ncert solutions for class 10 maths Chapter 2 Polynomials class 10th maths (बहुपद) Chapter 2 class 10 maths. Here We learn what is in class 10 maths solution Chapter 2 Polynomials and how to solve questions with easiest method. एनसीइआरटी कक्षा 10 गणित प्रश्नावली 2 बहुपद के सभी प्रश्न उत्तर सवालों के जवाब सम्मिलित है। In this chapter we solve the question of NCERT Solutions for Class 10 Maths Chapter 2 exercise 2.1, class 10 maths Chapter 2 exercise 2.2, class 10 maths Chapter 2 exercise 2.3 and class 10 maths Chapter 2 exercise 2.4. Chapter 2 class 10 maths
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ncert solutions for class 10 maths Chapter 2 Polynomials
class 10th maths

Exercise – 2
Polynomials
प्रश्नावली – 2
बहुपद

Ncert Solutions for Class 10 Maths
class 10th maths solution ex 2.1 प्रश्नावली 2.1

1. किसी बहुपद 𝑝(𝑥) के लिए, 𝑦 = 𝑝(𝑥) का ग्राफ नीचे आकृति 2.10 में दिया है। प्रत्येक स्थिति में, 𝑝(𝑥) के शुन्यकों की संख्या ज्ञात कीजिए।

हल : (i) शुन्यकों की संख्या 0 है।
क्योंकि ग्राफ की रेखा 𝑋 – अक्ष को किसी भी बिंदु पर प्रतिच्छेदित नहीं कर रही है।
(ii) शुन्यकों की संख्या 1 है।
क्योंकि ग्राफ़िय रेखा 𝑋 – अक्ष को एक ही बिंदु पर प्रतिच्छेदित कर रही है।
(iii) शुन्यकों की संख्या 3 है।
क्योंकि ग्राफ़िय रेखा 𝑋 – अक्ष को तीन बिंदुओं पर प्रतिच्छेदित कर रही है।
(iv) शुन्यकों की संख्या 2 है।
क्योंकि ग्राफ़िय रेखा 𝑋 – अक्ष को दो बिंदुओं पर प्रतिच्छेदित कर रही है।
(v) शुन्यकों की संख्या 4 है।
क्योंकि ग्राफ़िय रेखा 𝑋 – अक्ष को चार बिंदुओं पर प्रतिच्छेदित कर रही है।
(vi) शुन्यकों की संख्या 3 है।
क्योंकि ग्राफ़िय रेखा 𝑋 – अक्ष को तीन बिंदुओं पर प्रतिच्छेदित कर रही है।

Ncert Solutions for Class 10 Maths
class 10th maths solution
ex 2.2 प्रश्नावली 2.2

1. निम्न द्विघात बहुपदों के शून्यक ज्ञात कीजिए और शुन्यकों तथा गुणांकों के बीच के संबंध की सत्यता की जाँच कीजिए :
(i) 𝑥² – 2𝑥 – 8          (ii) 4𝑠² – 4𝑠 + 1          (iii) 6𝑥² – 3 – 7𝑥
(iv) 4𝑢² + 8𝑢          (v) 𝑡² – 15                   (vi) 3𝑥² – 𝑥 – 4
हल : (i) 𝑥² – 2𝑥 – 8
बहुपद के गुणनखंड से –
= 𝑥² – 4𝑥 + 2𝑥 – 8
= 𝑥 (𝑥 – 4) + 2 (𝑥 – 4)
= (𝑥 – 4) (𝑥 + 2)
अतः 𝑥² – 2𝑥 – 8 का मान शून्य होगा जब 𝑥 – 4 = 0 है या 𝑥 + 2 = 0 है, अर्थात जब 𝑥 = – 2 या 𝑥 = 4 हो।
अतः 𝑥² – 2𝑥 – 8 के शुन्यंक – 2 और 4 हैं। अब
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = 
शुन्यकों का गुणनफल = (- 2) × 4 = – 8 = \(\displaystyle \frac{{-8}}{1}\) =

(ii) 4𝑠² – 4𝑠 + 1
बहुपद के गुणनखंड से –
4𝑠 – 4𝑠 + 1
= 4𝑠² – 2𝑠 – 2𝑠 + 1
= 2𝑠 (2𝑠 – 1) – 1 (2𝑠 – 1)
= (2𝑠 – 1) (2𝑠 – 1)
अतः 4𝑠² – 4𝑠 + 1 का मान शून्य है, जब 2𝑠 – 1 = 0 है,
अर्थात जब 2𝑠 – 1 = 0
2𝑠 = 1
\(\displaystyle s=\frac{1}{2}\) हो।
अतः 4𝑠² – 4𝑠 + 1 के दोनो शून्यक ही \(\displaystyle \frac{1}{2}\) होंगे।
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = \(\displaystyle \frac{1}{2}+\frac{1}{2}\) = 1
\(\displaystyle \frac{{-\left( {-4} \right)}}{4}\) =

शुन्यकों का गुणनफल = αβ = \(\displ aystyle \frac{c}{a}\) = \(\displaystyle \frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\frac{1}{4}\) =

(iii) 6𝑥² – 3 – 7𝑥
बहुपद के गुणनखंड से –
6𝑥² – 7𝑥 – 3
= 6𝑥² – 9𝑥 + 2𝑥 – 3
= 3𝑥 (2𝑥 – 3) + 1(2𝑥 – 3)
= (3𝑥 + 1) (2𝑥 – 3)
अतः 6𝑥² – 3 – 7𝑥 का मान शून्य है, जब 3𝑥 + 1 = 0 है या 2𝑥 – 3 = 0 है,
अर्थात जब \(\displaystyle x=\frac{{-1}}{3}\) या \(\displaystyle x=\frac{{3}}{2}\)
अतः 6𝑥² – 7𝑥 – 3 के शून्यक \(\displaystyle \frac{{-1}}{3}\) और \(\displaystyle \frac{{3}}{2}\) हैं। अतः अब
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = \(\displaystyle \frac{{-1}}{3}+\frac{3}{2}=\frac{{-2+9}}{6}\)
\(\displaystyle =\frac{7}{6}=\frac{{-(-7)}}{6}\) =

शुन्यकों का गुणनफल = αβ = \(\displ aystyle \frac{c}{a}\) = \(\displaystyle \left( {-\frac{1}{3}} \right)\times \frac{3}{2}\)
\(\displaystyle -\frac{1}{2}=\frac{{-3}}{6}\) =

(iv) 4𝑢² + 8𝑢
बहुपद के गुणनखंड से –
4𝑢² + 8𝑢
= 4𝑢 (4 + 2)
इसलिए 4𝑢² + 8𝑢 का मान शून्य है, जब 4𝑢 = 0 है या 𝑢 + 2 =0 है,
अर्थात जब 𝑢 = 0 या 𝑢 = – 2 हो।
अतः 4𝑢² + 8𝑢 के शून्यक 0 और – 2 हैं। अतः अब
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = 0 + (- 2) = – 2
= \(\displaystyle \frac{{-(8)}}{4}\) =

शुन्यकों का गुणनफल = αβ = \(\displ aystyle \frac{c}{a}\) = 0 × (- 2) = 0
\(\displaystyle =\frac{0}{4}\) =

(v) 𝑡² – 15
𝑡² – 15 = 0
𝑡² = 15
𝑡 = √15
अतः 𝑡 = + √15 और 𝑡 = – √15
इसलिए 𝑡² – 15 का मान शून्य है, जब 𝑡 + √15 है या 𝑡 – √15 हो।
इसलिए 𝑡² – 15 के शून्यक + √15 या – √15 हैं। अब
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = – √15 + √15 = 0
\(\displaystyle =\frac{{-(0)}}{1}\) =

शुन्यकों का गुणनफल = αβ = \(\displ aystyle \frac{c}{a}\) = (- √15) × (+ √15) = – 15
\(\displaystyle =\frac{{-15}}{1}\) =

(vi) 3𝑥² – 𝑥 – 4
बहुपद के गुणनखंड से –
= 3𝑥² – 4𝑥 + 3𝑥 – 4
= 𝑥 (3𝑥 – 4) + 1 (3𝑥 – 4)
= (3𝑥 – 4) (𝑥 + 1)
इसलिए 3𝑥² – 𝑥 – 4 का मान शून्य है, जब 3𝑥 – 4 = 0 है या 𝑥 + 1 = 0 है,
अर्थात जब \(\displaystyle x=\frac{4}{3}\) या 𝑥 = -1 है। अब
शुन्यकों का योग = α + β = \(\displaystyle \frac{{-b}}{a}\) = \(\displaystyle \frac{4}{3}+(-1)=\frac{{4-3}}{3}=\frac{1}{3}\)
\(\displaystyle \frac{{-(-1)}}{3}\) =

शुन्यकों का गुणनफल = αβ = \(\displ aystyle \frac{c}{a}\) = \(\displaystyle \frac{4}{3}\times (-1)=-\frac{4}{3}\)
= \(\displaystyle \frac{{-4}}{3}\) =

2. एक द्विघात बहुपद ज्ञात कीजिए, जिनके शुन्यकों के योग तथा गुणनफल क्रमश: दी गई संख्याएँ हैं :
(i) \(\displaystyle \frac{1}{4},-1\)          (ii) \(\displaystyle \sqrt{2},\frac{1}{3}\)          (iii) 0, √5
(iv) 1, 1          (v) \(\displaystyle -\frac{1}{4},\frac{1}{4}\)          (vi) 4, 1
हल : (i) \(\displaystyle \frac{1}{4},-1\)
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः \(\displaystyle \alpha +\beta =\frac{1}{4}=\frac{{-b}}{a}\)
\(\displaystyle \alpha \beta =-1=\frac{{-4}}{4}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 4, b = – 1 और c = – 4
अतः वह द्विघात बहुपद 4𝑥² – 𝑥 – 4 होगा।

(ii) \(\displaystyle \sqrt{2},\frac{1}{3}\)
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः \(\displaystyle \alpha +\beta =\sqrt{2}=\frac{{3\sqrt{2}}}{3}=\frac{{-b}}{a}\)
\(\displaystyle \alpha \beta =\frac{1}{3}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 3, b = – 3√2 और c = 1
अतः वह द्विघात बहुपद 3𝑥² – 3√2𝑥 + 1 होगा।

(iii) 0, √5
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः शुन्यकों का योग = \(\displaystyle \alpha +\beta =0=\frac{0}{1}=\frac{{-b}}{a}\)
शुन्यकों का गुणनफल = \(\displaystyle \alpha \beta =\sqrt{5}=\frac{{\sqrt{5}}}{1}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 1, b = 0 और c = √5
अतः वह द्विघात बहुपद 𝑥² – 0.𝑥 + √5 होगा।

(iv) 1, 1
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः शुन्यकों का योग = \(\displaystyle \alpha +\beta =1=\frac{1}{1}=\frac{{-b}}{a}\)
शुन्यकों का गुणनफल = \(\displaystyle \alpha \beta =1=\frac{1}{1}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 1, b = 1 और c = 1
अतः वह द्विघात बहुपद 𝑥² – 𝑥 + 1 होगा।

(v) \(\displaystyle -\frac{1}{4},\frac{1}{4}\)
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः शुन्यकों का योग = \(\displaystyle \alpha +\beta =\frac{{-1}}{4}=\frac{{-b}}{a}\)
शुन्यकों का गुणनफल = \(\displaystyle \alpha \beta =\frac{1}{4}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 4, b = 1 और c = – 1
अतः वह द्विघात बहुपद 4𝑥² – 𝑥 + 1 होगा।

(vi) 4, 1
मानाकि कोई द्विघात बहुपद a𝑥² + b𝑥 + c है और इसके शून्यक α और β है।
अतः शुन्यकों का योग = \(\displaystyle \alpha +\beta =4=\frac{4}{1}=\frac{{-b}}{a}\)
शुन्यकों का गुणनफल = \(\displaystyle \alpha \beta =1=\frac{1}{1}=\frac{c}{a}\)
दोनों की तुलना करने पर –
a = 1, b = – 4 और c = 1
अतः वह द्विघात बहुपद 𝑥² – 4𝑥 + 1 होगा।

Ncert Solutions for Class 10 Maths
class 10th maths solution
ex 2.3 प्रश्नावली 2.3

1. विभाजन एल्गोरिथ्म का प्रयोग करके, निम्न में 𝑝(𝑥) को 𝑔(𝑥) से भाग देने पर भागफल तथा शेषफल ज्ञात कीजिए :
(i) 𝑝(𝑥) = 𝑥³ – 3𝑥² + 5𝑥 -3,          𝑔(𝑥) = 𝑥² – 2
(ii) 𝑝(𝑥) = 𝑥⁴ – 3𝑥² + 4𝑥 + 5,       𝑔(𝑥) = 𝑥² + 1 – 𝑥
(iii) 𝑝(𝑥) = 𝑥⁴ – 5𝑥 + 6,                𝑔(𝑥) = 2 – 𝑥²
हल :
(i) 𝑝(𝑥) = 𝑥³ – 3𝑥² + 5𝑥 -3,          𝑔(𝑥) = 𝑥² – 2
\(\displaystyle \begin{array}{l}{{x}^{2}}-2\overset{{x-3}}{\overline{\left){\begin{array}{l}{{x}^{3}}-3{{x}^{2}}+5x-3\\x3\,\,\,\,\,\,\,\,\,\,\,\,\,-2x\end{array}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,{\mathrm X}-3{{x}^{2}}+7x-3\\\,\,\,\,\,\,\,\,-3{{x}^{2}}\,\,\,\,\,\,\,\,\,\,+6\\\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x-9\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}\)
अतः भागफल = 𝑥 – 3
शेषफल = 7𝑥 – 9

(ii) 𝑝(𝑥) = 𝑥⁴ – 3𝑥² + 4𝑥 + 5,       𝑔(𝑥) = 𝑥² + 1 – 𝑥
\(\displaystyle \begin{array}{l}{{x}^{2}}-x+1\overset{{{{x}^{2}}+x-3}}{\overline{\left){{{{x}^{4}}+0.{{x}^{3}}-3{{x}^{2}}+4x+5}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{4}}-\,\,\,\,{{x}^{3}}+{{x}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,-\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}-4{{x}^{2}}+4x+5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}-{{x}^{2}}+x\\\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,-\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{2}}+3x+5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{2}}+3x-3\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}}\end{array}\)
अतः भागफल = 𝑥² + 𝑥 – 3
शेषफल = 8 है।

(iii) 𝑝(𝑥) = 𝑥⁴ – 5𝑥 + 6,                𝑔(𝑥) = 2 – 𝑥²
\(\displaystyle \begin{array}{l}-{{x}^{2}}+2\overset{{-{{x}^{2}}-2}}{\overline{\left){{{{x}^{4}}+0.{{x}^{2}}-5x+6}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{4}}-\,\,2{{x}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,2{{x}^{2}}-5x+6\\\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,+4\\\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-5x+10\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}\)
अतः भागफल = – 𝑥² – 2
शेषफल = – 5𝑥 + 10 है।

2. पहले बहुपद से दूसरे बहुपद को भाग करके, जाँच कीजिए कि क्या प्रथम बहुपद द्वितीय बहुपद का एक गुणनखंड है :
(i) 𝑡² – 3, 2𝑡⁴ + 3𝑡³ – 2𝑡² – 9𝑡 – 12
(ii) 𝑥² + 3𝑥 + 1, 3𝑥⁴ + 5𝑥³ – 7𝑥² + 2𝑥 + 2
(iii) 𝑥³ – 3𝑥 + 1, 𝑥⁵ – 4𝑥³ + 𝑥² + 3𝑥 + 1
हल : (i) 𝑡² – 3, 2𝑡⁴ + 3𝑡³ – 2𝑡² – 9𝑡 – 12
\(\displaystyle \begin{array}{l}{{t}^{2}}+0.t-3\overset{{2{{t}^{2}}+3t+4}}{\overline{\left){{2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{t}^{4}}+0.{{t}^{3}}-6{{t}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{t}^{3}}+4{{t}^{2}}-9t-12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{t}^{3}}+0.{{t}^{2}}-9t\\\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,+\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{{t}^{2}}+0.t-12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4t2+0.t-12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,+\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}}\end{array}\)

(ii) 𝑥² + 3𝑥 + 1, 3𝑥⁴ + 5𝑥³ – 7𝑥² + 2𝑥 + 2
\(\displaystyle \begin{array}{l}{{x}^{2}}+3x+1\overset{{3{{x}^{2}}-4x+2}}{\overline{\left){{3{{x}^{4}}+5{{x}^{3}}-7{{x}^{2}}+2x+2}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{4}}+9{{x}^{2}}+3{{x}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4{{x}^{3}}-10{{x}^{2}}+2x+2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4×3-12×2-4x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{x}^{2}}+6x+2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2×2+6x+2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,-\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}\)

3. 3𝑥⁴ + 6𝑥³ – 2𝑥² – 10𝑥 – 5 के अन्य सभी शून्यक ज्ञात कीजिए, यदि इसके दो शून्यक \(\displaystyle \sqrt{{\frac{5}{3}}}\) और \(\displaystyle -\sqrt{{\frac{5}{3}}}\) हैं।
हल : मानाकि 𝑝(𝑥) = 3𝑥⁴ + 6𝑥³ – 2𝑥² – 10𝑥 – 5
अतः 𝑝(𝑥) के शून्यक = \(\displaystyle \sqrt{{\frac{5}{3}}}\) और \(\displaystyle -\sqrt{{\frac{5}{3}}}\) होंगे।
𝑝(𝑥) के गुणनखंड = \(\displaystyle \left( {x-\sqrt{{\frac{5}{3}}}} \right)\) और \(\displaystyle \left( {x+\sqrt{{\frac{5}{3}}}} \right)\) हैं।
या 𝑝(𝑥) का गुणनखंड = \(\displaystyle {{x}^{2}}-\frac{5}{3}\) है।
\(\displaystyle \begin{array}{l}{{x}^{2}}+0.x-\frac{5}{3}\overset{{3{{x}^{2}}+6x+3}}{\overline{\left){{3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{4}}+0.{{x}^{3}}-5{{x}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{{6{{x}^{3}}+3{{x}^{2}}\,\,\,-10x-5}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6{{x}^{3}}+0.{{x}^{2}}-10x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+0.x-5\\\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+0.x-5\\\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,+\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}\)
3𝑥⁴ + 6𝑥³ – 2𝑥² – 10𝑥 – 5 = \(\displaystyle \left( {{{x}^{2}}-\frac{5}{3}} \right)\left( {3{{x}^{2}}+6x+3} \right)\)
3𝑥⁴ + 6𝑥³ – 2𝑥² – 10𝑥 – 5 = \(\displaystyle 3\left( {{{x}^{2}}-\frac{5}{3}} \right)\left( {{{x}^{2}}+x+1} \right)\)
इस प्रकार
\(\displaystyle \begin{array}{l}p(x)=3\left( {{{x}^{2}}-\frac{5}{3}} \right)\left( {{{x}^{2}}+2x+1} \right)\\=3\left( {{{x}^{2}}-\frac{5}{3}} \right)\left( {{{x}^{2}}+x+x+1} \right)\\=3\left( {{{x}^{2}}-\frac{5}{3}} \right)\left[ {x(x+1)+1(x+1)} \right]\\=3\left( {{{x}^{2}}-\frac{5}{3}} \right)(x+1)(x+1)\end{array}\)
अतः 3x⁴ + 6x³ – 2x² – 10x – 5 के शून्यक \(\displaystyle \sqrt{{\frac{5}{3}}},\,-\sqrt{{\frac{5}{3}}}\) , -1 और -1 है।

4. यदि 𝑥³ – 3𝑥² + 𝑥 + 2 को एक बहुपद 𝑔(𝑥) से भाग देने पर, भागफल और शेषफल क्रमश: 𝑥 – 2 और – 2𝑥 + 4 है तो 𝑔(𝑥) ज्ञात कीजिए।
हल : यहाँ भाजक 𝑔(𝑥)
भागफल = 𝑥 – 2
भाज्य = 𝑥³ – 3𝑥² + 𝑥 + 2
शेषफल = – 2𝑥 + 4
हम जानते है कि –
भाज्य = भाजक × भागफल + शेषफल
अतः 𝑥³ – 3𝑥² + 𝑥 + 2 = 𝑔(𝑥) × (𝑥 – 2) – (- 2𝑥 + 4)
= 𝑥³ – 3𝑥² + 𝑥 + 2 – (- 2𝑥 + 4) = 𝑔(𝑥) × (𝑥 – 2)
𝑔(𝑥) = \(\displaystyle \frac{{{{x}^{3}}-3{{x}^{2}}+x+2-(-2x+4)}}{{(x-2)}}\)
\(\displaystyle \begin{array}{l}\frac{{{{x}^{3}}-3{{x}^{2}}+x+2+2x-4}}{{(x-2)}}\\\frac{{{{x}^{3}}-3{{x}^{2}}+3x-2}}{{(x-2)}}\end{array}\)
\(\displaystyle \begin{array}{l}x-2\overset{{{{x}^{2}}-x+1}}{\overline{\left){{{{x}^{3}}-3{{x}^{2}}+3x-2}}\right.}}\\\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}-2{{x}^{2}}\\\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,+\\\,\,\,\,\,\,\,\,\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}+3x-2\\\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}+2x\\\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,-\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,+\\\overline{\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\\\overline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\end{array}}\end{array}}\end{array}}\end{array}\)
अतः 𝑔(𝑥) = 𝑥² – 𝑥 + 1

5. बहुपदों 𝑝(𝑥), 𝑔(𝑥), 𝑞(𝑥) और 𝑟(𝑥) के ऐसे उदाहरण दीजिए जो विभाजन अल्गोरिथम को संतुष्ट करते हों तथा
(i) घात 𝑝(𝑥) = घात 𝑞(𝑥)       (ii) घात 𝑞(𝑥) = घात 𝑟(𝑥)       (iii) घात 𝑟(𝑥) = 0
हल : यूक्लिड विभाजन एल्गोरिथ्म से 𝑝(𝑥) = 𝑔(𝑥) × 𝑔(𝑥) × 𝑟(𝑥), जहाँ 𝑞(𝑥) ≠ 0,
घात 𝑟(𝑥) = 0 या घात 𝑟(𝑥) < घात 𝑔(𝑥)
(i) घात 𝑝(𝑥) = घात 𝑞(𝑥)
भाज्य और भागफल की घात तभी बराबर हो सकती है जब भाजक एक अचर (घात 0 हो) संख्या हो।
अतः माना 𝑝(𝑥) = 3𝑥² – 6𝑥 + 5
माना 𝑔(𝑥) = 3
इस प्रकार 𝑞(𝑥) = 𝑥² – 2𝑥 + 1 और 𝑟(𝑥) = 2

(ii) घात 𝑞(𝑥) = घात 𝑟(𝑥)
अतः माना 𝑝(𝑥) = 2𝑥² – 4𝑥 + 3
माना 𝑔(𝑥) = 𝑥² – 2𝑥 + 1
इस प्रकार 𝑞(𝑥) = 2 और 𝑟(𝑥) = 1

(iii) घात r(x) = 0
माना p(x) = 2x² -4x + 3
माना g(x) = x² – 2x + 1
इस प्रकार q(x) = 2 और r(x) = 1