Polynomials

CBSE Class 9 · Mathematics · Notes, formulas and practice questions

Class 9 Polynomials covers expressions like 3x² − 2x + 1: how to name them by degree, find zeroes, use the Remainder and Factor Theorems, factorise, and apply standard algebraic identities.

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What this chapter covers

This chapter is about expressions made from a single variable raised to whole-number powers. You learn to identify the terms, coefficients, and degree of a polynomial, and to recognise what makes some expressions polynomials and others not. A polynomial can be a constant, a linear expression, a quadratic, a cubic, or of higher degree, and each has a standard way of being written and talked about.

The idea of a zero of a polynomial comes next: substituting a number for the variable makes the whole expression equal to zero. The Remainder Theorem says that when a polynomial is divided by x − a, the remainder is simply p(a). The Factor Theorem builds on this, telling you that x − a is a factor exactly when p(a) = 0. These two theorems save you from doing long division every time.

Factorisation means writing a polynomial as a product of simpler polynomials. For quadratics you can often compare with x² + (a + b)x + ab. For cubics, grouping and taking out common factors are useful. The standard algebraic identities, such as the squares and cubes of sums and differences, give quick shortcuts both for expanding and for factorising expressions.

The key to using these tools is recognising which pattern fits the expression you are looking at. Expansion and factorisation are reverse processes: one multiplies factors together, and the other separates a polynomial into factors like (x − 1) or (x + 3). The identities connect the two directions, and with practice you can switch between forms smoothly.

Key terms

Polynomial in one variable: An algebraic expression involving one variable with only non-negative whole-number powers, such as 4x³ − 2x + 1. The variable does not appear in a denominator, under a root, or inside a fractional power.
Term: A part of a polynomial separated by + or − sign. For example, in 3x² − 5x + 2 the terms are 3x², −5x, and 2. A term can be just a number.
Coefficient: The numerical factor multiplying the variable in a term. In the term −5x, the coefficient of x is −5; in 3x², the coefficient of x² is 3.
Degree of a polynomial: The highest power of the variable in the polynomial after simplifying. For 2x⁴ − 3x² + 1 the degree is 4. A non-zero constant has degree 0.
Zero of a polynomial: A number c such that p(c) = 0. It is also called a root of the polynomial. For example, x = 2 is a zero of p(x) = x − 2.
Remainder Theorem: If a polynomial p(x) is divided by x − a, the remainder is p(a). This lets you find the remainder without performing the full division.
Factor Theorem: x − a is a factor of p(x) if and only if p(a) = 0. It is used to check whether a linear expression divides the polynomial exactly or to find one factor as a starting point.
Factorisation: Writing a polynomial as a product of lower-degree polynomials. For instance, x² − 9 = (x + 3)(x − 3). The factors, when multiplied out, give back the original polynomial.
Algebraic identity: An equation that is true for every value of the variables involved. Identities like (a + b)² = a² + 2ab + b² are used to expand or factorise expressions quickly.

Formula sheet

What	Formula	Notes
Square of a sum	`(a + b)² = a² + 2ab + b²`	a and b are any numbers or algebraic expressions.
Square of a difference	`(a − b)² = a² − 2ab + b²`	a and b are any numbers or algebraic expressions.
Difference of squares	`a² − b² = (a + b)(a − b)`	a and b are any numbers or algebraic expressions; useful in factorising expressions of the form a² − b².
Product (x + a)(x + b)	`(x + a)(x + b) = x² + (a + b)x + ab`	x is a variable and a and b are constants or algebraic expressions. This identity is used when factorising quadratics such as x² + 7x + 12.
Square of a trinomial	`(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca`	a, b and c are any numbers or algebraic expressions.
Cube of a sum	`(a + b)³ = a³ + b³ + 3a²b + 3ab² = a³ + b³ + 3ab(a + b)`	a and b are any numbers or algebraic expressions.
Cube of a difference	`(a − b)³ = a³ − b³ − 3a²b + 3ab² = a³ − b³ − 3ab(a − b)`	a and b are any numbers or algebraic expressions.
Sum of cubes	`a³ + b³ = (a + b)(a² − ab + b²)`	a and b are any numbers or algebraic expressions.
Difference of cubes	`a³ − b³ = (a − b)(a² + ab + b²)`	a and b are any numbers or algebraic expressions. Often used with substitution, such as a = 2x.
Factorisation of x³ + y³ + z³ − 3xyz	`x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)`	x, y and z are any numbers or algebraic expressions; if x + y + z = 0 then x³ + y³ + z³ = 3xyz.

Practice questions with answers

1. What is the degree of the polynomial p(x) = 4x³ − 2x² + x − 7?

The degree is the highest power of x in the polynomial. The powers present are 3, 2, 1 and 0, so the degree is 3.

2. Check whether x = 2 is a zero of p(x) = x² − 3x + 2.

Substitute x = 2: p(2) = 2² − 3(2) + 2 = 4 − 6 + 2 = 0. Since p(2) = 0, x = 2 is a zero of the polynomial.

3. Find the remainder when x³ − 2x² + x − 5 is divided by x − 2.

By the Remainder Theorem, the remainder is p(2). Compute p(2) = 2³ − 2(2²) + 2 − 5 = 8 − 8 + 2 − 5 = −3. So the remainder is −3.

4. Check whether x − 1 is a factor of p(x) = x³ − x² + x − 1.

Use the Factor Theorem: x − 1 is a factor only if p(1) = 0. Now p(1) = 1 − 1 + 1 − 1 = 0, so x − 1 is indeed a factor.

5. Factorise x² + 7x + 12.

Find two numbers whose sum is 7 and product is 12. They are 3 and 4. Therefore x² + 7x + 12 = (x + 3)(x + 4).

6. Factorise x³ − x² − 9x + 9.

Group the terms: (x³ − x²) + (−9x + 9) = x²(x − 1) − 9(x − 1) = (x − 1)(x² − 9). Then factorise x² − 9 as a difference of squares, giving (x − 1)(x − 3)(x + 3).

7. Expand (2x + 3y)².

Using the identity (a + b)² = a² + 2ab + b² with a = 2x and b = 3y: (2x)² + 2(2x)(3y) + (3y)² = 4x² + 12xy + 9y².

8. Evaluate 103² without direct multiplication.

Use (a + b)² with a = 100 and b = 3: 103² = (100 + 3)² = 100² + 2(100)(3) + 3² = 10000 + 600 + 9 = 10609.

9. Factorise 8a³ − 27b³.

Notice 8a³ = (2a)³ and 27b³ = (3b)³. Using a³ − b³ = (a − b)(a² + ab + b²) with a = 2a and b = 3b gives (2a − 3b)(4a² + 6ab + 9b²).

10. If x + y = 5 and xy = 6, find the value of x³ + y³.

Use x³ + y³ = (x + y)³ − 3xy(x + y). Substituting gives 5³ − 3(6)(5) = 125 − 90 = 35. So x³ + y³ = 35.