C2

Class 10 Mathematics - Chapter 2: Polynomials
Chapter 2: Polynomials
1. Geometrical Meaning of the Zeroes of a Polynomial
Definition: An algebraic expression of the form $p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ (where $a_n \neq 0$) is called a polynomial in $x$ of degree $n$.

Zero of a Polynomial: A real number $k$ is said to be a zero of a polynomial $p(x)$ if $p(k) = 0$.

Geometrical Interpretation: The zeroes of a polynomial $p(x)$ are precisely the $x$-coordinates of the points where the graph of $y = p(x)$ intersects the x-axis.
Graphical Representation of Quadratic Polynomials

For a quadratic polynomial $p(x) = ax^2 + bx + c$, the graph is a parabola opening upwards (if $a > 0$) or downwards (if $a < 0$). The number of real zeroes depends on the Discriminant $D = b^2 - 4ac$:

D > 0 (2 Real Roots) D = 0 (1 Equal Root) D < 0 (No Real Roots)
Example 1.1: Determining Zeroes from a Graph

The graph of $y = p(x)$ intersects the x-axis at 3 distinct points. How many zeroes does $p(x)$ have?

Solution:
Since the number of zeroes of a polynomial is equal to the number of points where its graph intersects the x-axis, the polynomial $p(x)$ has 3 zeroes.

2. Relationship Between Zeroes and Coefficients
1. For a Quadratic Polynomial $p(x) = ax^2 + bx + c \quad (a \neq 0)$:
If $\alpha$ and $\beta$ are the zeroes of $p(x)$, then:
$$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$$
$$\text{Product of zeroes } (\alpha \cdot \beta) = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$

Forming a Quadratic Polynomial:
If sum ($S$) and product ($P$) of zeroes are given, the polynomial is:
$$k[x^2 - Sx + P] \quad \text{or} \quad k[x^2 - (\alpha + \beta)x + \alpha\beta]$$
2. For a Cubic Polynomial $p(x) = ax^3 + bx^2 + cx + d \quad (a \neq 0)$:
If $\alpha, \beta, \gamma$ are the zeroes, then:
  • $\alpha + \beta + \gamma = -\frac{b}{a}$
  • $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
  • $\alpha\beta\gamma = -\frac{d}{a}$
Example 2.1: Finding Zeroes and Verifying Relationship

Find the zeroes of the quadratic polynomial $p(x) = x^2 + 7x + 10$, and verify the relationship between the zeroes and the coefficients.

Solution:
Factorising by splitting the middle term:

$$x^2 + 7x + 10 = x^2 + 2x + 5x + 10$$
$$= x(x + 2) + 5(x + 2) = (x + 2)(x + 5)$$
To find zeroes, set $p(x) = 0$:
$(x + 2)(x + 5) = 0 \implies x = -2 \text{ or } x = -5$.
So, zeroes are $\alpha = -2$ and $\beta = -5$.

Verification:
Here $a = 1, b = 7, c = 10$.
Sum of zeroes: $\alpha + \beta = -2 + (-5) = -7$
Formula value: $-\frac{b}{a} = -\frac{7}{1} = -7$ (Verified)
Product of zeroes: $\alpha\beta = (-2)(-5) = 10$
Formula value: $\frac{c}{a} = \frac{10}{1} = 10$ (Verified)

Example 2.2: Difference of Squares Form

Find the zeroes of $p(x) = x^2 - 3$ and verify the relationship between zeroes and coefficients.

Solution:
Using identity $a^2 - b^2 = (a - b)(a + b)$:

$$x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$$
So, the zeroes are $\alpha = \sqrt{3}$ and $\beta = -\sqrt{3}$.

Comparing $x^2 + 0x - 3$ with $ax^2 + bx + c$, we get $a = 1, b = 0, c = -3$.
Sum of zeroes: $\alpha + \beta = \sqrt{3} + (-\sqrt{3}) = 0 = -\frac{0}{1} = -\frac{b}{a}$.
Product of zeroes: $\alpha\beta = (\sqrt{3})(-\sqrt{3}) = -3 = \frac{-3}{1} = \frac{c}{a}$. Verified.

Example 2.3: Constructing a Quadratic Polynomial

Find a quadratic polynomial whose sum and product of zeroes are $-\frac{1}{4}$ and $\frac{1}{4}$ respectively.

Solution:
Given: Sum $S = \alpha + \beta = -\frac{1}{4}$, Product $P = \alpha\beta = \frac{1}{4}$.
Required quadratic polynomial form: $k[x^2 - Sx + P]$

$$p(x) = k\left[x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4}\right] = k\left[x^2 + \frac{x}{4} + \frac{1}{4}\right]$$
Taking $k = 4$ to clear denominators:
$$p(x) = 4x^2 + x + 1$$