Algebra – ICSE Class 10 Mathematics Notes & Guide

Algebra forms the backbone of Class 10 mathematics. It helps you translate word problems into symbols, solve equations, and understand the behaviour of functions. Mastering the concepts below will give you confidence to tackle any algebraic question in the ICSE exam.

Key concepts

Polynomials – degree, zeros and factorisation

A polynomial is an expression of the form anxⁿ + a(n-1)xⁿ⁻¹+… + a_1x + a₀ where the exponents are non‑negative integers. The highest exponent with a non‑zero coefficient is the degree. A zero (or root) of a polynomial P(x) is a value α such that P(α)=0. By the Factor Theorem, x-α is a factor of P(x). Repeated zeros give repeated linear factors, e.g., (x-2)³.

Quadratic equations – factorisation, completing the square and formula method

A quadratic equation is ax²+bx+c=0 (a≠0). It can be solved by:

1. Factorisation – express the left side as (px+q)(rx+s)=0 and set each factor to zero.
2. Completing the square – rewrite as ax²+bx+c = a≤ft(x+b/2aright)² - (b²-4ac)/4a and then take square roots.
3. Quadratic formula – x = frac-b ± √(b²-4ac)2a.

Each method is useful depending on the coefficients.

Linear equations in two variables and their graphical representation

A linear equation in two variables has the form Ax + By + C = 0. Its graph is a straight line. The slope m = -A/B (if B≠0) and the y‑intercept is -C/B. When B=0 the line is vertical, x = constant.

Simultaneous linear equations – substitution and elimination methods

Two linear equations: begincases a1x + b1y = c₁ \\ a2x + b2y = c₂ endcases can be solved by:

• Substitution – solve one equation for one variable and substitute into the other.
• Elimination – multiply equations (if needed) so that adding or subtracting eliminates one variable, then solve for the remaining variable.

Both give the unique ordered pair (x, y) when the lines intersect at a point.

Algebraic identities and manipulation of expressions

Common identities:

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²
• (a+b)(a-b) = a² - b²
• a³ + b³ = (a+b)(a² - ab + b²)
• a³ - b³ = (a-b)(a² + ab + b²)

Using these, you can expand, factorise, or simplify complex expressions quickly.

Functions – domain and range

A function f assigns each element of a set called the domain to exactly one element of another set called the range. For a given algebraic expression y = f(x), the domain consists of all real numbers for which the expression is defined (e.g., denominator ≠ 0, radicand ≥ 0). The range is the set of possible output values. Finding domain and range often involves checking restrictions and analysing the behaviour of the graph.

Important definitions and formulas

• Degree of a polynomial: highest power of x with non‑zero coefficient.
• Zero (root): α such that P(α)=0.
• Factor Theorem: x-α is a factor of P(x) iff P(α)=0.
• Quadratic formula: x = frac-b ± √(b²-4ac)2a.
• Completing the square: ax²+bx+c = a≤ft(x+b/2aright)² - (b²-4ac)/4a.
• Slope‑intercept form: y = mx + c where m = -A/B, c = -C/B.
• Standard form of simultaneous equations: a1x + b1y = c₁, a2x + b2y = c₂.
• Key identities: see section above.
• Domain restrictions: denominator ≠ 0, radicand ≥ 0, even roots of negative numbers not allowed.

Solved examples

Example 1 – Finding zeros and factorising a cubic polynomial

Find the zeros and factorise P(x)=x³-6x²+11x-6.

1. Try possible rational zeros ±1, ±2, ±3, ±6.
2. Substitute x=1: 1-6+11-6=0 ⇒ x=1 is a zero.
3. By Factor Theorem, (x-1) is a factor. Divide polynomial by x-1: x³-6x²+11x-6 = (x-1)(x²-5x+6)
4. Factor the quadratic: x²-5x+6 = (x-2)(x-3).
5. Hence P(x) = (x-1)(x-2)(x-3).

Zeros are x=1,2,3.

Example 2 – Solving a quadratic by completing the square

Solve 2x² - 8x + 6 = 0 using the completing‑the‑square method.

1. Divide by 2: x² - 4x + 3 = 0.
2. Move constant: x² - 4x = -3.
3. Add (4/2)² = 4 to both sides: x² - 4x + 4 = 1.
4. Left side becomes (x-2)² = 1.
5. Take square roots: x-2 = ± 1.
6. Solutions: x = 2 ± 1 ⇒ x = 3 or x = 1.

Example 3 – Solving simultaneous equations by elimination

Solve: begincases 3x + 4y = 11 \\ 5x - 2y = 7 endcases

1. Multiply the second equation by 2 to make the y coefficients opposite: 10x - 4y = 14.
2. Add to the first equation: (3x+4y) + (10x-4y) = 11 + 14 13x = 25 x = 25/13.
3. Substitute x into the first equation: 3×25/13 + 4y = 11 75/13 + 4y = 11 4y = 11 - 75/13 = (143-75)/13 = 68/13 y = 68/13×1/4 = 68/52 = 17/13.

Solution: (x, y) = (25/13, 17/13).

Common mistakes to avoid

• Forgetting to check that a factor truly makes the polynomial zero (always substitute back).
• Mixing up signs while completing the square; the term added is (b/2a)² with positive sign.
• Using the quadratic formula without simplifying the discriminant; a negative discriminant means no real roots.
• Ignoring domain restrictions for rational or radical functions (e.g., dividing by zero or taking √ of a negative number).
• When eliminating variables, not multiplying both equations by the correct factor, leading to incorrect coefficients.

Exam tips

• Write the given equation exactly as in the question before starting any manipulation – it helps avoid transcription errors.
• For quadratics, quickly test factorisation before resorting to the formula; it saves time and earns marks for method.
• In simultaneous equations, choose the variable whose coefficients are smallest for elimination – fewer arithmetic steps.
• Always state the domain of a function before finding its range; many marks are awarded for this explicit statement.
• After obtaining a solution, substitute back to verify – a quick check can catch sign errors.

Frequently asked questions

How do I know which method to use for a quadratic?

If the coefficients are small and the constant term can be expressed as a product of two numbers that add up to the middle coefficient, factorisation is fastest. Otherwise, completing the square or the formula works for any quadratic.

What is the difference between a zero and a factor?

A zero α is a value that makes the polynomial zero. The corresponding factor is the linear expression x-α. Every zero gives a factor, but a factor can be a higher‑degree expression (e.g., x²+1) which does not correspond to a real zero.

How can I find the range of a rational function?

Identify the domain first (exclude values that make the denominator zero). Then examine the behaviour as x approaches the excluded points and as x tends to ±∞. Sketching or using calculus (if allowed) helps, but for ICSE you can often use symmetry and asymptotes to describe the range.