Algebraic Expressions — for Class 7-8

What Are Algebraic Expressions?

You already know arithmetic — adding, subtracting, multiplying numbers. Algebra is the same thing, except we allow some numbers to be unknown. We give those unknowns names: $x$ , $y$ , $a$ , $b$ , whatever we like.

An algebraic expression is a combination of numbers, variables (unknown quantities), and arithmetic operations. So $3x + 5$ is an algebraic expression — it has a variable ( $x$ ), a coefficient ( $3$ ), and a constant ( $5$ ).

Why does this matter? Because real problems often have unknowns. “I have some mangoes. After eating 3, I have 7 left.” That “some mangoes” is $x$ . The expression $x - 3$ describes the situation. This is exactly how Class 7 and Class 8 algebra begins — formalising what we already do intuitively.

By Class 8, we push further: multiplying expressions together, factorising them, and recognising patterns (like $(a+b)^2$ ) that appear everywhere in JEE and board exams. The foundation we build here carries us through Class 10 polynomials, Class 11 functions, and beyond.

Key Terms and Definitions

Variable: A letter that represents an unknown or changing quantity. Example: $x$ , $y$ , $n$ .

Constant: A fixed number in an expression. In $4x + 7$ , the constant is $7$ .

Coefficient: The number multiplied by a variable. In $4x$ , the coefficient of $x$ is $4$ . Students often miss this — if you see $-x$ , the coefficient is $-1$ , not $1$ .

Term: A single chunk separated by $+$ or $-$ signs. In $3x^2 - 5x + 2$ , there are three terms: $3x^2$ , $-5x$ , and $2$ .

Like Terms: Terms with exactly the same variable(s) raised to the same power(s). $3x$ and $7x$ are like terms. $3x$ and $3x^2$ are NOT like terms — the powers differ.

Monomial: One term. Example: $5xy$ .

Binomial: Two terms. Example: $3x - 2$ .

Trinomial: Three terms. Example: $x^2 + 5x + 6$ .

Polynomial: Any expression with one or more terms. Monomials, binomials, trinomials are all polynomials.

Degree of an expression: The highest power of the variable. In $4x^3 - x + 9$ , the degree is $3$ .

The degree of $x^2y$ is $2 + 1 = 3$ — you add the exponents of all variables in that term. This trips students up in Class 8 when two variables appear together.

Core Concepts and Methods

Addition and Subtraction of Algebraic Expressions

The rule is simple: only like terms can be combined.

Think of it like fruits — you can add apples to apples, but you can’t “add” apples to oranges and call the result a single fruit.

Method:

Identify like terms (same variable, same power).
Add/subtract their coefficients.
Keep unlike terms as they are.

Worked Example: Simplify $(3x^2 + 5x - 2) + (x^2 - 3x + 7)$

Group like terms:

= (3x^2 + x^2) + (5x - 3x) + (-2 + 7)

= 4x^2 + 2x + 5

Worked Example: Subtract $(2a - 3b + c)$ from $(5a + b - 2c)$

“Subtract A from B” means B $-$ A — many students reverse this.

(5a + b - 2c) - (2a - 3b + c)

= 5a + b - 2c - 2a + 3b - c

= 3a + 4b - 3c

When subtracting, the sign of every term in the bracket flips. $-(2a - 3b + c)$ becomes $-2a + 3b - c$ . Students often flip only the first term sign and leave the rest unchanged — this is the single most common Class 7 algebra error.

Multiplication of Algebraic Expressions

Monomial × Monomial

Multiply coefficients, add exponents of same variables.

3x^2 \times 4x^3 = 12x^5

2xy \times 5x^2y = 10x^3y^2

Monomial × Polynomial

Use the distributive law: multiply the monomial with each term of the polynomial.

2x(3x^2 - 5x + 1) = 6x^3 - 10x^2 + 2x

Binomial × Binomial

Multiply each term of the first binomial with each term of the second. This gives four products — remember the acronym FOIL (First, Outer, Inner, Last) if it helps.

(x + 3)(x + 5) = x \cdot x + x \cdot 5 + 3 \cdot x + 3 \cdot 5

= x^2 + 5x + 3x + 15 = x^2 + 8x + 15

\boxed{(a+b)^2 = a^2 + 2ab + b^2}

\boxed{(a-b)^2 = a^2 - 2ab + b^2}

\boxed{(a+b)(a-b) = a^2 - b^2}

\boxed{(x+a)(x+b) = x^2 + (a+b)x + ab}

These identities appear in every CBSE Class 8 annual exam and form the backbone of factorisation in Class 9-10. Memorise these before anything else.

Factorisation (Class 8)

Factorisation is the reverse of multiplication. We write an expression as a product of simpler expressions.

Method 1 — Common Factor:

6x^2y + 9xy^2 = 3xy(2x + 3y)

Find the HCF of all terms, then take it out.

Method 2 — Using Identities:

x^2 - 25 = x^2 - 5^2 = (x+5)(x-5)

4a^2 + 12ab + 9b^2 = (2a)^2 + 2(2a)(3b) + (3b)^2 = (2a + 3b)^2

Method 3 — Splitting the Middle Term (for $x^2 + bx + c$ ):

Find two numbers that multiply to $c$ and add to $b$ .

For $x^2 + 7x + 12$ : we need two numbers multiplying to $12$ and adding to $7$ . That’s $3$ and $4$ .

x^2 + 7x + 12 = x^2 + 3x + 4x + 12 = x(x+3) + 4(x+3) = (x+3)(x+4)

Evaluating Expressions

When a value is given for the variable, substitute and calculate.

If $p = 3$ , find $p^2 - 2p + 5$ :

= (3)^2 - 2(3) + 5 = 9 - 6 + 5 = 8

Always substitute inside brackets, especially for negative values. If $x = -2$ , then $x^2 = (-2)^2 = 4$ , not $-4$ . This is a very common sign error.

Solved Examples

Easy — CBSE Class 7

Q: Add $3x^2 - 4x + 2$ and $x^2 + 6x - 5$ .

Solution:

(3x^2 + x^2) + (-4x + 6x) + (2 - 5)

= 4x^2 + 2x - 3

Medium — CBSE Class 8

Q: Expand using identity: $(3x - 2y)^2$

Solution: Use $(a-b)^2 = a^2 - 2ab + b^2$ with $a = 3x$ , $b = 2y$ :

(3x)^2 - 2(3x)(2y) + (2y)^2

= 9x^2 - 12xy + 4y^2

Medium-Hard — CBSE Class 8

Q: Factorise $2x^2 + 5x + 3$ .

Solution: Multiply leading coefficient by constant: $2 \times 3 = 6$ . We need two numbers that multiply to $6$ and add to $5$ : that’s $2$ and $3$ .

2x^2 + 2x + 3x + 3 = 2x(x+1) + 3(x+1) = (x+1)(2x+3)

Hard — CBSE Class 8 / Class 9 Bridge

Q: Using identities, find $97 \times 103$ .

Solution: Rewrite as $(100 - 3)(100 + 3)$ — now it’s $(a-b)(a+b) = a^2 - b^2$ .

= 100^2 - 3^2 = 10000 - 9 = 9991

This technique appears frequently in mental maths questions and is a favourite for Class 8 board exams.

Exam-Specific Tips

CBSE Class 7 Marking Pattern: Algebraic expressions typically carry 6-8 marks in the annual exam — usually 1 question in each section (1-mark, 2-mark, 3-mark). The 3-mark question almost always involves subtracting one polynomial from another, or finding the value of an expression.

CBSE Class 8 Marking Pattern: Standard identities dominate Section C (3-mark) and Section D (4-mark). Expect: expand using identity (2 marks), factorise using identity (3 marks), and at least one word-problem that requires forming and simplifying an expression (4 marks). Practise all four standard identities both ways — expand and factorise.

JEE Connection: Every factorisation skill from Class 8 reappears in Class 11 algebra and Class 12 integration. Students who struggle with JEE Main factorisation questions almost always trace the gap back to shaky Class 8 foundations. Get these identities under your belt now.

For ICSE students: ICSE Class 8 tests identities more heavily than CBSE, often asking you to prove that an expression equals a specific value without direct computation. For example: “If $a + b = 5$ and $ab = 6$ , find $a^2 + b^2$ .” Method: use $(a+b)^2 = a^2 + 2ab + b^2 \Rightarrow 25 = a^2 + 12 \Rightarrow a^2 + b^2 = 13$ .

Common Mistakes to Avoid

Mistake 1 — Adding unlike terms: $3x + 2y \neq 5xy$ . These cannot be combined. The answer stays as $3x + 2y$ .

Mistake 2 — Wrong sign when subtracting: $(5x - 3) - (2x - 7)$ is NOT $3x - 10$ . Distribute the minus: $5x - 3 - 2x + 7 = 3x + 4$ .

Mistake 3 — Forgetting the middle term in $(a+b)^2$ : $(x+3)^2 \neq x^2 + 9$ . The correct answer is $x^2 + 6x + 9$ . The $2ab$ term is non-negotiable.

Mistake 4 — Squaring a negative number as negative: If $a = -4$ , then $a^2 = (-4)^2 = 16$ , not $-16$ . Always substitute with brackets around negative values.

Mistake 5 — Incomplete factorisation: $4x^2 - 16$ factorised as $4(x^2 - 4)$ is incomplete. The correct answer is $4(x+2)(x-2)$ . Always check if terms inside brackets can be factorised further.

Practice Questions

Q1 (Class 7): Simplify: $(4a + 3b - 2c) + (2a - b + 5c) - (a + 2b - 3c)$

Removing brackets: $4a + 3b - 2c + 2a - b + 5c - a - 2b + 3c$

Grouping: $(4+2-1)a + (3-1-2)b + (-2+5+3)c = 5a + 0b + 6c = 5a + 6c$

Q2 (Class 7): If $x = 2$ , $y = -1$ , find the value of $3x^2 - 2xy + y^2$ .

$= 3(2)^2 - 2(2)(-1) + (-1)^2 = 3(4) + 4 + 1 = 12 + 4 + 1 = 17$

Q3 (Class 7-8): Multiply: $(2x - 3)(x + 5)$

$= 2x(x) + 2x(5) - 3(x) - 3(5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$

Q4 (Class 8): Expand: $(2a + 3b)^2$

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 2a$ , $b = 3b$ :

$= (2a)^2 + 2(2a)(3b) + (3b)^2 = 4a^2 + 12ab + 9b^2$

Q5 (Class 8): Factorise: $x^2 - 64$

$= x^2 - 8^2 = (x+8)(x-8)$ using $a^2 - b^2 = (a+b)(a-b)$

Q6 (Class 8): Factorise: $x^2 + 5x + 6$

Need two numbers multiplying to $6$ and adding to $5$ : that’s $2$ and $3$ .

$x^2 + 2x + 3x + 6 = x(x+2) + 3(x+2) = (x+2)(x+3)$

Q7 (Class 8 — identity trick): Find $105^2$ using an identity.

$105 = 100 + 5$ , so use $(a+b)^2 = a^2 + 2ab + b^2$ :

$(100+5)^2 = 10000 + 1000 + 25 = 11025$

Q8 (Class 8): Factorise: $6a^2 + 7a - 3$

Multiply $6 \times (-3) = -18$ . Need two numbers multiplying to $-18$ and adding to $7$ : that’s $9$ and $-2$ .

$6a^2 + 9a - 2a - 3 = 3a(2a+3) - 1(2a+3) = (2a+3)(3a-1)$

Q9 (Class 8): If $x + \frac{1}{x} = 5$ , find $x^2 + \frac{1}{x^2}$ .

Square both sides: $\left(x + \frac{1}{x}\right)^2 = 25$

$x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 25$

$x^2 + 2 + \frac{1}{x^2} = 25 \Rightarrow x^2 + \frac{1}{x^2} = 23$

This type of question is a favourite in ICSE Class 8 exams and appears in Class 9 as well.

Frequently Asked Questions

What is the difference between an expression and an equation?

An expression has no equals sign: $3x + 2$ is an expression. An equation sets two expressions equal: $3x + 2 = 8$ . Equations have solutions; expressions are just simplified.

Can we add $x$ and $x^2$ ?

No. $x$ and $x^2$ are unlike terms — the powers differ. $x + x^2$ stays as $x + x^2$ and cannot be simplified further.

Why do we use letters like $x$ and $y$ ? Can we use any letter?

Absolutely any letter works. $p$ , $q$ , $n$ , $t$ — all fine. Convention: $x$ , $y$ , $z$ for unknowns; $a$ , $b$ , $c$ for constants in identities; $n$ for natural numbers and counting. But there’s no strict rule.

What is the degree of a constant, like $7$ ?

A constant like $7$ can be written as $7x^0$ , so its degree is $0$ . The degree of the zero polynomial ( $0$ itself) is undefined — some textbooks say $-1$ or $-\infty$ , but for Class 7-8 boards, just remember constants have degree $0$ .

How do I remember all four standard identities?

Memorise just two: $(a+b)^2$ and $(a-b)(a+b)$ . The third, $(a-b)^2$ , is $(a+b)^2$ with $b$ replaced by $-b$ . The fourth, $(x+a)(x+b)$ , comes straight from the FOIL method. Two memorised, two derived.

Is factorisation reversible? How do I verify?

Yes — always multiply your factors back and check you get the original expression. If $(x+3)(x-2) = x^2 + x - 6$ , then the factorisation of $x^2 + x - 6$ is $(x+3)(x-2)$ . Verification is a 1-mark check in CBSE board exams and saves you from silly errors.

What if no factor method works?

For Class 7-8, the question will always factorise using one of the methods above. If you’re stuck, first check for a common factor (often missed), then try identities. In higher classes you’ll learn the quadratic formula for expressions that don’t factorise neatly.

Why does $(a+b)^2 \neq a^2 + b^2$ ?

Because squaring a sum means multiplying $(a+b)$ by $(a+b)$ — you must use the distributive law, which generates the cross term $2ab$ . Thinking of it geometrically helps: a square of side $(a+b)$ has area $a^2 + 2ab + b^2$ — there are two rectangles of area $ab$ in the middle.