Welcome to your complete resource for Chapter 2, Polynomials. This guide covers all fundamental concepts, graphical meanings, and problem-solving techniques. Use the navigation above to move to specific sections.
CBSE Class 10 Maths
Updated till October 2025
Welcome to your complete resource for Chapter 2, Polynomials. This guide covers all fundamental concepts, graphical meanings, and problem-solving techniques. Use the navigation above to move to specific sections.
A polynomial in one variable, x, is an algebraic expression where the exponents of the variable are all non-negative integers (0, 1, 2, 3, …). This is the most important rule for identifying a polynomial.
The general form is: p(x) = a_n*x^n + a_(n-1)*x^(n-1) + … + a_1*x + a_0
Polynomials are classified in two primary ways: by the number of terms they contain, and by their degree.
The degree of a polynomial is the highest exponent of the variable. This property determines the graph’s shape and the maximum number of zeroes.
| Degree | Name | General Form | Max. Zeroes | Graphical Shape |
|---|---|---|---|---|
| 1 | Linear | ax + b, a != 0 | 1 | Straight Line |
| 2 | Quadratic | ax^2 + bx + c, a != 0 | 2 | Parabola |
| 3 | Cubic | ax^3 + bx^2 + cx + d, a != 0 | 3 | Continuous Curve |
Polynomials are also classified by terms: **Monomial** (1 term), **Binomial** (2 terms), and **Trinomial** (3 terms).
The value of a polynomial p(x) at a specific number x = k is found by replacing x with k in the expression. This value is denoted as p(k).
If p(x) = 4x^2 – 1, find the value at x = 1.
p(1) = 4(1)^2 – 1 = 4 – 1 = 3.
So, the value of p(x) at x=1 is 3.
A real number k is called a zero (or root) of the polynomial p(x) if the value of the polynomial at k is zero. In other words, p(k) = 0.
For the same polynomial p(x) = 4x^2 – 1, let’s check if x = 1/2 is a zero.
p(1/2) = 4(1/2)^2 – 1 = 4(1/4) – 1 = 1 – 1 = 0.
Since p(1/2) = 0, the number 1/2 is a zero of the polynomial. (Another zero is -1/2).
The real zeroes of a polynomial are the **x-coordinates** of the points where its graph intersects the **x-axis**.
Click the buttons below to see how a parabola’s position relative to the x-axis determines its number of real zeroes. This visualization plots an equation like y = x² + k.
Drag the two red circles (zeroes) along the x-axis. The display below will update in real-time to show the quadratic polynomial formed from those zeroes. This shows how p(x) = (x – z1)(x – z2) changes.
There is a direct algebraic relationship between the zeroes of a polynomial and its coefficients. These formulas are useful for verifying zeroes quickly.
| Polynomial Type | Zeroes | Sum of Zeroes | Product of Zeroes |
|---|---|---|---|
|
Quadratic
ax^2 + bx + c
|
alpha, beta | alpha + beta = -b/a | alpha*beta = c/a |
|
Cubic
ax^3 + bx^2 + cx + d
|
alpha, beta, gamma | alpha + beta + gamma = -b/a | alpha*beta*gamma = -d/a |
For a cubic, there is one more relationship: alpha*beta + beta*gamma + gamma*alpha = c/a.
For a cubic polynomial p(x) = ax^3 + bx^2 + cx + d, with zeroes alpha, beta, and gamma, the relationships are more detailed:
1. Sum of Zeroes:
alpha + beta + gamma = -b/a
2. Sum of Products of Zeroes (taken two at a time):
alpha*beta + beta*gamma + gamma*alpha = c/a
3. Product of Zeroes:
alpha*beta*gamma = -d/a
These three formulas are used to verify the zeroes of a cubic polynomial or to find unknown coefficients.
If you are given the zeroes alpha and beta, you can form the polynomial with this formula:
p(x) = k[x^2 – (alpha + beta)x + alpha*beta]
(where k is any non-zero real number)
The division algorithm allows us to divide one polynomial, p(x), by another, g(x).
$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$
p(x) = g(x) * q(x) + r(x)
The division stops when the remainder r(x) is 0 or its degree is less than the degree of the divisor g(x).
Let’s divide p(x) = -x^3 + 3x^2 – 3x + 5 by g(x) = -x^2 + x – 1.
Here, q(x) = x – 2 and r(x) = 3. This is a key tool for finding remaining zeroes of a polynomial.
Q.1: A graph cuts the x-axis at three points. What is the number of zeroes?
Q.2: What is the maximum number of zeroes a cubic polynomial can have?
Q.3: Find the zeroes of 6x^2 – 7x – 3 and verify the relationship with coefficients.
1. Factoring: 6x^2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1)(2x – 3).
The zeroes (alpha, beta) are x = -1/3 and x = 3/2.
2. Verification: Here a = 6, b = -7, c = -3.
Sum: alpha + beta = (-1/3) + (3/2) = 7/6. Formula: -b/a = -(-7)/6 = 7/6. (Verified)
Product: alpha*beta = (-1/3) * (3/2) = -1/2. Formula: c/a = -3/6 = -1/2. (Verified)
Q.4: Find a quadratic polynomial whose sum of zeroes is 2 and product of zeroes is -15.
Using the formula p(x) = k[x^2 – (alpha + beta)x + alpha*beta].
p(x) = k[x^2 – (2)x + (-15)] = k[x^2 – 2x – 15].
Letting k=1, a valid polynomial is x^2 – 2x – 15.
Q.5: If 4 is a zero of p(x) = x^3 – 3x^2 – 10x + 24, find its other two zeroes.
1. Divide (x^3 – 3x^2 – 10x + 24) by (x – 4) using long division.
2. The division gives a quotient q(x) = x^2 + x – 6 and a remainder of 0.
3. Find the zeroes of the quotient by setting q(x) = 0:
x^2 + x – 6 = 0
(x + 3)(x – 2) = 0
The other two zeroes are x = -3 and x = 2.
Q.6: Obtain all zeroes of p(x) = 3x^4 + 6x^3 – 2x^2 – 10x – 5, if two zeroes are sqrt(5/3) and -sqrt(5/3).
1. The factor from the given zeroes is g(x) = (x – sqrt(5/3))(x + sqrt(5/3)) = x^2 – 5/3.
2. We can use 3g(x) = 3x^2 – 5 as the divisor.
3. Dividing p(x) = 3x^4 + 6x^3 – 2x^2 – 10x – 5 by g(x) = 3x^2 – 5 gives the quotient q(x) = x^2 + 2x + 1.
4. Find the zeroes of q(x): x^2 + 2x + 1 = 0 (x + 1)(x + 1) = 0
The remaining zeroes are x = -1 and x = -1 (a repeated root).
The four zeroes are {sqrt(5/3), -sqrt(5/3), -1, -1}.
