CBSE Class 9 Maths Notes: Polynomials

Definition of Polynomials in One Variable

Definitions: A polynomial in one variable, say *x*, is an expression of the form:

$P(x) = a_nx^n + a_{n-1}x^{n-1} + … + a_1x + a_0$

where:

$a_0, a_1, …, a_n$ are real numbers (coefficients).
*n* is a non-negative integer (the degree of the polynomial).
*x* is the variable.

Core Principle: The key is that the exponents of the variable must be whole numbers.

Coefficients, Terms, and Zero Polynomial

Definitions:

Coefficients: The real numbers ($a_0, a_1, …, a_n$) in the polynomial.
Terms: The individual parts of the polynomial separated by addition or subtraction signs (e.g., $a_nx^n$, $a_{n-1}x^{n-1}$, etc.).
Zero Polynomial: A polynomial where all coefficients are zero. It’s degree is not defined. It is represented as $P(x) = 0$.

Example: In the polynomial $3x^2 – 5x + 2$:

Coefficients: 3, -5, 2
Terms: $3x^2$, $-5x$, 2

Degree-Based Types: Constant, Linear, Quadratic, Cubic

Definitions:

Constant Polynomial: Degree 0 (e.g., 5, -2). Form: $P(x) = c$, where *c* is a constant.
Linear Polynomial: Degree 1 (e.g., $2x + 1$, $-x + 3$). Form: $P(x) = ax + b$, where $a \neq 0$.
Quadratic Polynomial: Degree 2 (e.g., $x^2 + 2x – 1$, $3x^2$). Form: $P(x) = ax^2 + bx + c$, where $a \neq 0$.
Cubic Polynomial: Degree 3 (e.g., $x^3 – 4x^2 + x + 1$). Form: $P(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$.

Monomials, Binomials, Trinomials

Definitions: These classifications are based on the number of terms.

Monomial: A polynomial with one term (e.g., $5x^2$, $7x$, -3).
Binomial: A polynomial with two terms (e.g., $2x + 1$, $x^3 – 4$).
Trinomial: A polynomial with three terms (e.g., $x^2 + 2x + 1$, $x^3 – x + 5$).

Finding Zeroes of Polynomials

Definitions: A zero (or root) of a polynomial $P(x)$ is a value of *x* for which $P(x) = 0$.

Methods:

By Substitution: Substitute different values of *x* into the polynomial until you find a value that makes the polynomial equal to zero.
By Factoring: Factor the polynomial and set each factor equal to zero to solve for *x*.

Example: To find the zero of $P(x) = x – 2$, set $x – 2 = 0$. Therefore, $x = 2$ is the zero.

Remainder Theorem

Core Principles:

Statement: If a polynomial $P(x)$ is divided by $(x – a)$, the remainder is $P(a)$.
Explanation: When dividing $P(x)$ by $(x – a)$, we can write it as: $P(x) = (x – a)Q(x) + R$, where Q(x) is the quotient, and R is the remainder. Substituting $x = a$ we get $P(a) = (a – a)Q(a) + R$ or $P(a) = R$.

Example: If $P(x) = x^2 + 2x + 1$ is divided by $(x – 1)$, the remainder is $P(1) = 1^2 + 2(1) + 1 = 4$.

Factor Theorem (Statement & Proof)

Core Principles:

Statement: $(x – a)$ is a factor of the polynomial $P(x)$ if and only if $P(a) = 0$.
Proof: This theorem is a direct consequence of the Remainder Theorem.
Proof:
- If $(x – a)$ is a factor of $P(x)$, then $P(x) = (x – a)Q(x)$. Substituting $x = a$, we get $P(a) = (a – a)Q(a) = 0$.
- If $P(a) = 0$, then according to the Remainder Theorem, the remainder when $P(x)$ is divided by $(x – a)$ is 0. This means $(x – a)$ is a factor of $P(x)$.

Factorization of $ax^2 + bx + c$ and Simple Cubics

Methods:

Factoring $ax^2 + bx + c$: Split the middle term (*bx*) into two terms whose sum is *b* and whose product is $ac$. Group and factor.
Factoring Simple Cubics: Use factor theorem to find one factor (often by trial and error). Then, divide the cubic by that factor to find the quadratic factor. Factor the quadratic further if possible. Use identities like $x^3 – a^3 = (x-a)(x^2 + ax + a^2)$ and $x^3 + a^3 = (x+a)(x^2 – ax + a^2)$.

Example (Quadratic): Factor $x^2 + 5x + 6$. We need two numbers that add up to 5 and multiply to 6 (2 and 3). $x^2 + 5x + 6 = x^2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3)$.

Example (Cubic): Factor $x^3 – 8$. Recognize that 8 is $2^3$. Then apply the formula $x^3 – a^3 = (x-a)(x^2 + ax + a^2)$, where $a=2$: $x^3 – 8 = (x – 2)(x^2 + 2x + 4)$.