Class 9 — Mathematics (NCERT)

Polynomials

Class 9

✓By the end of this lesson students will be able to define polynomials and identify their degree.
✓By the end of this lesson students will be able to find the zeroes of a polynomial.
✓By the end of this lesson students will be able to state and apply the Remainder Theorem.
✓By the end of this lesson students will be able to state and apply the Factor Theorem.
✓By the end of this lesson students will be able to recall and use algebraic identities to factorise and expand polynomials.

Key concepts

Polynomial

A polynomial is an algebraic expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication and non-negative integer exponents of variables. For example, 2x^2 + 5x - 3 is a polynomial. The general form of a polynomial in one variable x is P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where a_n, a_{n-1}, ..., a_0 are coefficients and n is a non-negative integer.

P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0

Degree of a Polynomial

The highest power of the variable in a polynomial is called the degree of the polynomial. For example, in 5x^3 - 2x^2 + 7x - 1, the highest power of x is 3, so its degree is 3. A non-zero constant polynomial has degree 0. The degree of the zero polynomial is not defined.

Types of Polynomials

Polynomials can be classified based on the number of terms or their degree:\n\nBased on number of terms:\n* Monomial: A polynomial with one term (e.g., 5x).\n* Binomial: A polynomial with two terms (e.g., 2x + 3).\n* Trinomial: A polynomial with three terms (e.g., x^2 + 2x - 1).\n\nBased on degree:\n* Linear Polynomial: A polynomial of degree 1 (e.g., ax + b, where a ≠ 0).\n* Quadratic Polynomial: A polynomial of degree 2 (e.g., ax^2 + bx + c, where a ≠ 0).\n* Cubic Polynomial: A polynomial of degree 3 (e.g., ax^3 + bx^2 + cx + d, where a ≠ 0).

Zeroes of a Polynomial

A real number 'k' is said to be a zero of a polynomial P(x) if P(k) = 0. In other words, when 'k' is substituted for the variable 'x' in the polynomial, the value of the polynomial becomes zero. Geometrically, the zeroes are the x-coordinates of the points where the graph of y = P(x) intersects the x-axis.

P(k) = 0

Remainder Theorem

If P(x) is any polynomial of degree greater than or equal to one and P(x) is divided by the linear polynomial x - a, then the remainder is P(a). This theorem provides a shortcut to find the remainder without actually performing long division.

Remainder = P(a) when P(x) is divided by x - a.

Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem. It states:\n1. If P(a) = 0, then (x - a) is a factor of the polynomial P(x).\n2. Conversely, if (x - a) is a factor of the polynomial P(x), then P(a) = 0.\nThis theorem is very useful for factorising polynomials.

P(a) = 0 ⇔ (x - a) is a factor of P(x).

Algebraic Identities

Algebraic identities are equalities that hold true for all values of the variables involved. These are fundamental tools for factorisation, expansion, and simplification of algebraic expressions. Some important identities for Class 9 are:

(a + b)^2 = a^2 + 2ab + b^2\n(a - b)^2 = a^2 - 2ab + b^2\na^2 - b^2 = (a + b)(a - b)\n(x + a)(x + b) = x^2 + (a + b)x + ab\n(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\n(a + b)^3 = a^3 + b^3 + 3ab(a + b) = a^3 + b^3 + 3a^2b + 3ab^2\n(a - b)^3 = a^3 - b^3 - 3ab(a - b) = a^3 - b^3 - 3a^2b + 3ab^2\na^3 + b^3 = (a + b)(a^2 - ab + b^2)\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\nIf a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.

Key facts to remember

1A polynomial is an algebraic expression with non-negative integer exponents for its variables.
2The degree of a polynomial is the highest power of the variable in the polynomial.
3A real number 'k' is a zero of a polynomial P(x) if P(k) = 0.
4Remainder Theorem: When a polynomial P(x) is divided by (x - a), the remainder is P(a).
5Factor Theorem: (x - a) is a factor of a polynomial P(x) if and only if P(a) = 0.
6Important algebraic identities include (a+b)^2, (a-b)^2, a^2-b^2, (x+a)(x+b), (a+b+c)^2, (a+b)^3, (a-b)^3, a^3+b^3, a^3-b^3, and a^3+b^3+c^3-3abc.

Worked examples

Example 1

Find the degree of the polynomial P(x) = 5x^4 - 3x^2 + 7x - 9. Also, check if x = 1 is a zero of the polynomial Q(x) = x^2 - 2x + 1.

IFor P(x) = 5x^4 - 3x^2 + 7x - 9, the highest power of the variable x is 4.

IITherefore, the degree of P(x) is 4.

IIIFor Q(x) = x^2 - 2x + 1, to check if x = 1 is a zero, we substitute x = 1 into Q(x).

IVQ(1) = (1)^2 - 2(1) + 1

VQ(1) = 1 - 2 + 1

VIQ(1) = 0

Answer

The degree of P(x) is 4. Yes, x = 1 is a zero of Q(x).

Example 2

Find the remainder when P(x) = x^3 + 3x^2 + 3x + 1 is divided by x + 1.

IAccording to the Remainder Theorem, if P(x) is divided by x - a, the remainder is P(a).

IIHere, the divisor is x + 1, which can be written as x - (-1). So, a = -1.

IIISubstitute x = -1 into P(x).

IVP(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1

VP(-1) = -1 + 3(1) - 3 + 1

VIP(-1) = -1 + 3 - 3 + 1

VIIP(-1) = 0

Answer

The remainder is 0.

Since the remainder is 0, (x + 1) is a factor of P(x) by the Factor Theorem.

Example 3

Check whether (x - 2) is a factor of P(x) = x^3 - 3x^2 + 4x - 4.

IAccording to the Factor Theorem, (x - a) is a factor of P(x) if P(a) = 0.

IIHere, the potential factor is (x - 2), so we take a = 2.

IIISubstitute x = 2 into P(x).

IVP(2) = (2)^3 - 3(2)^2 + 4(2) - 4

VP(2) = 8 - 3(4) + 8 - 4

VIP(2) = 8 - 12 + 8 - 4

VIIP(2) = 16 - 16

VIIIP(2) = 0

Answer

Since P(2) = 0, (x - 2) is a factor of P(x).

Example 4

Expand (3x + 4y)^2 using a suitable algebraic identity.

IWe use the algebraic identity: (a + b)^2 = a^2 + 2ab + b^2.

IIHere, we identify a = 3x and b = 4y.

IIISubstitute these values into the identity:

IV(3x + 4y)^2 = (3x)^2 + 2(3x)(4y) + (4y)^2

V= 9x^2 + 24xy + 16y^2

Answer

9x^2 + 24xy + 16y^2

Example 5

Factorise 49a^2 - 36b^2 using a suitable algebraic identity.

IWe observe that 49a^2 can be written as (7a)^2 and 36b^2 can be written as (6b)^2.

IIThe expression is in the form a^2 - b^2.

IIIWe use the algebraic identity: a^2 - b^2 = (a + b)(a - b).

IVHere, we identify a = 7a and b = 6b.

VSubstitute these values into the identity:

VI49a^2 - 36b^2 = (7a)^2 - (6b)^2

VII= (7a + 6b)(7a - 6b)

Answer

(7a + 6b)(7a - 6b)

Always look for common factors first before applying identities.

Common mistakes

✗Confusing the degree of a polynomial with the number of terms it has.
✗Making sign errors when substituting negative values into polynomials or applying algebraic identities.
✗Incorrectly expanding or factorising expressions, especially with identities involving subtraction like (a-b)^2 or (a-b)^3.
✗Not understanding that for a divisor (x - a), we substitute 'a', and for (x + a), we substitute '-a' in the Remainder/Factor Theorem.
✗Forgetting to check if the condition a + b + c = 0 is met before applying the special case of the identity a^3 + b^3 + c^3 - 3abc.

Exam tips

★Memorise all algebraic identities thoroughly, as they are fundamental for solving many problems efficiently.
★Practice polynomial division and the application of Remainder and Factor Theorems with various types of polynomials to build confidence.
★Always show all steps clearly in your solutions, especially for worked examples, to avoid losing marks for incomplete working.
★When finding zeroes or using the Remainder/Factor Theorem, double-check your substitution and arithmetic carefully.
★For factorisation problems, first look for common factors, then try to apply suitable algebraic identities.

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