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ICSE • Class 8 • Mathematics • Ch 12
Estimated Time: 45 Mins
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Algebraic Expressions

In ICSE Class 8 Mathematics, "Algebraic Expressions" provides an authoritative, mathematically rigorous master study guide investigating the terminology, operations, and polynomial arithmetic of algebraic expressions. This comprehensive chapter explores Basic Concepts (Constants, Variables, Terms, Numerical and Literal coefficients; Like terms vs Unlike terms; Degree of a polynomial in one variable and multi-variables), Classification of Polynomials (Monomials, Binomials, Trinomials, Multinomials), Addition and Subtraction of Algebraic Expressions (Column / Vertical method and Horizontal method; Grouping like terms with sign precision), Multiplication of Algebraic Expressions (1. Monomial by monomial: multiply numerical coefficients and apply product law of exponents to literals, 2. Polynomial by monomial: Distributive law $a(b + c + d) = ab + ac + ad$, 3. Polynomial by polynomial: multiplying every term of the first polynomial by every term of the second polynomial and combining like terms; Binomial by binomial and trinomial by trinomial expansions), Division of Algebraic Expressions (1. Monomial by monomial, 2. Polynomial by monomial: dividing each term individually, 3. Polynomial by polynomial: Long Division Algorithm; Writing polynomials in standard descending powers of variable; Quotient and Remainder verification: $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$), and Evaluation of Expressions by substitution aligned with the 2026–27 CISCE ICSE curriculum.

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How Did a Medieval Persian Scholar's Book of "Restoring Broken Balances" Give Humanity the Universal Language of Digital Algorithms?

In the 9th century in Baghdad, mathematician Muhammad ibn Musa al-Khwarizmi sat in the House of Wisdom writing a revolutionary manuscript titled Al-Kitab al-mukhtasar fi hisab al-jabr wal-muqabala. The Arabic word Al-Jabr meant "the reunion of broken parts" or "bone-setting"—restoring balance to a fractured equation by shifting terms from one side to the other! From that word Al-Jabr, the world got ALGEBRA, and from Al-Khwarizmi's name, we got ALGORITHM! Algebra replaced rigid arithmetic with flexible symbols: instead of solving a problem for "5 apples" or "10 apples", we solve for $x$, discovering an eternal truth that works for five, five billion, or infinite apples! What is the degree of a multi-variable term like $5x^3y^4$? Why must polynomials be strictly arranged in descending powers before starting Long Division? Let's master algebraic expressions.

Why This Chapter Matters

Algebraic expressions are the universal symbolic code of all STEM disciplines: writing computer software algorithms, engineering aerodynamics, financial econometrics, and physics equations of motion. Mastering polynomial multiplication and long division is an indispensable pillar of ICSE mathematics.

Before You Begin (Prerequisites)

  • Operations on integers (rules of signs for $+$ and $-$).
  • Basic algebraic terms and like terms from Class 7.
  • Laws of exponents (Product law $x^a \times x^b = x^{a+b}$).

What You Will Learn (Core Objectives)

  • Identify terms, coefficients, like/unlike terms, and determine the degree of any polynomial.
  • Add and subtract complex polynomials using both horizontal and vertical column methods.
  • Multiply polynomials (binomial by binomial, trinomial by trinomial) using the distributive law.
  • Divide a polynomial by a monomial and execute polynomial Long Division with remainders.
  • Verify the Division Algorithm: $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$.
  • Evaluate algebraic expressions by numerical value substitution.

Chapter Roadmap & Progression

1 1. Fundamentals & Classification of...
2 2. Addition & Subtraction of Polyno...
3 3. Multiplication of Polynomials
4 4. Polynomial Long Division Algorit...

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamentals & Classification of Polynomials

Understand
A. Anatomy of an Expression:
  • Constant: A symbol having a fixed numerical value (e.g., $7, -12, \frac{3}{4}$).
  • Variable: A symbol whose value can change, represented by letters ($x, y, z$).
  • Term: A constant, variable, or product/quotient of constants and variables separated by $+$ or $-$ signs (e.g., in $5x^2 - 3xy + 8$, terms are $5x^2, -3xy,$ and $8$).
  • Coefficient: In $-7x^2y$, the numerical coefficient is $-7$, and the literal coefficient of $x^2$ is $-7y$.
  • Like Terms: Terms having the exact same literal variables raised to the exact same powers (e.g., $4x^2y$ and $-9x^2y$). Only like terms can be added or subtracted!
B. Degree of a Polynomial:
  • In one variable: The highest exponent of the variable (e.g., in $4x^5 - 2x^3 + 7$, degree is 5).
  • In multiple variables: The maximum sum of the exponents of all variables in any single term (e.g., in $7x^2y^3 - 5x^4y^2 + 8$, term $7x^2y^3$ has sum $2+3=5$; term $-5x^4y^2$ has sum $4+2=6$. Hence, degree is 6!).

2. Addition & Subtraction of Polynomials

Addition & Subtraction
A. Horizontal & Column Methods:

To add or subtract, group like terms together and add/subtract their numerical coefficients while leaving literal variables untouched.

  • The Subtraction Sign Rule: To subtract expression $B$ from $A$ ($A - B$), enclose $B$ in brackets and change the sign of every single term inside $B$ when removing brackets: $$(5x^2 - 3x + 4) - (2x^2 - 7x - 8) = 5x^2 - 3x + 4 - 2x^2 + 7x + 8 = \mathbf{3x^2 + 4x + 12}$$

3. Multiplication of Polynomials

Multiplication
A. Monomial by Monomial:

Multiply numerical coefficients together, then apply the product law of exponents ($x^m \times x^n = x^{m+n}$):

$$(-4x^2y) \times (3xy^3) = (-4 \times 3) \times (x^2 \cdot x) \times (y \cdot y^3) = \mathbf{-12x^3y^4}$$
B. Polynomial by Polynomial:

Multiply every term of the first polynomial by every term of the second (the Distributive Law):

$$(2x - 3)(4x^2 + 5x - 6) = 2x(4x^2 + 5x - 6) - 3(4x^2 + 5x - 6)$$ $$= (8x^3 + 10x^2 - 12x) - (12x^2 + 15x - 18)$$ $$= 8x^3 + 10x^2 - 12x - 12x^2 - 15x + 18 = \mathbf{8x^3 - 2x^2 - 27x + 18}$$

4. Polynomial Long Division Algorithm

Long Division
Step-by-Step Procedure:
  1. Arrange both the dividend and divisor in descending powers of the variable.
  2. Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
  3. Multiply the entire divisor by this quotient term, write it below the dividend matching like terms, and subtract (reversing signs).
  4. Bring down the next term and repeat until the remainder degree is strictly less than the divisor degree.
  5. Verification Identity: $$\mathbf{\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}}$$

Key Formulas, Identities & Theorems

Polynomial Division Algorithm
$$P(x) = D(x) \cdot Q(x) + R(x) \quad [\deg R(x) < \deg D(x)]$$
Dividend = Divisor x Quotient + Remainder.
Multi-Variable Term Degree
$$\deg(a x^p y^q z^r) = p + q + r$$
Sum of exponents of all variables in a monomial term.

Algebra: Multi-Variable Term Anatomy & Long Division

Algebraic Expressions: Term Anatomy & Division Algorithm ANATOMY OF AN ALGEBRAIC TERM -7 x3 y4 • Numerical Coefficient: -7 • Literal Variables: x3 and y4 • Degree of Term: 3 + 4 = 7 (Sum of powers!) • Like Terms: Same variables with same powers (Add/Sub) • Unlike Terms: Cannot be combined! POLYNOMIAL LONG DIVISION Dividend = Divisor × Quotient + Remainder 1. Arrange in Descending Powers: x3 → x2 → x → Constant 2. Subtract Step: FLIP ALL SIGNS! (+ becomes - and - becomes +) 3. Stop when: Degree of Remainder < Degree of Divisor ONLY LIKE TERMS CAN BE ADDED • MULTIPLY POWERS ADD • DIVIDEND = DIVISOR × Q + R

Chapter Summary & 10 Key Takeaways

Takeaway 1
An algebraic expression combines constants and variables using arithmetic operations (+, -, *, /).
Takeaway 2
Only like terms (identical variables and exponents) can be added or subtracted.
Takeaway 3
The degree of a polynomial in one variable is the highest exponent; in multiple variables, it is the highest sum of exponents in any single term.
Takeaway 4
When subtracting an expression, flip the sign of every term inside the subtracted expression.
Takeaway 5
Monomials are multiplied by multiplying numerical coefficients and adding powers of like variables.
Takeaway 6
Polynomial multiplication applies the distributive law: every term in the first polynomial multiplies every term in the second.
Takeaway 7
In polynomial long division, always arrange dividend and divisor in descending powers of the variable.
Takeaway 8
In each subtraction step of long division, invert the signs of the bottom terms.
Takeaway 9
Division algorithm: Dividend = Divisor * Quotient + Remainder.
Takeaway 10
To evaluate an expression, substitute given numerical values for variables and follow order of operations (PEMDAS).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Subtract $(2x^3 - 4x^2 + 3x + 5)$ from $(4x^3 + x^2 - 2x + 7)$.
Reveal Answer & Explanation
Answer: Step 1: Set up the subtraction: $(\text{Second Expression}) - (\text{First Expression})$:
$$(4x^3 + x^2 - 2x + 7) - (2x^3 - 4x^2 + 3x + 5)$$
Step 2: Remove brackets by changing the sign of every term in the subtracted expression:
$$= 4x^3 + x^2 - 2x + 7 - 2x^3 + 4x^2 - 3x - 5$$
Step 3: Group and combine like terms:
$$= (4x^3 - 2x^3) + (x^2 + 4x^2) + (-2x - 3x) + (7 - 5)$$
$$= \mathbf{2x^3 + 5x^2 - 5x + 2}$$.
Change signs of the subtracted expression: $(4x^3 - 2x^3) + (x^2 + 4x^2) + (-2x - 3x) + (7 - 5) = 2x^3 + 5x^2 - 5x + 2$.
2
Multiply: $(3x^2 - 5x + 2)$ by $(2x - 3)$.
Reveal Answer & Explanation
Answer: Apply the Distributive Law:
$$(2x - 3)(3x^2 - 5x + 2) = 2x(3x^2 - 5x + 2) - 3(3x^2 - 5x + 2)$$
Step 1: Multiply by $2x$:
$$2x(3x^2) - 2x(5x) + 2x(2) = 6x^3 - 10x^2 + 4x$$
Step 2: Multiply by $-3$:
$$-3(3x^2) - 3(-5x) - 3(2) = -9x^2 + 15x - 6$$
Step 3: Combine all terms:
$$= 6x^3 + (-10x^2 - 9x^2) + (4x + 15x) - 6$$
$$= \mathbf{6x^3 - 19x^2 + 19x - 6}$$.
Expand $2x(3x^2 - 5x + 2) - 3(3x^2 - 5x + 2) = 6x^3 - 10x^2 + 4x - 9x^2 + 15x - 6 = 6x^3 - 19x^2 + 19x - 6$.
3
Divide $(2x^2 + 11x + 15)$ by $(x + 3)$ using the Long Division method. State the quotient and remainder.
Reveal Answer & Explanation
Answer:

Step 1: Divide first term: $\frac{2x^2}{x} = 2x$. Multiply divisor $(x + 3)$ by $2x$: $2x^2 + 6x$.
Subtract from $(2x^2 + 11x)$: $(2x^2 + 11x) - (2x^2 + 6x) = 5x$.
Step 2: Bring down $+15$ to form $5x + 15$.
Step 3: Divide: $\frac{5x}{x} = 5$. Multiply divisor $(x + 3)$ by $5$: $5x + 15$.
Subtract: $(5x + 15) - (5x + 15) = 0$.
• Quotient:

$$\mathbf{2x + 5}$$


• Remainder:

$$\mathbf{0}$$


• Verification: $(x + 3)(2x + 5) + 0 = 2x^2 + 5x + 6x + 15 = 2x^2 + 11x + 15$. Correct!


First quotient term is $2x$, second is $+5$. Quotient is $2x + 5$, remainder is $0$.
4
What is the degree of the polynomial: $3x^4y^2 - 5x^2y^5 + 7x^3y^3 - 12$?
Reveal Answer & Explanation
Answer:

Calculate the degree of each term by summing the powers of variables in that term:
• Term 1 ($3x^4y^2$): $4 + 2 = 6$
• Term 2 ($-5x^2y^5$): $2 + 5 = \mathbf{7}$
• Term 3 ($7x^3y^3$): $3 + 3 = 6$
• Term 4 ($-12$): Constant term has degree $0$.
The degree of a polynomial is the highest degree among all its terms.
• Degree of Polynomial:

$$\mathbf{7}$$

.


The term $-5x^2y^5$ has exponent sum $2 + 5 = 7$, which is the highest.
5
Simplify: $2x - [3y - \{2x - (y - x)\}]$.
Reveal Answer & Explanation
Answer: Work from the innermost parentheses outward:
Step 1: Remove parentheses $(y - x)$:
$$y - x \implies 2x - (y - x) = 2x - y + x = 3x - y$$
Step 2: Substitute into curly braces $\{ \}$:
$$3y - \{3x - y\} = 3y - 3x + y = 4y - 3x$$
Step 3: Substitute into square brackets $[ \]$:
$$2x - [4y - 3x] = 2x - 4y + 3x = \mathbf{5x - 4y}$$.
Innermost: $2x - y + x = 3x - y$. Next: $3y - 3x + y = 4y - 3x$. Final: $2x - 4y + 3x = 5x - 4y$.
6
If $x = 2$ and $y = -1$, evaluate the expression: $3x^2y - 2xy^2 + 4xy - 5$.
Reveal Answer & Explanation
Answer: Substitute $x = 2$ and $y = -1$:
$$= 3(2)^2(-1) - 2(2)(-1)^2 + 4(2)(-1) - 5$$
$$= 3(4)(-1) - 2(2)(1) + 4(-2) - 5$$
$$= -12 - 4 - 8 - 5 = \mathbf{-29}$$.
$3(4)(-1) - 2(2)(1) + 4(-2) - 5 = -12 - 4 - 8 - 5 = -29$.
7
What must be added to $(x^3 - 3x^2 + 5x - 2)$ to get $(2x^3 + x^2 - 4x + 6)$?
Reveal Answer & Explanation
Answer: Let the required expression to be added be $P$.
$$(x^3 - 3x^2 + 5x - 2) + P = (2x^3 + x^2 - 4x + 6)$$
$$P = (2x^3 + x^2 - 4x + 6) - (x^3 - 3x^2 + 5x - 2)$$
$$P = 2x^3 + x^2 - 4x + 6 - x^3 + 3x^2 - 5x + 2$$
Combine like terms:
$$P = (2x^3 - x^3) + (x^2 + 3x^2) + (-4x - 5x) + (6 + 2)$$
$$\mathbf{P = x^3 + 4x^2 - 9x + 8}$$.
Subtract first from second: $(2x^3 - x^3) + (x^2 + 3x^2) + (-4x - 5x) + (6 + 2) = x^3 + 4x^2 - 9x + 8$.
8
Divide $(x^3 - 1)$ by $(x - 1)$ using long division and show that the remainder is $0$.
Reveal Answer & Explanation
Answer:

Write $x^3 - 1$ with place-holder zero terms in descending order: $x^3 + 0x^2 + 0x - 1$.
Step 1: $\frac{x^3}{x} = x^2$. Multiply $(x - 1)$ by $x^2$: $x^3 - x^2$.
Subtract: $(x^3 + 0x^2) - (x^3 - x^2) = x^2$.
Step 2: Bring down $0x$ to get $x^2 + 0x$. $\frac{x^2}{x} = x$. Multiply $(x - 1)$ by $x$: $x^2 - x$.
Subtract: $(x^2 + 0x) - (x^2 - x) = x$.
Step 3: Bring down $-1$ to get $x - 1$. $\frac{x}{x} = 1$. Multiply $(x - 1)$ by $1$: $x - 1$.
Subtract: $(x - 1) - (x - 1) = 0$.
• Quotient:

$$\mathbf{x^2 + x + 1}$$


• Remainder:

$$\mathbf{0}$$

.


Use placeholder zero terms $x^3 + 0x^2 + 0x - 1$. Quotient is $x^2 + x + 1$, remainder is 0.
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