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RBSE Solutions for Class 9 Maths Chapter 3 बहुपद Ex 3.3

February 20, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 9 Maths Chapter 3 बहुपद Ex 3.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Chapter 3 बहुपद Exercise 3.3.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 9
Subject Maths
Chapter Chapter 3
Chapter Name बहुपद
Exercise Exercise 3.3
Number of Questions Solved 4
Category RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Chapter 3 बहुपद Ex 3.3

प्रश्न 1.
बहुपद x4 + x3 – 3x2 + 3x + 1 में निम्नलिखित एक घातीय व्यंजक से भाग देने पर शेषफल ज्ञात कीजिए।
(i) x – 1
(ii) x – \(\frac { 1 }{ 2 }\)
(iii) x + π
(iv) 3 + 2x
(v) x
हल
माना p (x) = x4 + x3 – 3x2 + 3x + 1
(i) p (3) में (x – 1) का भाग देने पर शेषफल p (1) के बराबर होगा।
p (1) = 14 + 13 – 3(1)2 + 3(1) + 1 = 1 + 1 – 3 + 3 + 1 = 3 = शेषफल
(ii) p (x) में (x – \(\frac { 1 }{ 2 }\) का भाग देने(RBSESolutions.com)पर शेषफल p(\(\frac { 1 }{ 2 }\)) के बराबर होगी।
RBSE Solutions for Class 9 Maths Chapter 3 बहुपद Ex 3.3

RBSE Solutions

प्रश्न 2.
2x3 + 2ax2 – 5x + a को x + a से भाग देने पर शेषफल ज्ञात कीजिए।
हल
माना, p (x) = 2x3 + 2ax2 – 5x + a
p (x) में (x + a) का भाग देने पर शेषफल p (-a) के बराबर होगा।
p(-a) = 2(-a)3 + 2a (-a)2 – 5(-a) + a = -2a3 + 2a3 + 5a + a = 6a = शेषफल

प्रश्न 3.
जाँच कीजिए कि x + 1, x3 + 3x2 + 3x + 1 का एक गुणनखण्ड है या नहीं।
हल
(x + 1) दिए गए व्यजंक का एक(RBSESolutions.com)गुणनखण्ड होगा यदि व्यजंक को (x + 1) से भाग देने पर शेषफल शून्य हो।
x + 1 = 0. = x = -1
माना p (x) = x3 + 3x2 + 3x + 1
p(-1) = (-1)3 + 3 (-1)2 + 3 (-1) + 1 = -1 + 3 – 3 + 1 = 0
चूंकि, शेषफल शून्य है, अतः (x + 1), दिए गए।
व्यजंक का एक गुणनखण्ड है।

RBSE Solutions

प्रश्न 4.
यदि बहुपद x3 + x2 – 4x + a और 2x3 + ax2 + 3x – 3 को -2 से भाग देने पर समान शेषफल प्राप्त होता है, तो a का मान ज्ञात कीजिए।
हल
माना p(x) = x3 + x2 – 4x + a
तथा q(x) = 2x3 + ax2 + 3x – 3
p(x) तथा q (3) में (x – 2) से भाग देने पर शेषफल p (2) तथा q (2) के बराबर होगा।
p (2) = 23 + 22 – 4(2) + a = 8 + 4 – 8 + 2 = 4 + 2 = शेषफल
पुनः q(2) = 2 (2)3 + a (2)2 +3 (2) – 3
= 2 x 8 + a x 4 + 6 – 3
= 16 + 4a + 3
= 19 + 4a = शेषफल
अतः p(2) = q(2)
⇒ 4 + a = 19 + 4a
⇒4a – a = 4 – 19
⇒ 3a = -15
⇒ a = -5

RBSE Solutions

We hope the given RBSE Solutions for Class 9 Maths Chapter 3 बहुपद Ex 3.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Chapter 3 बहुपद Exercise 3.3, drop a comment below and we will get back to you at the earliest.

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