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Q5(iv):
Classify the following as linear, quadratic and cubic polynomials: (iv) $1 + x$

Solution :

Initial Setup & Given Expression

We are given the algebraic expression:

$P(x) = 1 + x$

Our objective is to classify this polynomial into one of three categories: linear, quadratic, or cubic. This classification is strictly determined by the degree of the polynomial.

Step 1: Standardizing the Polynomial

First, we rewrite the polynomial in its standard form, where the terms are ordered in descending powers of the variable $x$.

$P(x) = x + 1$

We can explicitly write the exponents for every term to ensure absolute clarity:

  • The term $x$ is equivalent to $x^1$.
  • The constant term $1$ is equivalent to $1 \cdot x^0$ [Per the Zero Exponent Rule, where $x^0 = 1$ for $x \neq 0$].

Thus, the fully expanded standard form is:

$P(x) = 1 \cdot x^1 + 1 \cdot x^0$

Step 2: Determining the Degree of the Polynomial

The degree of a polynomial in one variable is defined as the highest exponent (power) of the variable present in the expression with a non-zero coefficient.

Analyzing the exponents in $P(x) = x^1 + 1 \cdot x^0$:

  • The exponent of the first term is $1$.
  • The exponent of the second term is $0$.

Comparing these values, the maximum exponent is $1$. Therefore, the degree of the polynomial $P(x) = 1 + x$ is exactly $1$.

Step 3: Applying the Classification Theorem

Polynomials are classified by their degree according to the following universally accepted algebraic definitions:

Degree Classification Name 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 $a \neq 0$)

Since we have rigorously determined that the degree of $1 + x$ is $1$, it maps directly to the definition of a linear polynomial.

Visual Verification: Geometric Interpretation

A linear polynomial, when graphed on a Cartesian coordinate system as a function $y = P(x)$, will always produce a perfectly straight line [Per the geometric definition of first-degree equations]. Below is the precise graphical representation of $y = x + 1$, demonstrating its constant slope ($m = 1$) and y-intercept ($c = 1$).

x y 0 1 2 -1 1 2 -1 (0, 1) (-1, 0) y = x + 1

Final Solution: The polynomial $1 + x$ has a highest degree of $1$. Therefore, it is classified as a linear polynomial.


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