Table of Contents

- 1 Introduction to Algebra
- 2 Algebra Formula
- 3 Important Formulas in Algebra
- 4 Algebra Formulas for Class 8
- 5 Algebra Formulas for Class 9
- 6 Algebra Formulas for Class 10
- 7 Algebra Formulas for Class 11
- 8 Algebra Formulas for Class 12
- 9 Algebra Formulas – Function
- 10 Algebra Formulas – Fractions
- 11 Examples using Algebra Formulas
- 12 FAQs on Algebra Formulas
- 12.1 What are Algebra Formulas in Math?
- 12.2 What Are the Basic Algebraic Formulas in Math?
- 12.3 How do I Learn Algebra Formulas?
- 12.4 What Is The Formula For a2– b2 in Algebra Formulas?
- 12.5 How to Solve Algebra Formulas?
- 12.6 What is the Basis Of Algebra Formulas?
- 12.7 What are Algebra Expressions?
- 12.8 What are the Algebra Formulas for Triangular Numbers?
- 12.9 What Are Algebraic Expressions Formulas?
- 12.10 How To Derive the Algebraic Expressions Formula (x + y)3 = x3 + y3 + 3xy (x + y)?
- 12.11 What Are the Applications of Algebraic Expressions Formulas?
- 12.12 How to Use Algebraic Expressions Formulas While Solving Problems?

- 13 MATHEMATICAL FORMULAE

## Introduction to Algebra

Mathematics is a vast field. It is impossible for one person to know everything there is to know in mathematics, even after a lifetime of study. And while it can be cumbersome, mathematics is also one of the most important fields of study. Right from how much to tip the waiter to when the universe began, all answers can be found due to the application of maths.

As we approach the higher classes, we see our introduction to algebra. In algebra, we substitute numbers with letters or alphabets to arrive at a solution. We use these letters like (x, a, b etc.) to represent unknown quantities in an equation. Then we solve the equation or algebra formula to arrive at a definite answer.

Algebra itself is divided into two major fields. The more basic functions that we learn in school are known as elementary algebra. Then the more advanced algebra formula, which is more abstract in nature fall under modern algebra, sometimes even known as abstract algebra.

## Algebra Formula

Algebra includes both numbers and letters. Numbers are fixed, i.e. their value is known. Letters or alphabets are used to represent the unknown quantities in the algebra formula. Now, a combination of numbers, letters, factorials, matrices etc. is used to form an equation or formula. This is essentially the methodology for algebra.

As students study for their exams, there are certain very important algebra formulas and equations that they must learn. These formulas are the cornerstone of basic or elementary algebra. Only learning the formulas is not sufficient. The students must also understand the concept behind the formula and learn to apply them correctly.

Here, we will provide a list of all the important algebra formulas. The comprehensive list will allow the students to have a quick look before exams or refer to whenever they wish. Remember, only rote learning is not enough. You must also know how to effectively apply these formulas to a problem.

- a² – b² = (a-b)(a+b)
- (a+b)² = a² + 2ab + b²
- (a-b)² = a² – 2ab + b²
- a² + b² = (a-b)² +2ab
- (a+b+c)² = a²+b²+c²+2ab+2ac+2bc
- (a-b-c)² = a²+b²+c²-2ab-2ac+2bc
- a³-b³ = (a-b) (a² + ab + b²)
- a³+b³ = (a+b) (a² – ab + b²)
- (a+b)³ = a³+ 3a²b + 3ab² + b³
- (a-b)³ = a³- 3a²b + 3ab² – b³
- “n” is a natural number, a
^{n}– b^{n}= (a-b) (a^{n-1}+ a^{n-2}b +….b^{n-2}a + b^{n-1}) - “n” is a even number, a
^{n}+ b^{n}= (a+b) (a^{n-1 }– a^{n-2}b +….+ b^{n-2}a – b^{n-1}) - “n” is an odd number a
^{n}+ b^{n}= (a-b) (a^{n-1}– a^{n-2}b +…. – b^{n-2}a + b^{n-1}) - (a
^{m})(a^{n}) = a^{m+n}(ab)^{m}= a^{mn}

## Important Formulas in Algebra

**Here is a list of Algebraic formulas** –

- a
^{2}– b^{2}= (a – b)(a + b) - (a + b)
^{2}= a^{2}+ 2ab + b^{2} - a
^{2}+ b^{2}= (a + b)^{2}– 2ab - (a – b)
^{2}= a^{2}– 2ab + b^{2} - (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca - (a – b – c)
^{2}= a^{2}+ b^{2}+ c^{2}– 2ab + 2bc – 2ca - (a + b)
^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}; (a + b)^{3}= a^{3}+ b^{3}+ 3ab(a + b) - (a – b)
^{3}= a^{3}– 3a^{2}b + 3ab^{2}– b^{3 }= a^{3}– b^{3}– 3ab(a – b) - a
^{3}– b^{3}= (a – b)(a^{2}+ ab + b^{2}) - a
^{3}+ b^{3}= (a + b)(a^{2}– ab + b^{2}) - (a + b)
^{4}= a^{4}+ 4a^{3}b + 6a^{2}b^{2}+ 4ab^{3}+ b^{4} - (a – b)
^{4}= a^{4}– 4a^{3}b + 6a^{2}b^{2}– 4ab^{3}+ b^{4} - a
^{4}– b^{4}= (a – b)(a + b)(a^{2}+ b^{2}) - a
^{5}– b^{5}= (a – b)(a^{4}+ a^{3}b + a^{2}b^{2}+ ab^{3}+ b^{4}) **If n is a natural number**a^{n}– b^{n}= (a – b)(a^{n-1}+ a^{n-2}b+…+ b^{n-2}a + b^{n-1})**If n is even**(n = 2k), a^{n}+ b^{n}= (a + b)(a^{n-1}– a^{n-2}b +…+ b^{n-2}a – b^{n-1})**If n is odd**(n = 2k + 1), a^{n}+ b^{n}= (a + b)(a^{n-1}– a^{n-2}b +a^{n-3}b^{2}…- b^{n-2}a + b^{n-1})- (a + b + c + …)
^{2}= a^{2}+ b^{2}+ c^{2}+ … + 2(ab + ac + bc + ….) **Laws of Exponents**(a^{m})(a^{n}) = a^{m+n}; (ab)^{m}= a^{m}b^{m }; (a^{m})^{n}= a^{mn}**Roots of Quadratic Equation**- For a quadratic equation ax
^{2}+ bx + c = 0 where a ≠ 0, the roots will be given by the equation as

- Δ = b
^{2}− 4ac is called the discriminant - For real and distinct roots, Δ > 0
- For real and coincident roots, Δ = 0
- For non-real roots, Δ < 0
- If α and β are the two roots of the equation ax
^{2}+ bx + c = 0 then, α + β = (-b / a) and α × β = (c / a). - If the roots of a quadratic equation are α and β, the equation will be (x − α)(x − β) = 0
**Factorials**

- n! = (1).(2).(3)…..(n − 1).n
- n! = n(n − 1)! = n(n − 1)(n − 2)! = ….
- 0! = 1

**Algebra Formulas for Class 8**** **

The algebra formulas for three variables a, b, and c and for a maximum degree of 3 can be easily derived by multiplying the expression by itself, based on the exponent value of the algebraic expression. The below formulas are for class 8.

- (a + b)
^{2}= a^{2}+ 2ab + b^{2} - (a – b)
^{2}= a^{2}– 2ab + b^{2} - (a + b)(a – b) = a
^{2}– b^{2} - (a + b)
^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3} - (a – b)
^{3}= a^{3}– 3a^{2}b + 3ab^{2}– b^{3} - a
^{3}+ b^{3}= (a + b)(a^{2}+ ab + b^{2}) - a
^{3}+ b^{3}= (a + b)(a^{2}+ ab + b^{2}) - (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca

Some of the common laws of exponents with the same bases having different powers, and different bases having the same power, are useful to solve complex exponential terms. The higher exponential values can be easily solved without any expansion of the exponential terms. These exponential laws are further useful to derive some of the logarithmic laws.

- a
^{m}. a^{n}= a^{m + n} - a
^{m}/a^{n}= a^{m – n} - (a
^{m})^{n}= an - (ab)
^{m}= a^{m}. b^{m} - a
^{0}= 1 - a
^{-m}= 1/a^{m}

**Algebra Formulas for Class 9**

Logarithms are useful for the computation of highly complex multiplication and division calculations. The normal exponential form of 2^{5} = 32 can be transformed to a logarithmic form as Log 32 to the base of 2 = 5. Further, the multiplication and division between two mathematic expressions can be easily transformed into addition and subtraction, after converting them to logarithmic form. The below properties of logarithms formulas, which are applicable in logarithmic calculations.

## Algebra Formulas for Class 10

An important algebra formula introduced in class 10 is the “quadratic formula”. The general form of the quadratic equation is ax^{2} + bx + c = 0, and there are two methods of solving this quadratic equation. The first method is to solve the quadratic equation by the algebraic method, and the second method is to solve through the use of the quadratic formula. The below formula is helpful to quickly find the values of the variable x with the least number of steps.

In the above expression, the value b^{2} – 4ac is called the determinant and is useful to find the nature of the roots of the given equation. Based on the value of the determinant, the three types of roots are given below.

- If b
^{2}– 4ac > 0, then the quadratic equation has two**distinct real roots**. - If b
^{2}– 4ac = 0, then the quadratic equation has two**equal real roots**. - If b
^{2}– 4ac < 0, then the quadratic equation has two**imaginary roots**.

Apart from this, we have a few other formulas related to progressions. Progressions include some of the basic sequences such as arithmetic sequence and geometric sequence. The arithmetic sequence is obtained by adding a constant value to the successive terms of the series. The terms of the arithmetic sequence is a, a + d, a + 2d, a + 3d, a + 4d, …. a + (n – 1)d. The geometric sequence is obtained by multiplying a constant value to the successive terms of the series. The terms of the geometric sequence are a, ar, ar^{2}, ar^{3}, ar^{4}, …..ar^{n-1}. The below formulas are helpful to find the nth term and the sum of the terms of the arithmetic, and geometric sequence.

**Algebra Formulas for Class 11**

The important topics of Class 11 which have extensive use of algebraic formulas are permutations and combinations. Permutations help in finding the different arrangement of r things from the n available things, and combinations help in finding the different groups of r things from the available n things. The following formulas help in finding the permutations and combination values.

Apart from the permutations and combinations, there is another important topic of “Binomial Theorem” as well which is used to evaluate the large exponents of algebraic expressions with two terms. Here the coefficients of the binomial terms are calculated from the formula of combinations. The below expression provides the complete formula for binomial expansion, and it can be termed as the algebraic expression of the binomial theorem.

**Algebra Formulas for Class 12**

The vector algebra formulas that are involved in class 12 are as follows.

**Algebra Formulas – Function**

An algebraic function is of the form y=f(x). Here, x is the input and y is the output of this function. Here, each input corresponds to exactly one output. But multiple inputs may correspond to a single output. For example: f(x)= x^{2} is an algebraic function. Here, when x=2, f(2)= 2^{2} =4. Here, x=2 is the input, and f(2)=4 is the output of the function.

**Algebra Formulas – Fractions**

We can perform numerous arithmetic operations such as addition, subtraction, multiplication, and the dividing of fractions in algebra just the same way we do with fractions involving numbers. Further, it only has the unknown variables and involved the same rules of working across fractions. The below four expressions are useful for working with algebraic fractions. **Adding Fractions: **x/y + z/w = (x.w + y.z)/(y.w)** **

**Subtracting Fractions:**x/y – z/w = (x.w – y.z)/(y.w)**Multiplying Fractions:**x/y × z/w = xz/yw**Dividing Fractions:**x/y ÷ z/w = x/y × w/z = xw/yz

### Algebra Formulas Tips and Tricks:

The following quick tips and tricks would be useful to easily understand the memorize algebraic identities.

- You can memorize the algebraic identities by understanding how they are derived. For example, (a+b)
^{2}=(a+b)(a+b)= a^{2}+ab+ab+b^{2 }= a^{2}+2ab+b^{2.} - In the same way, you can try to derive the other algebraic identities as well.

## Examples using Algebra Formulas

**Example 1: Using algebra formulas, find (2x-3y) ^{2}.**

**Solution:**

Here, we use the identity (a-b)^{2} = a^{2} – 2ab + b^{2} to expand this. Here, a= 2x and b=3y. Then we get: (2x-3y)^{2} = (2x)^{2} -2(2x)(3y)+(3y)^{2 }= 4x^{2} -12xy + 9y^{2}. Therefore, (2x – 3y)^{2} = 4x^{2} -12xy + 9y^{2}.

**Example 2: Using algebra formulas – identities, evaluate 297 × 303. **

**Solution:**

The above product can be written as (300-3) × (300+3). We will find this product using the formula: (a-b)(a+b)=a^{2}– b^{2} Here a=300 and b=3. Then we get: (300-3) × (300+3) =300^{2} – 3^{2} = 90000-9 = 89991. Therefore, 297 × 303 = 89991.

**Example 3: Find the roots of the quadratic equation x ^{2}+5x+6=0 using algebra formulas for quadratic equations.**

**Solution:**

The given equation is x^{2}+5x+6=0. Comparing this with ax^{2}+bx+c=0, we get: a=1; b=5; c=6. Substituting these values in the quadratic formula:

## FAQs on Algebra Formulas

### What are Algebra Formulas in Math?

Here are some most commonly used formulas of algebra:

- a
^{2}– b^{2}=(a-b)(a+b) - (a+b)
^{2}=a^{2}+ 2ab + b^{2} - (a-b)
^{2}=a^{2}– 2ab+b^{2} - (x+a)(x+b)=x
^{2}+x(a+b)+a b - (a+b+c)
^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca - (a+b)
^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} - (a – b)
^{3}=a^{3}– 3a^{2}b+3ab^{2}– b^{3}

Further, there are algebraic formulas for other topics of maths such as exponents, logarithms, permutations, sequences, and vector algebra.

### What Are the Basic Algebraic Formulas in Math?

Here are some basic math formulas:

- a
^{2}– b^{2}=(a-b)(a+b) - (a+b)
^{2}=a^{2}+ 2ab + b^{2} - (a-b)
^{2}= a^{2}– 2ab + b^{2} - (x+a)(x+b)=x
^{2}+ x(a+b) + ab - (a+b+c)
^{2 }= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca - (a+b)
^{3}=a^{3}+3a^{2}b + 3ab^{2 }+ b^{3} - (a-b)
^{3}=a^{3}– 3a^{2}b + 3ab^{2}– b^{3}

### How do I Learn Algebra Formulas?

Algebra formulas can be easily memorized by visualizing the formulas as squares or rectangles. Further, the understanding of the factorized forms of the formulas helps to easily learn and remember the algebraic formulas.

### What Is The Formula For a^{2}– b^{2 }in Algebra Formulas?

The formula for a^{2}– b^{2} is (a+b)(a-b)= a^{2}– b^{2}. It is called the difference of squares formula.

### How to Solve Algebra Formulas?

The solving of algebra formulas is to aim at equalizing the left-hand side of the expression with the right-hand side of the expression. Further, the terms can be transferred from the left to the right side of the expression, based on the laws of algebra.

### What is the Basis Of Algebra Formulas?

The basis of algebra formulas is that the resultant numeric value of the expressions on either side of the equals to sign is equal. Further, algebraically the terms are modified on either side to match up with the algebraic formulas.

### What are Algebra Expressions?

For each of the algebra formulas, the equations with variables, powers, and arithmetic operations, and on either side of the equals to sign are called algebraic expressions. In the algebraic formula (a+b)(a-b)= a^{2}– b^{2}, the terms on either side of the equals to sign are called algebraic expressions.

### What are the Algebra Formulas for Triangular Numbers?

The algebra formula for triangular numbers is H^{2} = B^{2} + A^{2} and it helps to relate the length of the sides of the triangle. It is applicable for a right triangle and has been derived from the Pythagoras theorem. The alphabets H represents the hypotenuse, B represents the base of the right triangle, and A represents the altitude of the triangle. Applying this same formula an example of triangular numbers is (6, 8, 10).

### What Are Algebraic Expressions Formulas?

The algebraic expression formulas are formulas that are used to simplify the algebraic expressions. Some important algebraic expressions formulas are:

(a + b)

^{2}= a^{2}+ 2ab + b^{2}(a – b)

^{2}= a^{2}– 2ab + b^{2}(a + b) (a – b) = a

^{2}– b^{2}(a + b)

^{3}= a^{3}+ b^{3}+ 3ab (a + b)(a – b)

^{3}= a^{3}– b^{3}– 3ab (a – b)(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2cax

^{3}+ y^{3}= (x + y) (x^{2 }– xy + y^{2})x

^{3}– y^{3}= (x – y) (x^{2 }+ xy + y^{2})

### How To Derive the Algebraic Expressions Formula (x + y)^{3} = x^{3} + y^{3} + 3xy (x + y)?

Let us start with the left-hand side of this formula and reach the right-hand side at the end.

(x + y)^{3} = (x + y)^{2} (x + y)

= (x^{2} + 2xy + y^{2}) (x + y)

= x^{3} + 2x^{2}y + xy^{2} + x^{2}y + 2xy^{2} + y^{3}

= x^{3} + y^{3} + 3x^{2}y + 3xy^{2} (or)

= x^{3} + y^{3} + 3xy (x + y)

### What Are the Applications of Algebraic Expressions Formulas?

The algebraic expressions formulas are used to simplify the complex algebraic expressions such as (3x + 4y)^{2}, (a – 3b + 2c)^{2}, etc. These formulas are also used to factorize the polynomials.

### How to Use Algebraic Expressions Formulas While Solving Problems?

We have multiple algebraic expressions formulas and some of them have to be used according to the need while solving the problems. For example, to factorize the expression, 8x^{3} + 27, we apply the a^{3} + b^{3} formula as follows.

a^{3 }+ b^{3 }= (a + b) (a^{2} – ab + b^{2})

Substitute a = 2x and b = 3 on both sides,

(2x)^{3 }+ 3^{3} = (2x + 3) ( (2x)^{2 }– (2x)(3) + 3^{2})

8x^{3} + 27 = (2x + 3) (4x^{2} – 6x + 9).

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