Class 9  >  Mathematics (Maths) Class 9  >  Ex 2.2 NCERT Solutions - Polynomials

Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9

Document Description: Ex 2.2 NCERT Solutions - Polynomials for Class 9 2022 is part of Mathematics (Maths) Class 9 preparation. The notes and questions for Ex 2.2 NCERT Solutions - Polynomials have been prepared according to the Class 9 exam syllabus. Information about Ex 2.2 NCERT Solutions - Polynomials covers topics like and Ex 2.2 NCERT Solutions - Polynomials Example, for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Ex 2.2 NCERT Solutions - Polynomials.

Introduction of Ex 2.2 NCERT Solutions - Polynomials in English is available as part of our Mathematics (Maths) Class 9 for Class 9 & Ex 2.2 NCERT Solutions - Polynomials in Hindi for Mathematics (Maths) Class 9 course. Download more important topics related with notes, lectures and mock test series for Class 9 Exam by signing up for free. Class 9: Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9
1 Crore+ students have signed up on EduRev. Have you?

Q.1. Find the value of the polynomial 5x – 4x2 + 3 at 
(i) x = 0 
(ii) x = –1 
(iii) x = 2
Solution: 
Let f(x) = 5x−4x2+3
(i) When x = 0
f(0) = 5(0)-4(0)2+3

= 3

(ii) When x = -1

f(x) = 5x−4x2+3
f(−1) = 5(−1)−4(−1)2+3
= −5–4+3
= −6

(iii) When x = 2
f(x) = 5x−4x2+3
f(2) = 5(2)−4(2)2+3
= 10–16+3
= −3

Q.2. Find p(0), p(1) and p(2) for each of the following polynomials: 
(i) p(y) = y2 – y + 1 
(ii) p(t) = 2 + t + 2t2 – t3 
(iii) p(x) = x3 
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) p(y)=y2−y+1
p(y) = y2–y+1
∴ p(0) = (0)2−(0)+1=1
p(1) = (1)2–(1)+1=1
p(2) = (2)2–(2)+1=3

(ii) p(t)=2+t+2t2−t3
p(t) = 2+t+2t2−t3
∴ p(0) = 2+0+2(0)2–(0)3=2
p(1) = 2+1+2(1)2–(1)3=2+1+2–1=4
p(2) = 2+2+2(2)2–(2)3=2+2+8–8=4

(iii) p(x)=x3
∴ p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8

(iv) P(x) = (x−1)(x+1)
∴ p(0) = (0–1)(0+1) = (−1)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3

Q.3. Verify whether the following are zeros of the polynomial, indicated against them. 

(i) p(x) = 3x + 1, x =-1/3
(ii) p(x) = 5x – p , x =4/5
(iii) p(x) = x2 – 1, x = 1, – 1
(iv) p(x) = (x + 1) (x – 2), x = –1, 2
(v) p(x) = x2, x = 0
(vi) p(x) = lx + m, x = -m/l
(vii) p(x) = 3x2 – 1, x = -1/√3, 2/√3
(viii) p(x) = 2x + 1, x = 1/2

Solution:

(i) p(x)=3x+1, x=−1/3
For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0
∴ -1/3 is a zero of p(x).

(ii) p(x)=5x–π, x = 4/5
For, x = 4/5, p(x) = 5x–π
∴ p(4/5) = 5(4/5)- = 4-
∴ 4/5 is not a zero of p(x).

(iii) p(x)=x2−1, x=1, −1
For, x = 1, −1;
p(x) = x2−1
∴ p(1)=12−1=1−1 = 0
p(−1)=(-1)2−1 = 1−1 = 0
∴ 1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2
For, x = −1,2;
p(x) = (x+1)(x–2)
∴ p(−1) = (−1+1)(−1–2)
= (0)(−3) = 0
p(2) = (2+1)(2–2) = (3)(0) = 0
∴ −1,2 are zeros of p(x).

(v) p(x) = x2, x = 0
For, x = 0 p(x) = x2
p(0) = 02 = 0
∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l
For, x = -m/l ; p(x) = lx+m
∴ p(-m/l)= l(-m/l)+m = −m+m = 0
∴ -m/l is a zero of p(x).

(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3

For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴ p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0
∴ p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0
∴ -1/√3 is a zero of p(x) but 2/√3  is not a zero of p(x).

(viii) p(x) =2x+1, x = 1/2
For, x = 1/2 p(x) = 2x+1
∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0
∴ 1/2 is not a zero of p(x).

Zero PolynomialZero Polynomial

Q.4. Find the zero of the polynomial in each of the following cases: 
(i) p(x) = x + 5 
(ii) p(x) = x – 5 
(iii) p(x) = 2x + 5 
(iv) p(x) = 3x – 2 
(v) p(x) = 3x 
(vi) p(x) = ax, a ≠ 0 
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Solution:

(i) p(x) = x+5 
p(x) = x+5
⇒ x+5 = 0
⇒ x = −5
∴ -5 is a zero polynomial of the polynomial p(x).

(ii) p(x) = x–5
p(x) = x−5
⇒ x−5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).

(iii) p(x) = 2x+5
p(x) = 2x+5
⇒ 2x+5 = 0
⇒ 2x = −5
⇒ x = -5/2
∴ x = -5/2 is a zero polynomial of the polynomial p(x).

(iv) p(x) = 3x–2 
p(x) = 3x–2
⇒ 3x−2 = 0
⇒ 3x = 2
⇒x = 2/3
∴ x = 2/3  is a zero polynomial of the polynomial p(x).

(v) p(x) = 3x 
p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).

(vi) p(x) = ax, a0
p(x) = ax
⇒ ax = 0
⇒ x = 0
∴ x = 0 is a zero polynomial of the polynomial p(x).

(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.
p(x) = cx + d
⇒ cx+d =0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x).

The document Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Related Searches

Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9

,

mock tests for examination

,

Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9

,

Extra Questions

,

ppt

,

shortcuts and tricks

,

Important questions

,

Exam

,

Summary

,

practice quizzes

,

pdf

,

past year papers

,

MCQs

,

video lectures

,

Viva Questions

,

Objective type Questions

,

Previous Year Questions with Solutions

,

Sample Paper

,

study material

,

Free

,

Ex 2.2 NCERT Solutions - Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9

,

Semester Notes

;