Vedic Mathematics is an ancient system of mathematics as it provides us the shortcut tricks to solve the mathеmatical operation quickly. There are many tricks of Vedic mathematics that can help us solve complex math problems in less than a few seconds. Vedic mathematics was founded between AD 1911 and 1918 by Jagadguru Shri Bharathi Krishna Tirthaji. It consists of 16 formulas and 13 sub-formulas, It can be used to solve arithmetic, algebra, geometry trigonometry, etc. In this article, we will explore special multiplication methods for multiplying binomials.

Multiplying Binomials: Step-by-Step Procedure

Here, we will use the Urdhva-Tiryagbhyam Sutra to solve binomial multiplication. This method uses vertical and crosswise formulas. It breaks down the multiplication process into distinct steps, making it easier to perform mental calculations and arrive at accurate results in a few seconds. Let’s deep dive into each step in detail.

Urdhva-Tiryagbhyam Sutra (Vertically and Crosswise):

Let’s consider two binomials: (2x + 3) and (4x + 5).

Step 1: Write one binomial under the other as shown below.

2x + 3
4x + 5

First multiply vertically on right-hand side column,

3 * 5 = 15

Step 2: Then multiply crosswise and add the products,

(5 * 2x) + (4x * 3) = 10x + 12x = 22x

Step 3: At last multiply vertically on left-hand side column,

2x * 4x = 8x2

So, the answer is = 8x² + 22x + 15

Vedic Maths: Multiplying Binomials

Subtopics

1. Special Cases:

In this case, we’ll see some special situations where multiplying binomials can be simpler due to common terms or coefficients in a expression.

• Common Factors: When both the binomials have something in common, like numbers or variables, then multiplying becomes much easier. For example, think about (2x + 3)(2x + 5). We’ll learn how to use these shared elements to make calculations smoother. The same goes for expressions like (3a – 4)(a – 4).
• Using Distribution: By spreading out each term of one binomial across the terms of the other, we can simplify multiplication. Imagine (x + 2)(x – 3) or (2a – 5)(3a + 4). We’ll solve this in further examples.

2. Difference of Squares:

Sometimes, we’ll see expressions like (a + b)(a – b), which have a special pattern. Understanding how to work with this pattern using Vedic Mathematics makes multiplication much easier.

• Square Differences: We’ll dig into multiplying binomials that fit the “difference of squares” pattern, like (x + 3)(x – 3) or (2a + 5)(2a – 5). You’ll see how the Vedic technique directly applies to these situations.

3. Application in Algebraic Expressions:

This subtopic demonstrates how the Vedic Mathematics technique can help simplify complex algebraic expressions that involved in real-world problem.

• Real-world Problem Solving: We’ll use real-world examples to demonstrate how this technique helps solve problems involving algebraic expressions with binomials. You might see how it works when finding the area of rectangles with binomial side lengths or analyzing financial investments with binomial terms.

Practice Questions

Here are some examples to understand the topic well with their solution.

Que 1: Multiply (x + 4)(x + 6).

Solution:

Step 1: Write one binomial under the other as shown below.
x + 4
x + 6
First multiply vertically on right-hand side column,
4 * 6 = 24
Step 2: Then multiply crosswise and add the products, 
(6 * x) + (x * 4) = 6x + 4x = 10x

Step 3: At last multiply vertically on left-hand side column,
x * x = x2

So, the answer is = x2 + 10x + 24

Que 2: Multiply (2a + 3)(2a - 3).

Solution:

Step 1: Write one binomial under the other as shown below.
2a + 3
2a - 3
First multiply vertically on right-hand side column,
3 * -3 = -9
Step 2: Then multiply crosswise and add the products, 
(-3 * 2a) + ( 2a * 3) =  -6a + 6a = 0

Step 3: At last multiply vertically on left-hand side column,
2a * 2a = 4a2

So, the answer is = 4a2 - 9

Que 3: Multiply (3x + 2)(2x + 5).

Solution:

Step 1: Write one binomial under the other as shown below.
3x + 2
2x + 5
First multiply vertically on right-hand side column,
2 * 5 = 10
Step 2: Then multiply crosswise and add the products, 
(5 * 3x) + (2x * 2) = 15x + 4x = 19x

Step 3: At last multiply vertically on left-hand side column,
3x * 2x = 6x2

So, the answer is = 6x2 + 19x + 10

Que 4: Multiply (a + 7)(a - 7).

Solution:

Step 1: Write one binomial under the other as shown below.
a + 7
a - 7
First multiply vertically on right-hand side column,
7 * -7 = -49
Step 2: Then multiply crosswise and add the products, 
(-7 * a) + (a * 7) = -7a + 7a = 0

Step 3: At last multiply vertically on left-hand side column,
a* a = a2

So, the answer is = a2 - 49

Que 5: Multiply (5a + 2)(a + 3).

Solution:

Step 1: Write one binomial under the other as shown below.
5a + 2
a + 3
First multiply vertically on right-hand side column,
2 * 3 = 6
Step 2: Then multiply crosswise and add the products, 
(3 * 5a) + (a * 2) = 15a + 2a = 17a

Step 3: At last multiply vertically on left-hand side column,
5a * a = 5a2

So, the answer is = 5a2 + 17a + 6

Que 6: You’re planning a flower bed with length 3x+2 feet and width x+1 feet. Find the total area of bed using Vedic Mathematics in terms of ‘x’?

Solution:

Given:
Length = (3x + 2) 
Width = (x + 1)

Total area = length * width
Total area = (3x + 2)(x + 1)

Step 1: Write one binomial under the other as shown below.
3x + 2
x + 1
First multiply vertically on right-hand side column,
2 * 1 = 2
Step 2: Then multiply crosswise and add the products, 
(1 * 3x) + (x * 2) = 3x + 2x = 5x

Step 3: At last multiply vertically on left-hand side column,
3x * x = 3x2

So, the answer is = 3x2 + 5x + 2