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178 LearnersLast updated on December 15, 2025

In mathematics, algebra formulas are important as they form the foundation for polynomials, calculus, trigonometry, and quadratic equations. These formulas help solve and simplify algebraic expressions. In this article, algebraic formulas will be discussed in detail.
Algebra formulas are rules or equations that help with factoring, expanding, and simplifying expressions. We can use these formulas to solve complex algebraic equations efficiently. Here are some algebraic formulas:
\(1. \ (a + b)^2 = a^2 + 2ab + b^2 \\[1em]
2. \ (a - b)^2 = a^2 - 2ab + b^2\\[1em]
3. \ (a + b)(a - b) = a^2 - b^2\\[1em]
4. \ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\\[1em]
5. \ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\\[1em]
6. \ a^3 + b^3 = (a + b) (a^2 - ab + b^2)\\[1em]
7. \ a^3 - b^3 = (a - b) (a^2 + ab + b^2)\\[1em]
8. \ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
To solve the expressions involving powers or exponents, we use the exponent rules. These rules are used to simplify expressions with powers. The exponent rules are:
| Rule | Formula |
| Product rule | \(a^m × a^n = a^{m + n}\) |
| Quotient rule | \(a^m \div a^n = a^{m - n}\) |
| Power of a power rule | \((a^m)^n = a^{mn}\) |
| Power of a product rule | \((ab)^m = a^mb^m\) |
| Power of a quotient rule | \(\left(\frac{a}{b}\right)^m = \frac{(a^m)}{(b^m)}\) |
| Zero exponent rule | \(a^0 = 1\) |
| Negative exponent rule | \(a^{-m} = \frac{1}{a^m}\) |
Logarithms are used to solve multiplication and division of numbers with powers in simple ways. This makes them an effective tool to work with algebraic formulas with exponents. The relationship between the exponent and logarithm is: \(x^m = a ⇒ log_x a = m\)
Some commonly used log algebraic formulas are:


The quadratic formula is one of the two methods to solve a quadratic equation.
The standard form of a quadratic equation is given as;
\(ax^2 + bx + c = 0. \)
The value of the variable x can be found by using the formula:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
In algebra, permutations and combinations are formulas that help us identify the number of ways something can be arranged. Permutations refer to arrangements of items where the order matters, and combinations are the selection of items where order does not matter.
Factorial formula: n! = n × (n - 1) × (n - 2) × …. × 3 × 2 × 1
Permutations formula: \({}^{n}P_{r} = \frac{n!}{(n - r)!} \)
Combination formula: \({}^{n}C_{r} = \frac{n!}{r!(n - r)!} \)
Binomial theorem:
\((x + y)^n = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \cdots + \binom{n}{n-1}x^1 y^{n-1} + \binom{n}{n}x^0 y^n \)
The vector algebra formula is used to solve problems related to directions and magnitude. Some important vector formulas are:
For any three vectors a, b, and c in a 3D space
Algebraic identities are the equations that hold true for all values of the variables involved. It means LHS = RHS of the equation. Some common algebraic identities are -
\((a + b)^2 = a^2 + 2ab + b^2\\[1em] (a - b)^2 = a^2 - 2ab + b^2\\[1em] (a + b)(a - b) = a^2 - b^2\\[1em] (x + a) (x + b) = x^2 + (a + b)x + ab\)
Algebra formulas of functions
The algebraic function expresses a relationship between two variables. It is written in the form y = f(x). Where x is the input and y is the output.
For example, if x = 4, then
f(x) = f(4) = 42 = 16.
Algebra formulas of fractions
In algebra, fractions that contain variables are called rational expressions. We can add, subtract, multiply, and divide fractions. The algebraic expressions of fractions are:
\(\frac{x}{y}+ \frac{z}{w} = \frac{(xw + yz)} {(yw)} \\[1em] \frac{x}{y} - \frac{z}{w}= \frac{(xw - yz)}{(yw)} \\[1em] \frac{x}{y} × \frac{z}{w} = \frac{xz}{yw} \\[1em] \frac{x}{y} ÷ \frac{z}{w} = \frac{xw}{yz}\)
Students often find algebra formulas challenging to memorize. Here are some tips and tricks to help with memorization:
Students often make mistakes when applying algebra formulas. Here are some common errors and ways to avoid them to master algebra.
Expand the expression (3x + 4)².
The expanded form is 9x² + 24x + 16.
Using the identity (a + b)² = a² + 2ab + b², where
a = 3x and b = 4:
(3x + 4)² = (3x)² + 2(3x)(4) + 4²
(3x + 4)² = 9x² + 24x + 16
Solve the quadratic equation 2x² - 4x - 6 = 0.
The solutions are x = 3 and x = -1.
Using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\),
Where a = 2, b = -4, c = -6:
\(x = \frac{-4 \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2 \cdot 2} \)
\(x = \frac{4 \pm \sqrt{16 + 48}}{4} \)
\(x = \frac{4 \pm \sqrt{64}}{4} \)
\(x = \frac{4 \pm 8}{4} \)
x = 3 and x = -1
Find the solution for the linear equation 3x + 2y = 12 and x - y = 1.
The solution is x = 2, y = 1
Solving by substitution: From x - y = 1, we get x = y + 1
Substitute in 3x + 2y = 12: 3(y + 1) + 2y = 12
3y + 3 + 2y = 12
5y = 9 y = 1.8
Substitute y = 1.8 in x - y = 1: x - 1.8 = 1 x = 2.8
Simplify the expression 4a² - 9b².
The simplified form is (2a + 3b)(2a - 3b).
Using the identity a² - b² = (a + b)(a - b),
Where a = 2a and b = 3b:
4a² - 9b² = (2a)² - (3b)²
4a² - 9b² = (2a + 3b)(2a - 3b)
Verify the identity (x + y)² = x² + 2xy + y² for x = 2 and y = 3.
The identity is verified.
LHS: (x + y)² = (2 + 3)² = 5² = 25
RHS: x² + 2xy + y² = 2² + 2(2)(3) + 3² = 4 + 12 + 9 = 25
LHS = RHS
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.






