Chapter 2: Polynomials (Detailed Guide)

1. Core Concepts

Definition: A polynomial is an algebraic expression consisting of variables and coefficients. For Class 10, we primarily focus on Quadratic Polynomials ($ax^2 + bx + c$, where $a \neq 0$).

Degree of Polynomial: The highest power of the variable $x$ in a polynomial $P(x)$ is called the degree.

Linear: Degree 1 (e.g., $3x + 5$)

Quadratic: Degree 2 (e.g., $x^2 - 5x + 6$)

Cubic: Degree 3 (e.g., $x^3 - 1$)

Geometrical Meaning of Zeroes: The zeroes of a polynomial $P(x)$ are the $x$-coordinates of the points where the graph of $y = P(x)$ intersects the X-axis.

Relationship Between Zeroes and Coefficients (For Quadratic $ax^2 + bx + c$):

If $\alpha$ and $\beta$ are the zeroes:

Sum of Zeroes ($\alpha + \beta$): $-b/a$ (i.e., $-\text{coefficient of } x / \text{coefficient of } x^2$).

Product of Zeroes ($\alpha \cdot \beta$): $c/a$ (i.e., $\text{constant term} / \text{coefficient of } x^2$).


2. Important Examples

Example 1: Find the zeroes of $x^2 + 7x + 10$ and verify the relationship.

Solution:

Factorizing by splitting the middle term: $x^2 + 5x + 2x + 10 = 0$.

$x(x+5) + 2(x+5) = 0 \Rightarrow (x+2)(x+5) = 0$.

Zeroes are $\alpha = -2$ and $\beta = -5$.

Verification:

Sum: $-2 + (-5) = -7$. Formula: $-b/a = -7/1 = -7$. (Matched!)

Product: $(-2) \times (-5) = 10$. Formula: $c/a = 10/1 = 10$. (Matched!)

Example 2: Find a quadratic polynomial whose sum and product of zeroes are -3 and 2 respectively.

Solution:

Form: $x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes})$.

$x^2 - (-3)x + 2 = x^2 + 3x + 2$.


3. Previous Year Questions (PYQs) with Solutions

Q1. (1 Mark) If one zero of the quadratic polynomial $x^2 + 3x + k$ is 2, then find the value of $k$.

Solution: Since 2 is a zero, putting $x = 2$ must make the polynomial zero.

$(2)^2 + 3(2) + k = 0$

$4 + 6 + k = 0 \Rightarrow 10 + k = 0$

Result: $k = -10$.

Q2. (2 Marks) Find the zeroes of $4x^2 - 4x + 1$.

Solution:

This is $(2x - 1)^2 = 0$.

$2x - 1 = 0 \Rightarrow x = 1/2$.

Result: The zeroes are $1/2, 1/2$.

Q3. (3 Marks) If $\alpha$ and $\beta$ are the zeroes of $x^2 - 5x + q$ such that $\alpha - \beta = 1$, find the value of $q$.

Solution:

$\alpha + \beta = -(-5)/1 = 5$ (Equation 1).

$\alpha - \beta = 1$ (Given - Equation 2).

Adding both: $2\alpha = 6 \Rightarrow \alpha = 3$.

Put $\alpha$ in (1): $3 + \beta = 5 \Rightarrow \beta = 2$.

Now, Product $\alpha \cdot \beta = q/1 \Rightarrow 3 \times 2 = q$.

Result: $q = 6$.

Q4. (5 Marks) Verify that 3, -1, and -1/3 are the zeroes of the cubic polynomial $3x^3 - 5x^2 - 11x - 3$, and then verify the relationship between zeroes and coefficients.

Solution Hint:

Substitute each value into the polynomial to show $P(x) = 0$.

Use the cubic formulas: $\text{Sum} = -b/a$, $\text{Sum of product in pairs} = c/a$, and $\text{Product} = -d/a$.