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Q2(iii):
Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iii) $p(x) = x^3$

Solution :

Initial Setup & Given Variables

We are given a single-variable polynomial function of degree 3 (a cubic polynomial). The polynomial is defined as:

$p(x) = x^3$

Our objective is to evaluate this polynomial at three specific discrete points in its domain: $x = 0$, $x = 1$, and $x = 2$. [Per the Substitution Principle of Functions, evaluating a polynomial $p(x)$ at a specific value $x = a$ requires replacing every instance of the variable $x$ in the expression with the constant $a$].

Step 1: Evaluating the Polynomial at $x = 0$

To find $p(0)$, we substitute $x = 0$ into the polynomial equation.

  • $p(0) = (0)^3$
  • $p(0) = 0 \times 0 \times 0$
  • $p(0) = 0$

[Theoretical Justification: The zero product property and the definition of exponentiation state that zero raised to any positive real power is strictly zero. Geometrically, this indicates that the graph of the polynomial passes exactly through the origin $(0,0)$].

Step 2: Evaluating the Polynomial at $x = 1$

To find $p(1)$, we substitute $x = 1$ into the polynomial equation.

  • $p(1) = (1)^3$
  • $p(1) = 1 \times 1 \times 1$
  • $p(1) = 1$

[Theoretical Justification: The number 1 is the multiplicative identity. Any real number multiplied by 1 remains unchanged, hence 1 raised to any real power is always 1].

Step 3: Evaluating the Polynomial at $x = 2$

To find $p(2)$, we substitute $x = 2$ into the polynomial equation.

  • $p(2) = (2)^3$
  • $p(2) = 2 \times 2 \times 2$
  • $p(2) = 4 \times 2$
  • $p(2) = 8$

[Theoretical Justification: Exponentiation represents repeated multiplication. The rapid growth from $p(1)=1$ to $p(2)=8$ demonstrates the non-linear, cubic expansion characteristic of degree-3 polynomials].

Graphical Analysis & Verification

Below is the precise Cartesian mapping of the polynomial $p(x) = x^3$. The specific points we evaluated—$(0,0)$, $(1,1)$, and $(2,8)$—are plotted to visually confirm the cubic curve's trajectory.

x p(x) 0 p(0) = 0 p(1) = 1 p(2) = 8

Summary Table of Evaluations

Input Variable ($x$) Substitution ($x^3$) Output Value ($p(x)$)
$0$ $0^3$ $0$
$1$ $1^3$ $1$
$2$ $2^3$ $8$

Final Solution: For the polynomial $p(x) = x^3$, the evaluated values are $p(0) = 0$, $p(1) = 1$, and $p(2) = 8$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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