Let’s dive into Algebra, a section you’re definitely going to encounter in exams like the SAT, ACT, GRE, and even the MCAT. Algebra can seem tricky at first because it involves a lot of formulas, but don’t worry—once you get the hang of it, it’ll be much easier to tackle. It’s a key part of the quantitative section, and understanding it is essential not only for doing well in school but also for various standardized tests.

In this article, we’ll cover the key concepts of Algebra that you’ll need to know for exams like the SAT and GRE. We’ll walk through common types of questions, give you important formulas, and share some helpful tips and tricks for solving problems efficiently. Plus, we’ve got some solved examples and practice questions for you to explore so you can really master this section. Keep reading to clear up any confusion and get ready to ace Algebra in your upcoming tests!

What is Algebra?

The term “algebra” was derived from the word al-jabr. Algebra can be defined as the branch of mathematics in which we perform mathematical operations with the help of numbers (constants) and alphabets (variables). Let us understand different types of Algebraic questions one by one from below.

History of Algebra

Algebra is a branch of Mathematics that dates back centuries, to the Middle East. It was invented by a famous mathematician, astronomer, and geographer Abu Ja’far Muhammad ibn Musa al-Khwarizmi who was born about 780 in Baghdad. His book on Algebra is known as The Compendious Book on Calculation by Completion and Balancing. His book was translated into several languages and became very popular in the west.

Algebra uses letters and numbers. Algebra tends to find the unknown by putting the real-life variables into equations and then solving them by using various formulas or equations. Some of the sub-topics under Algebra are real numbers, complex numbers, matrices, vectors and so on.

Types of Questions from Algebra

Let us see different types of questions that may come in the Algebra section one by one from below.

1. Algebraic Expressions

The combination of constants, variables and elementary arithmetic operations (+, -, ×, ÷) is called algebraic expression.

Example: \(3x + 5y, 4x^3 -5y \frac{1}{3} , 5x – 6y + 7z ÷ 8y\frac{1}{3}\) etc.

Algebraic Expressions can be further divided into two main categories. The types of algebraic expressions are as follows:

(a) On the basis of number of terms

These are the following types of algebraic expressions on the basis of number of terms:

(i) Monomial

The algebraic expression that consists of only one term is called a monomial.

Example: \(4x, 4\frac{y}{3}, -6.33z\) etc.

(ii) Binomial

The algebraic expression that consists of two terms is called binomial.

Example: \(2x + 3y, 4x + x^2\) etc.

(iii) Trinomial

The algebraic expression that consists of three terms is called trinomial.

Example: \(4xy + 3x + 5y^2\)

(iv) Polynomial

The algebraic expression that consists of two or more than two terms is called polynomial.

The general form of polynomial is given by P(x) = \(a_{0} + a_{1}x + a_{2}x^{2} + … +a_{n}x^{n}\)

Where, \(a_{0}, a_{1}, a_{2}, a_{3} … and a_{n}\), are real numbers and n is a whole number (non-negative integer).

Example: \(1 + 2x – 3x^2 + 5x^4, 6x^3 – 2x^2 + 4\) etc.

(b) On the basis of degree of a polynomial

(i) Linear Polynomial

The polynomial in which the highest power of the variables is one or, the polynomial of degree 1 is called linear polynomial.

Example: 2x + 3

(ii) Quadratic Polynomial

The polynomial of degree two is called a quadratic polynomial.

Example: \(x^2 – 4x + 5, x^2 – 5\) etc.

(iii) Cubic Polynomial

The polynomial of degree three is called a cubic polynomial.

Example: \(x^3 + 4x^2 – 3x + 4, x^3 – 4x – 3\) etc.

(iv) Quartic Polynomial

The polynomial of degree four is called the quartic equation.

Example: \(5x^4 – 4x + 2, 5x^4 + 4x^3 – 3x\) etc.

2. Equations

When we equate an expression or polynomial with a number, then the resultant is called an equation.

There are following types of equations:

(a) Linear Equation in One Variable

The equation having only one variable with highest power 1 is called a linear equation in one variable.

The general form of linear equation in one variable is ax + b = 0, where a and b are real constants and a ≠ 0

Example: \(23 + x = 30\)

(b) Linear Equation in Two Variables

The equation having two variables with the highest power of the variable 1 is called linear equation in two variables.

The general form of the linear equation in two variables is given by ax + by + c = 0, where a, b and c are real numbers and (a, b) ≠ (0, 0).

Example: \(12x + 23y = 30\)

(c) Quadratic Equation

An equation of the form \(ax^2 + bx + c = 0\) is called a quadratic equation.

The standard form of the quadratic equation is \(ax^2 + bx + c = 0\), where a, b and c are real numbers and a ≠ 0

Example: \(4x^2 + 2x + 1 = 0, 5x^2 – 2x + 4 = 0 \) etc.

(d) Cubic Equation

An equation of the form \(ax^3 + bx^2 + cx + d = 0\) is called a cubic equation. Where a, b, c, and d are real constants and a ≠ 0.

How to Solve Algebra Questions – Tips and Tricks

Candidates can find different tips and tricks below for solving the questions related to Algebra.

Tip # 1: Definition of Coefficient – The number or the fixed value multiplied to the variable in a term of an algebraic expression is called the coefficient of the term.

Example: In polynomial \(4x^3 + 5x^2 – 7x – 2\), the coefficient of \(x^3\) is 4, coefficient of \(x^2\(\) is 5, coefficient of x is -7 and the constant term is -2.

Tip # 2: Definition of Degree of Polynomial – The highest power of the variables in a polynomial is called the degree of polynomial.

Example: The degree of polynomial \(\)4x^3 + 5x^2 – 7x – 2\) is 3, the degree of polynomial \(1 + 2x – 3x^2 + 5x^4\) is 4.

Tip # 3: Zeros of Polynomial – Those value(s) of x(variable) which satisfy the polynomial is/ are called the zero/zeros of the polynomial.

Example: since x = 2 satisfies the polynomial \(x^2 – 5x + 6\), so we can say that x = 2 is a zero of polynomials.

Tip # 4: Remainder Theorem – When P(x) is divided by (x – a), then remainder will be P(a).

Example: when P(x) = \(x^2 -3x + 6\) is divided by (x – 2), then remainder will be P(2)

⇒ P(2) = 22 – 3 × 2 + 6 = 4

Tip # 5: Factor Theorem – If (x – a) is a factor of P(x), then remainder will be zero i.e. P(x) = 0

Example: Find the value of k if x – 2 is a factor of \(x^2 – k x + 6 \)

Solution: according to question P(2) = 0 ⇒ 22 – k × 2 + 6 = 0

Hence, k = 5

Tip # 6: The value of the variables that satisfies the equation is called the solution of the equation. The graph of the linear equation ax + by + c = 0 is a straight line.

Example: since, x = 2 and y = 1 satisfy the linear equation in two variables x – 4y + 2 = 0, so we can say that x = 2 and y = 1 is a solution of the above equation.

Tip # 7: A system of a pair of linear equations in two variables is said to be consistent if it has at least one solution. A system of a pair of linear equations in two variables is said to be inconsistent if it has no solution.

Tip # 8: The system of a pair of linear equations \(a_{1}x + b_{1}y + c_{1} = 0\) and \(a_{2}x + b_{2}y + c_{2} = 0\) has:

A unique solution (i.e. consistent) if: \(\frac{a_1}{a_2}\) is not equals to \(\frac{b_1}{b_2}\)

In this case the graphs of the pair of linear equations intersect at only one point and the coordinate of points is the solution of the pair of linear equations.

No solution (i.e. inconsistent) if:

\(\frac{a_1}{a_2}\) is equals to \(\frac{b_1}{b_2}\) not equals to \(\frac{c_1}{c_2}\)

In this case the graphs of the linear equations are parallel to each other, that is the lines do not intersect each other.

III. An infinite number of solution (i.e. consistent) if:

\(\frac{a_1}{a_2}\) is equals to \(\frac{b_1}{b_2}\) is also equals to \(\frac{c_1}{c_2}\)

In this case the graphs of the linear equations coincide with each other, that is the lines overlap each other.

Tip # 9: Root of Equation – The value(s) of x which satisfy the equation is/are called the root of the equation.

Maximum number of real roots of a linear, quadratic, cubic and a quartic equation are 1, 2, 3 and 4 respectively.

Example: x = 3, satisfy the equation \(x^2 – 6x + 9 = 0\), so we can say that x = 3 is a root of the equation.

Tip # 10: To solve a quadratic equation by the help of quadratic formula method, we first need to find the discriminant (D), where, \(D = b^2 – 4ac\).

• If D > 0, then the roots of the quadratic equation are real and distinct.
• If D = 0, the roots of the quadratic equation are real and equal.
• If D < 0, the roots of the quadratic equation are imaginary.

Tip # 11: Usually we denote the roots of quadratic equations by α and β, where α + β = \(-\frac{b}{a}\)and α × β = \(\frac{c}{a}\)

Tip # 12: If it is given that α and β are the roots of the quadratic equation, then we can find the quadratic equation by the help of the formula \(x^2 – (α + β)x + αβ = 0\)

Example: if x = 3 and x = 4 are the roots of a quadratic equation then we can find the equation by taking α = 2 and β = 4.

α + β = 7, α.β = 3 × 4 = 12

So, the required quadratic equation is x2 – 7x + 12 = 0

Tip # 13: If α, β and δ are the roots of equation ax3 + bx2 + cx + d = 0, then α + β + δ = -b/a, αβ + βδ + δα = c/a and αβδ = -d/a

Tip # 14: If α, β and δ are the roots of a cubic equation, then we can find the equation by using the formula x3 – (α + β + δ)x2 + (αβ + βδ + δα)x – αβδ = 0

Tip # 15: We can find the roots of quadratic equations by the help of factorization method, and quadratic formula method.

Uses of Algebra in Real life

Algebra is used in various fields such as DBMS (Database Management System), Medicine, Architecture and Accounting. Algebra can also be used for everyday problem-solving and developing critical thinking such as logic, patterns, and deductive and inductive reasoning or understanding the core concepts of algebra can prove to be beneficial to everyone to solve the complex problems which involve numbers.

It plays a major role in solving various problems in the workplace where real-life scenarios of unknown variables involve expenses and profits calculation related to employees. Algebraic equations can be used to determine the missing factors in these scenarios.

Some Important Basic Algebra Formulas

Below candidates can find some of the frequently used formulas in Algebra to solve questions.

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

\((a – b)^2 = a^2 – 2ab + b^2\)

\((a + b)^2 = (a – b)^2 + 4ab\)

\((a – b)^2 = (a + b)^2 – 4ab\)

\(a^2 + b^2 = (a + b)^2 – 2ab\)

\(a^2 + b^2 = (a – b)^2 + 2ab\)

\(a^2 – b^2 = (a + b) (a – b)\)

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

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

\((a + b)^3 = a^3 + b^3 + 3ab (a + b)\)

\((a – b)^3 = a^3 – b^3 – 3ab (a – b)\)

\(a^3 + b^3 = (a + b)^3 – 3ab (a + b)\)

\(a^3 – b^3 = (a – b)^3 + 3ab (a – b)\)

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

\(a^3 + b^3 + c^3 – 3abc = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ac) = ½ (a + b + c)[(a – b)^2 + (b – c)^2 + (c – a)^2]\)

\(a^3 + b^3 + c^3 = 3abc If (a + b + c) = 0\)

\((a + b + c)3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)\)

\(a^4 + b^4 + a^2b^2 = (a^2 – ab + b^2)(a^2 + ab + b^2)\)

If \(x^2 + y^2 + z^2 = 0, then x = y = z = 0\)

If \(a^3 + b^3 + c^3 = 3abc Then (a + b + c) = 0\)

If xy = 1, then 1/(x+1) + 1/(y+1) = 1

Other Important Formulas

Apart from all the formulas mentioned above candidates can find an Algebra formulas cheat sheet filled with shortcut formulas which candidates can use to solve the questions related to Algebra quickly.

Notes: For solving questions related to long roots, candidates can follow the below-mentioned formulas.

For the two forms mentioned below, if the difference between the factors of x is 1, then

the factor which is greater in value will be the answer of the given question.

Algebra Sample Questions with Answers

Question 1: If (a – 2)2 + (b + 3)2 + (c – 5)2 = 0, then find the value of a – b – c.

Solution: Given: (a – 2)2 + (b + 3)2 + (c – 5)2 = 0

So, a – 2 = 0 ⇒ a = 2

And, b + 3 = 0 ⇒ b = -3

And, c – 5 = 0 ⇒ c = 5

Therefore, a – b – c = 2 – (-3) – 5 = 0

 

Question 2: If (2x + 3)3 + (x – 8)3 + (x + 13)3 = (2x + 3) (3x – 24) (x + 13), then what is the value of x?

Solution: As we know,

If x3 + y3 + z3 – 3xyz = 0, then x + y + z = 0

⇒ (2x + 3)3 + (x – 8)3 + (x + 13)3 = (2x + 3) (3x – 24) (x + 13)

⇒ (2x + 3)3 + (x – 8)3 + (x + 13)3 – 3 (2x + 3) (x – 8) (x + 13) = 0

⇒ (2x + 3) + (x – 8) + (x + 13) = 0

⇒ 4x + 8 = 0

⇒ 4x = -8

⇒ x = -8/4 = -2

Question 3: If x + 1/x = 2 then find the value of x^5 + 1/x^5

Solution: Given: x + 1/x = 2

So, x = 1

So, x^5 + 1/x^5 = 1 + 1/1 = 2

Question 4: If x + 1/x = -2 then find the value of x^5 + 1/x^5 and x^4 + 1/x^4

Solution: Given: x + 1/x = 2

So, x = -1

So, x^5 + 1/x^5 = 1 + 1/1 = -2 and x^4 + 1/x^4 = 2

Question 5: If x + 1/x = 2, then find the value of x^2 + 1/x^2

Solution: x^2 + 1/x^2 = 2

Question 6: If x + 1/x = 5, then find the value of x^3 – 1/x^3

Solution: x^3 – 1/x^3 = 5^3 + 3 x 5 = 140

Question 7: If x + 1/x = 1, then find the value of x^20 – 1/x^20

Solution: Given: x + 1/x = 1

So, x = -1

x^20 – 1/x^20 = 2

Question 8: If x + 1/x = -1, then find the value of x^20 – 1/x^20

Solution: Given: x + 1/x = -1

So, X = +1

x^20 – 1/x^20 = 0

Question 9: If x + 1/x = 5, then find the value of x^3 + 1/x^3

Solution: x^3 + 1/x^3 = 5^3 – 3 x 5 = 110

Question 10: Find the value of

Solution: = 3

Conclusion

In conclusion, Algebra is an essential math skill that you'll encounter on exams like the SAT, ACT, and GRE. By understanding key concepts such as algebraic expressions, equations, and polynomials, and practicing strategies like factorization and the quadratic formula, you’ll be well-prepared. Mastering these topics will help you solve complex problems efficiently and boost your confidence when tackling algebra-related questions on any standardized test. Keep practicing!