Algebra Formulas: A Comprehensive Overview

Algebra is a fundamental branch of mathematics that deals with the study of symbols and the rules for manipulating them. It provides a method to express real-world problems in a mathematical format. In algebra, letters or symbols (called variables) are used to represent numbers. These symbols can form algebraic expressions, equations, and identities, which are widely used to solve mathematical problems.

In this guide, we go through the algebra formulas that are important in your maths classes and exams, covering examples for each of the formulas.

Key Algebraic Formulas

Here are some important algebra formulas that every student should know:

Basic Algebraic Identities:

\[ (a + b)^2 = a^2 + 2ab + b^2 \]

\[ (a – b)^2 = a^2 – 2ab + b^2 \]

\[ a^2 – b^2 = (a – b)(a + b) \]

Cube Identities:

\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]

\[ (a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3 \]

Factoring Formulas:

\[ a^3 – b^3 = (a – b)(a^2 + ab + b^2) \]

\[ a^3 + b^3 = (a + b)(a^2 – ab + b^2) \]

General Polynomial Identity:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]

Laws of Exponents

Algebra often involves working with exponents. Here are some essential exponent laws:

\(a^m \times a^n = a^{m+n}\)

\( (a^m)^n = a^{mn} \)

\( \frac{a^m}{a^n} = a^{m-n} \)

\(a^0 = 1\)

Examples for Basic Algebraic Formulas

In this section, we’ll provide examples for each of the algebra formulas discussed. These examples will help you better understand how to apply the formulas in solving algebraic problems.

1. Examples for \( (a + b)^2 = a^2 + 2ab + b^2 \)

This formula is used to expand the square of a binomial expression, and it’s helpful for calculating squares of numbers near round figures.

Problem: Find \( 103^2 \)

We can rewrite \( 103^2 \) as \( (100 + 3)^2 \). Follow these steps to calculate:

  1. Identify \( a = 100 \) and \( b = 3 \).
  2. Apply the formula \( (a + b)^2 = a^2 + 2ab + b^2 \).
  3. Calculate each part: \( a^2 = 100^2 = 10000 \), \( 2ab = 2(100)(3) = 600 \), and \( b^2 = 3^2 = 9 \).
  4. Add the results: \( 10000 + 600 + 9 = 10609 \).

2. Examples for \( (a – b)^2 = a^2 – 2ab + b^2 \)

This formula is used to expand the square of a difference, ideal for calculating squares of numbers just below a round number.

Problem: Find \( 98^2 \)

We can rewrite \( 98^2 \) as \( (100 – 2)^2 \). Follow these steps:

  1. Identify \( a = 100 \) and \( b = 2 \).
  2. Apply the formula \( (a – b)^2 = a^2 – 2ab + b^2 \).
  3. Calculate each part: \( a^2 = 100^2 = 10000 \), \( 2ab = 2(100)(2) = 400 \), and \( b^2 = 2^2 = 4 \).
  4. Subtract and add the results: \( 10000 – 400 + 4 = 9604 \).

3. Examples for \( a^2 – b^2 = (a – b)(a + b) \)

This formula is used to factorize the difference of two squares, useful for large numbers that are close to each other.

Problem: Find \( 102^2 – 98^2 \)

We can apply the difference of squares formula. Follow these steps:

  1. Identify \( a = 102 \) and \( b = 98 \).
  2. Apply the formula \( a^2 – b^2 = (a – b)(a + b) \).
  3. Calculate the differences and sums: \( a – b = 4 \) and \( a + b = 200 \).
  4. Multiply the results: \( 4 \times 200 = 800 \).

4. Examples for \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)

This formula is used to expand the cube of a binomial sum.

Problem: Find \( 101^3 \)

We can rewrite \( 101^3 \) as \( (100 + 1)^3 \). Follow these steps:

  1. Identify \( a = 100 \) and \( b = 1 \).
  2. Apply the formula \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).
  3. Calculate each part: \( a^3 = 100^3 = 1000000 \), \( 3a^2b = 3(100^2)(1) = 30000 \), \( 3ab^2 = 3(100)(1^2) = 300 \), and \( b^3 = 1^3 = 1 \).
  4. Add the results: \( 1000000 + 30000 + 300 + 1 = 1030301 \).

5. Examples for \( a^3 – b^3 = (a – b)(a^2 + ab + b^2) \)

This formula is used to factorize the difference of two cubes.

Problem: Factorize \( 125 – 64 \)

We can apply the cube difference formula. Follow these steps:

  1. Identify \( a = 5 \) and \( b = 4 \).
  2. Apply the formula \( a^3 – b^3 = (a – b)(a^2 + ab + b^2) \).
  3. Calculate each part: \( a^2 + ab + b^2 = 25 + 20 + 16 = 61 \).
  4. Multiply the terms: \( (5 – 4)(61) = 61 \).

6. Examples for \( (a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3 \)

This formula is used to expand the cube of a binomial difference.

Problem: Find \( 99^3 \)

We can rewrite \( 99^3 \) as \( (100 – 1)^3 \). Follow these steps:

  1. Identify \( a = 100 \) and \( b = 1 \).
  2. Apply the formula \( (a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3 \).
  3. Calculate each part: \( a^3 = 100^3 = 1000000 \), \( 3a^2b = 3(100^2)(1) = 30000 \), \( 3ab^2 = 3(100)(1^2) = 300 \), and \( b^3 = 1^3 = 1 \).
  4. Subtract and add the results: \( 1000000 – 30000 + 300 – 1 = 970269 \).

Examples for Laws of Exponents

The laws of exponents help simplify expressions that involve powers. Let’s go over each law with an example to better understand how they work.

1. \( a^m \times a^n = a^{m+n} \)

This law states that when multiplying two expressions with the same base, you add their exponents.

Problem: Simplify \( 2^3 \times 2^4 \):

Using the law \( a^m \times a^n = a^{m+n} \), where \( a = 2 \), \( m = 3 \), and \( n = 4 \):

\[ 2^3 \times 2^4 = 2^{3+4} = 2^7 = 128 \]

2. \( \frac{a^m}{a^n} = a^{m-n} \)

This law states that when dividing two expressions with the same base, you subtract the exponent of the denominator from the exponent of the numerator.

Problem: Simplify \( \frac{5^6}{5^2} \):

Using the law \( \frac{a^m}{a^n} = a^{m-n} \), where \( a = 5 \), \( m = 6 \), and \( n = 2 \):

\[ \frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625 \]

3. \( (a^m)^n = a^{mn} \)

This law states that when raising a power to another power, you multiply the exponents.

Problem: Simplify \( (3^2)^3 \):

Using the law \( (a^m)^n = a^{mn} \), where \( a = 3 \), \( m = 2 \), and \( n = 3 \):

\[ (3^2)^3 = 3^{2 \times 3} = 3^6 = 729 \]

4. \( a^0 = 1 \) (where \( a \neq 0 \))

This law states that any non-zero number raised to the power of zero equals 1.

Problem: Simplify \( 7^0 \):

Using the law \( a^0 = 1 \), where \( a = 7 \):

\[ 7^0 = 1 \]

5. \( a^{-n} = \frac{1}{a^n} \)

This law states that a negative exponent means the reciprocal of the base raised to the positive exponent.

Example:

Simplify \( 4^{-2} \):

Using the law \( a^{-n} = \frac{1}{a^n} \), where \( a = 4 \) and \( n = 2 \):

\[ 4^{-2} = \frac{1}{4^2} = \frac{1}{16} \]