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Q4(vii):
Find the zero of the polynomial in each of the following cases: (vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.

Solution :

Given Variables & Initial Setup

We are given a linear polynomial in one variable:

$p(x) = cx + d$

Where the given conditions are:

  • $c$ and $d$ are real numbers ($c, d \in \mathbb{R}$).
  • $c \neq 0$ (This ensures the polynomial is strictly of degree 1, i.e., a linear polynomial).

Step 1: Applying the Definition of a Zero of a Polynomial

The "zero" (or root) of a polynomial is defined as the specific value of the variable $x$ for which the value of the entire polynomial becomes zero. [Per the Fundamental Theorem of Algebra and polynomial root definitions].

Therefore, to find the zero of $p(x)$, we must set the polynomial equal to zero:

$p(x) = 0$

Step 2: Setting up the Equation

Substitute the given expression for $p(x)$ into the equation:

$cx + d = 0$

Step 3: Algebraic Manipulation to Isolate $x$

To solve for $x$, we perform inverse operations to isolate the variable on one side of the equation.

First, subtract $d$ from both sides of the equation [By the Subtraction Property of Equality]:

$cx + d - d = 0 - d$

$cx = -d$

Next, divide both sides by $c$ [By the Division Property of Equality]. We are mathematically permitted to divide by $c$ because the initial problem explicitly stated the condition $c \neq 0$, thereby avoiding the undefined operation of division by zero:

$\frac{cx}{c} = \frac{-d}{c}$

$x = -\frac{d}{c}$

Graphical Interpretation (Visualizing the Zero)

Geometrically, the polynomial $p(x) = cx + d$ represents a straight line on the Cartesian coordinate plane. The "zero" of the polynomial corresponds to the $x$-intercept of this line—the exact point where the graph crosses the $x$-axis (where $y = 0$).

x p(x) O (0,0) p(x) = cx + d (-d/c, 0) (0, d)

As demonstrated in the coordinate geometry above, the line intersects the horizontal axis precisely at the coordinate $(-\frac{d}{c}, 0)$.

Verification

To verify the accuracy of our derived zero, we substitute $x = -\frac{d}{c}$ back into the original polynomial $p(x)$:

$p\left(-\frac{d}{c}\right) = c\left(-\frac{d}{c}\right) + d$

$p\left(-\frac{d}{c}\right) = -d + d$

$p\left(-\frac{d}{c}\right) = 0$

Since the polynomial evaluates to $0$, the root is mathematically verified.

Final Solution: The zero of the polynomial $p(x) = cx + d$ is $x = -\frac{d}{c}$.


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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