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Q5(v):
Classify the following as linear, quadratic and cubic polynomials: (v) $3t$

Solution :

Given Polynomial & Initial Setup

We are given the following algebraic expression:

$P(t) = 3t$

Our objective is to classify this polynomial into one of three distinct algebraic categories: linear, quadratic, or cubic. The classification of a polynomial is strictly determined by its degree.

Step 1: Identify the Variable and its Exponents

In the given polynomial $P(t) = 3t$, the single independent variable present is $t$. To determine the nature of the polynomial, we must inspect the exponents attached to this variable.

The term $3t$ can be explicitly rewritten by revealing its hidden exponent:

$P(t) = 3t^1$

[Per the Fundamental Laws of Exponents, any variable $x$ written without a visible power is mathematically understood to have an exponent of $1$, i.e., $x = x^1$].

Step 2: Determine the Degree of the Polynomial

The degree of a polynomial in one variable is defined as the highest power (exponent) of the variable in that expression.

  • The only term containing the variable is $3t^1$.
  • The exponent of $t$ in this term is $1$.

Therefore, the highest power of the variable $t$ is $1$. This establishes that the degree of the polynomial $P(t) = 3t$ is exactly $1$.

Step 3: Apply Polynomial Classification Criteria

Polynomials are universally classified based on their degree [By the Fundamental Theorem of Algebra and standard algebraic nomenclature]. The classification matrix is as follows:

Degree Classification Standard Form
$1$ Linear $ax + b$ (where $a \neq 0$)
$2$ Quadratic $ax^2 + bx + c$ (where $a \neq 0$)
$3$ Cubic $ax^3 + bx^2 + cx + d$ (where $a \neq 0$)

Because the degree of $3t$ is $1$, it perfectly satisfies the condition for a linear polynomial. It matches the standard linear form $at + b$ where the leading coefficient $a = 3$ and the constant term $b = 0$.

Step 4: Geometric Verification (Visualizing Linearity)

A defining characteristic of a linear polynomial is that its graph forms a perfectly straight line [Per Cartesian Coordinate Geometry]. Below is the precise geometric representation of $P(t) = 3t$, demonstrating a constant rate of change (slope $m = 3$).

t P(t) 0 (1, 3) (-1, -3) P(t) = 3t

Final Solution: The given polynomial $3t$ has a highest exponent (degree) of $1$. Therefore, it is classified as a linear polynomial.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.1


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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