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Q1(iii):
Find the value of the polynomial $5x – 4x^2 + 3$ at (iii) $x = 2$

Solution :

Given Variables & Initial Setup

We are given a polynomial in a single variable, $x$. Let us define the polynomial as a function $P(x)$:

$P(x) = 5x - 4x^2 + 3$

The objective is to evaluate this polynomial at the specific domain value $x = 2$. [Per the fundamental theorem of polynomial evaluation, finding the value of a polynomial at a given point requires the direct substitution of that point into the variable, followed by simplification according to the standard order of operations (PEMDAS/BODMAS)].

Step 1: Substitution of the Variable

Substitute $x = 2$ into every instance of $x$ within the polynomial $P(x)$:

$P(2) = 5(2) - 4(2)^2 + 3$

Step 2: Resolution of Exponents

According to the order of operations, exponentiation must be resolved before multiplication. We evaluate the quadratic term $(2)^2$:

$2^2 = 2 \times 2 = 4$

Substituting this back into the equation yields:

$P(2) = 5(2) - 4(4) + 3$

Step 3: Execution of Multiplication

Next, perform the scalar multiplications for both the linear and quadratic terms:

  • Linear term: $5 \times 2 = 10$
  • Quadratic term: $4 \times 4 = 16$

Updating the polynomial expression:

$P(2) = 10 - 16 + 3$

Step 4: Sequential Addition and Subtraction

Finally, evaluate the arithmetic expression from left to right:

$P(2) = (10 - 16) + 3$

$P(2) = -6 + 3$

$P(2) = -3$

Graphical Verification

The polynomial $P(x) = -4x^2 + 5x + 3$ represents a downward-opening parabola [since the leading coefficient $a = -4$ is less than zero]. Evaluating the polynomial at $x = 2$ corresponds to finding the $y$-coordinate of the point on this parabola where the $x$-coordinate is $2$. As calculated, this point exists exactly at $(2, -3)$ on the Cartesian plane.

x P(x) 0 1 2 -3 (2, -3) P(x) = -4x² + 5x + 3

Final Solution: The value of the polynomial $5x - 4x^2 + 3$ at $x = 2$ is $-3$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2


CBSE Solutions for Class 9 Mathematics Polynomials


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