---
source: ncert
topic: algebra
class_level: class_10
chapter: ch2
difficulty: intermediate
---

Class 10 | Ch2: Polynomials
ZEROES OF POLYNOMIAL:
Zero of polynomial p(x): value of x where p(x)=0.
Geometrically: x-coordinates where graph of y=p(x) crosses x-axis.
Linear polynomial (degree 1): has exactly 1 zero.
Quadratic polynomial (degree 2): has at most 2 zeroes.
Cubic polynomial (degree 3): has at most 3 zeroes.
RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS:
For quadratic p(x)=ax^2+bx+c with zeroes alpha and beta:
Sum of zeroes: alpha+beta = -b/a.
Product of zeroes: alpha×beta = c/a.
For cubic p(x)=ax^3+bx^2+cx+d with zeroes alpha, beta, gamma:
alpha+beta+gamma = -b/a.
alpha*beta+beta*gamma+gamma*alpha = c/a.
alpha*beta*gamma = -d/a.
FORMING POLYNOMIAL FROM ZEROES:
Quadratic with zeroes alpha, beta: p(x)=k[x^2-(alpha+beta)x+alpha*beta].
DIVISION ALGORITHM FOR POLYNOMIALS:
p(x) = g(x)×q(x) + r(x).
Degree of remainder < degree of divisor.
SOLVED EXAMPLES:
Example 1: Find zeroes of p(x)=6x^2-3-7x=6x^2-7x-3.
6x^2-7x-3=0. Split: 6x^2-9x+2x-3=3x(2x-3)+1(2x-3)=(3x+1)(2x-3).
Zeroes: x=-1/3 and x=3/2.
Verify: sum=-1/3+3/2=(-2+9)/6=7/6=-(-7)/6=b/a. Wait: -b/a=-(-7)/6=7/6. Correct.
Product=(-1/3)(3/2)=-1/2=c/a=-3/6=-1/2. Correct.
Example 2: Form quadratic with zeroes 2 and -3.
Sum=2+(-3)=-1. Product=2×(-3)=-6. p(x)=x^2-(-1)x+(-6)=x^2+x-6.
Example 3: Cubic zeroes 2,-1,-3.
Sum=2-1-3=-2=-b/a → b/a=2. Sum of products=2(-1)+(-1)(-3)+(-3)(2)=-2+3-6=-5=c/a.
Product=2×(-1)×(-3)=6=-d/a → d/a=-6. p(x)=x^3+2x^2-5x-6 (taking a=1).
COMMON MISTAKES:
Sum of zeroes=-b/a (NEGATIVE b over a, not b over a).
Product of zeroes=c/a (positive, not negative).
Zeroes of polynomial: solve p(x)=0 (set equal to zero, NOT to y).