equation, we know a = 5.

From the last equation:

-2b = -160

b = 80

We can verify using the middle equation:

b – 2a = 70

80 – 2(5) = 70

80 – 10 = 70

70 = 70 (This checks out)

Therefore, the other factor is (5x + 80)

So, we can factor 5x² + 70x – 160 as:

(x – 2)(5x + 80)

You can verify this by multiplying these factors:

(x – 2)(5x + 80) = 5x² + 80x – 10x – 160 = 5x² + 70x – 160

Therefore, the other factor is (5x + 80).

9).x² + 5x + 6 = 0. (Factor Trinomials)

Solution

(x + 2)(x + 3) = 0

This factored form allows us to find the roots of the equation by setting each factor to zero:

x + 2 = 0 or x + 3 = 0

Solving these, we get:

x = -2 or x = -3

10).If the factorization of a number is 22 × 32 × 5. Find the number by using the factorization formula.

Solution:

To find: The factorized number

Given:

The factorization of a number = 2 × 2 × 3 × 3 × 5.

Using the Factorization Formula,

Factorization Formula for any number, N = Xa × Yb × Zc

= (2 × 2 × 3 × 3 × 5)

= 22 × 32 × 5

= 180

Practice Questions

1).Factor: x^2 + 11x + 30

2).Factor completely: 24a^2 – 54b^2

3).Factor: 2y^3 – 54y

4).Factorise 11×2 + 33x − 110

5).Factor the difference of squares: 49m^2 – 1

6).Factor completely: x^4 – 16

7).Factor: 3z^2 – 7z – 6

8).Factor the sum of cubes: x^3 + 27

9).Find the factors of 9×2 +4y2 +12xy.

10).Factor completely: 6a^2 + 13a – 5

FAQs

How to solve a problem in factorization?

Some problems require you to solve quadratic equations. To solve these problems, you set the coefficients to 0 and solve for x. To solve this problem, you need to set all the coefficients to 0 and solve for x. There are three solutions

What is the important formula of factorisation?

Some important factoring formulas are given as, (a + b)2 = a2 + 2ab + b. (a – b)2 = a2 – 2ab + b. (a + b) (a – b) = a2 – b.

What are the factors of 36?

36 has a total of 9 factors: 1, 2, 3, 4, 6, 9, 12, 18, and 36. There is a trick to calculate the total number of factors of a number. For example, 36 = 2 × 2 × 3 × 3 = 2² × 3². So the answer is 2² × 3²