Class 9 Chapter 8 উৎপাদকে বিশ্লেষণ (Factorisation)

নবম শ্রেণী – অষ্টম অধ্যায় : উৎপাদকে বিশ্লেষণ সম্পূর্ণ সমাধান

কষে দেখি – 8.1

নীচের বহুপদী সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি –

 

1. x3 – 3x + 2                                 

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}-3x+2

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0পাই।

\[f\left( 1 \right)={{1}^{3}}-3\times 1+2=1-3+2=0\]

দেখছি f\left( 1 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( x-1 \right),\,f\left( x \right) এর একটি উৎপাদক।

\[{{x}^{3}}-3x+2\]

\[={{x}^{3}}-{{x}^{2}}+{{x}^{2}}-x-2x+2\]

\[={{x}^{2}}\left( x-1 \right)+x\left( x-1 \right)-2\left( x-1 \right)\]

\[=\left( x-1 \right)\left( {{x}^{2}}+x-2 \right)\]

\[=\left( x-1 \right)\left\{ {{x}^{2}}+\left( 2-1 \right)x-2 \right\}\]

\[=\left( x-1 \right)\left\{ {{x}^{2}}+2x-x-2 \right\}\]

\[=\left( x-1 \right)\left\{ x\left( x+2 \right)-1\left( x+2 \right) \right\}\]

\[=\left( x-1 \right)\left( x+2 \right)\left( x-1 \right)\]

\[={{\left( x-1 \right)}^{2}}\left( x+2 \right)\]

 

2. x3 + 2x + 3                                 

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}+2x+3

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ –1, +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0পাই।

f\left( -1 \right)={{\left( -1 \right)}^{3}}+2\times \left( -1 \right)+3=-1-2+3=0

দেখছি f\left( -1 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( x+1 \right),\,f\left( x \right)এর একটি উৎপাদক।

\[{{x}^{3}}+2x+3\]

\[={{x}^{3}}+{{x}^{2}}-{{x}^{2}}-x+3x+3\]

\[={{x}^{2}}\left( x+1 \right)-x\left( x+1 \right)+3\left( x+1 \right)\]

\[=\left( x+1 \right)\left( {{x}^{2}}-x+3 \right)\]

 

3. a3 – 12a – 16

উত্তর-

মনেকরি, f\left( a \right)={{a}^{3}}-12a-16

প্রথমে f\left( a \right)-এর একটি উৎপাদক খুঁজি।

f\left( a \right)–এর x –এ –1, +1, –2, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( a \right)=0পাই।

\[f\left( -2 \right)={{\left( -2 \right)}^{3}}-12\left( -2 \right)-16=-8+24-16=0\]

দেখছি f\left( -2 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( a+2 \right),\,f\left( a \right)এর একটি উৎপাদক।

\[{{a}^{3}}-12a-16\]

\[={{a}^{3}}+2{{a}^{2}}-2{{a}^{2}}-4a-8a-16\]

\[={{a}^{2}}\left( a+2 \right)-2a\left( a+2 \right)-8\left( a+2 \right)\]

\[=\left( a+2 \right)\left( {{a}^{2}}-2a-8 \right)\]

\[=\left( a+2 \right)\left\{ {{a}^{2}}-\left( 4-2 \right)a-8 \right\}\]

\[=\left( a+2 \right)\left\{ {{a}^{2}}-4a+2a-8 \right\}\]

\[=\left( a+2 \right)\left\{ a\left( a-4 \right)+2\left( a-4 \right) \right\}\]

\[=\left( a+2 \right)\left( a-4 \right)\left( a+2 \right)\]

\[={{\left( a+2 \right)}^{2}}\left( a-4 \right)\]

 

4. x3 – 6x + 4                                 

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}-6x+4

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0পাই।

\[f\left( 2 \right)={{\left( 2 \right)}^{3}}-6\times \left( 2 \right)+4=8-12+4=0\]

দেখছি f\left( 2 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( x-2 \right),\,f\left( x \right)এর একটি উৎপাদক।

\[{{x}^{3}}-6x+4\]

\[={{x}^{3}}-2{{x}^{2}}+2{{x}^{2}}-4x-2x+4\]

\[={{x}^{2}}\left( x-2 \right)+2x\left( x-2 \right)-2\left( x-2 \right)\]

\[=\left( x-2 \right)\left( {{x}^{2}}+2x-2 \right)\]

 

5. x3 – 19x – 30                            

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}-19x-30

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ +1, –2, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0 পাই।

\[f\left( -2 \right)={{\left( -2 \right)}^{3}}-19\times \left( -2 \right)-30=-8+38-30=0\]

দেখছি f\left( -2 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( x+2 \right),\,f\left( x \right)এর একটি উৎপাদক।

\[{{x}^{3}}-19x-30\]

\[={{x}^{3}}+2{{x}^{2}}-2{{x}^{2}}-4x-2x+30\]

\[={{x}^{2}}\left( x+2 \right)-2x\left( x+2 \right)-15\left( x+2 \right)\]

\[=\left( x+2 \right)\left( {{x}^{2}}-2x-15 \right)\]

\[=\left( x+2 \right)\left\{ {{x}^{2}}-\left( 5-3 \right)x-15 \right\}\]

\[=\left( x+2 \right)\left\{ {{x}^{2}}-5x+3x-15 \right\}\]

\[=\left( x+2 \right)\left\{ x\left( x-5 \right)+3\left( x-5 \right) \right\}\]

\[=\left( x+2 \right)\left( x-5 \right)\left( x+3 \right)\]

 

6. 4a3 – 9a2 + 3a + 2

উত্তর-

মনেকরি, f\left( a \right)=4{{a}^{3}}-9{{a}^{2}}+3a+2

প্রথমে f\left( a \right)-এর একটি উৎপাদক খুঁজি।

f\left( a \right)–এর x –এ –1, +1, –2, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( a \right)=0 পাই।

\[f\left( 1 \right)=4{{\left( 1 \right)}^{3}}-9{{\left( 1 \right)}^{2}}+3\left( 1 \right)+2=4-9+3+2=0\]

দেখছি f\left( 1 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( a-1 \right),\,f\left( a \right)এর একটি উৎপাদক।

\[4{{a}^{3}}-9{{a}^{2}}+3a+2\]

\[=4{{a}^{3}}-4{{a}^{2}}-5{{a}^{2}}+5a-2a+2\]

\[=4{{a}^{2}}\left( a-1 \right)-5a\left( a-1 \right)-2\left( a-1 \right)\]

\[=\left( a-1 \right)\left( 4{{a}^{2}}-5a-2 \right)\]

 

7. x3 – 9x2 + 23x – 15                  

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}-9{{x}^{2}}+23x-15

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0পাই।

\[f\left( 1 \right)={{\left( 1 \right)}^{3}}-9{{\left( 1 \right)}^{2}}+23\left( 1 \right)-15=1-9+23-15=0\]

দেখছি f\left( 1 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( x-1 \right),\,f\left( x \right)এর একটি উৎপাদক।

\[{{x}^{3}}-9{{x}^{2}}+23x-15\]

\[={{x}^{3}}-{{x}^{2}}-8{{x}^{2}}+8x+15x-15\]

\[={{x}^{2}}\left( x-1 \right)-8x\left( x-1 \right)+15\left( x-1 \right)\]

\[=\left( x-1 \right)\left( {{x}^{2}}-8x+15 \right)\]

\[=\left( x-1 \right)\left\{ {{x}^{2}}-\left( 5+3 \right)x+15 \right\}\]

\[=\left( x-1 \right)\left\{ {{x}^{2}}-5x-3x+15 \right\}\]

\[=\left( x-1 \right)\left\{ x\left( x-5 \right)-3\left( x-5 \right) \right\}\]

\[=\left( x-1 \right)\left( x-5 \right)\left( x-3 \right)\]

 

8. 5a3 + 11a2 + 4a – 2                 

উত্তর-

মনেকরি, f\left( a \right)=5{{a}^{3}}+11{{a}^{2}}+4a-2

প্রথমে f\left( a \right)-এর একটি উৎপাদক খুঁজি।

f\left( a \right)–এর x –এ –1, +1, –2, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( a \right)=0পাই।

\[f\left( -1 \right)=5{{\left( -1 \right)}^{3}}+11{{\left( -1 \right)}^{2}}+4\left( -1 \right)-2=-5+11-4-2=0\]

দেখছি f\left( -1 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( a+1 \right),\,f\left( a \right)এর একটি উৎপাদক।

\[5{{a}^{3}}+11{{a}^{2}}+4a-2\]

\[=5{{a}^{3}}+5{{a}^{2}}+6{{a}^{2}}+6a-2a+2\]

\[=5{{a}^{2}}\left( a+1 \right)+6a\left( a+1 \right)-2\left( a+1 \right)\]

\[=\left( a+1 \right)\left( 5{{a}^{2}}+6a-2 \right)\]

 

9. 2x3 – x2 + 9x + 5

উত্তর-

মনেকরি, f\left( x \right)=2{{x}^{3}}-{{x}^{2}}+9x+5

প্রথমে f\left( x \right)-এর একটি উৎপাদক খুঁজি।

f\left( x \right)–এর x –এ -\frac{1}{2}, +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( x \right)=0পাই।

\[f\left( -\frac{1}{2} \right)=2{{\left( -\frac{1}{2} \right)}^{3}}-{{\left( -\frac{1}{2} \right)}^{2}}+9\left( -\frac{1}{2} \right)+5=-\frac{1}{4}-\frac{1}{4}-\frac{9}{2}+5=-5+5=0\]

দেখছি f\left( -\frac{1}{2} \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( 2x+1 \right),\,f\left( x \right)এর একটি উৎপাদক।

\[2{{x}^{3}}-{{x}^{2}}+9x+5\]

\[=2{{x}^{3}}+{{x}^{2}}-2{{x}^{2}}-x+15x+5\]

\[={{x}^{2}}\left( 2x+1 \right)-x\left( 2x+1 \right)+5\left( 2x+1 \right)\]

\[=\left( 2x+1 \right)\left( {{x}^{2}}-x+5 \right)\]

 

10. 2y3 – 5y2 – 19y + 42

উত্তর-

মনেকরি, f\left( y \right)=2{{y}^{3}}-5{{y}^{2}}-19y+42

প্রথমে f\left( y \right)-এর একটি উৎপাদক খুঁজি।

f\left( y \right)–এর x –এ  +1, +2, +3, বসিয়ে দেখি –এর কোন মানে f\left( y \right)=0পাই।

\[f\left( 2 \right)=2{{\left( 2 \right)}^{3}}-5{{\left( 2 \right)}^{2}}-19\left( 2 \right)+42=16-20-38+42=0\]

দেখছি f\left( 2 \right)=0

গুণনীয়ক উপপাদ্য থেকে বলতে পারি, \left( y-2 \right),\,f\left( y \right)এর একটি উৎপাদক।

\[2{{y}^{3}}-5{{y}^{2}}-19y+42\]

\[=2{{y}^{3}}-4{{y}^{2}}-{{y}^{2}}+2y-21y+42\]

\[=2{{y}^{2}}\left( y-2 \right)-y\left( y-2 \right)-21\left( y-2 \right)\]

\[=\left( y-2 \right)\left( 2{{y}^{2}}-y-21 \right)\]

\[=\left( y-2 \right)\left\{ 2{{y}^{2}}-\left( 7-6 \right)y-21 \right\}\]

\[=\left( y-2 \right)\left\{ 2{{y}^{2}}-7y+6y-21 \right\}\]

\[=\left( y-2 \right)\left\{ y\left( 2y-7 \right)+3\left( 2y-7 \right) \right\}\]

\[=\left( y-2 \right)\left( 2y-7 \right)\left( y+3 \right)\]

কষে দেখি – 8.2

নীচের বীজগাণিতিক সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি –

 

1.     \frac{x^4}{16}-\frac{y^4}{81}                               

উত্তর-

\[\frac{{{x}^{4}}}{16}-\frac{{{y}^{4}}}{81}\]

\[={{\left( \frac{{{x}^{2}}}{4} \right)}^{2}}-{{\left( \frac{{{y}^{2}}}{9} \right)}^{2}}\]

\[=\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)\left( \frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{9} \right)\]

\[=\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)\left\{ {{\left( \frac{x}{2} \right)}^{2}}-{{\left( \frac{y}{3} \right)}^{2}} \right\}\]

\[=\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)\left( \frac{x}{2}+\frac{y}{3} \right)\left( \frac{x}{2}-\frac{y}{3} \right)\]

 

2.     m^2+\frac1{m^2}+2-2m-\frac2m

উত্তর-

\[{{m}^{2}}+\frac{1}{{{m}^{2}}}+2-2m-\frac{2}{m}\]

\[={{\left( m+\frac{1}{m} \right)}^{2}}-2\times m\times \frac{1}{m}+2-2\left( m+\frac{1}{m} \right)\]

\[={{\left( m+\frac{1}{m} \right)}^{2}}-2+2-2\left( m+\frac{1}{m} \right)\]

\[={{\left( m+\frac{1}{m} \right)}^{2}}-2\left( m+\frac{1}{m} \right)\]

\[=\left( m+\frac{1}{m} \right)\left( m+\frac{1}{m}-2 \right)\]

 

3. 9p2 – 24pq + 16q2 + 3ap – 4aq

উত্তর-

\[9{{p}^{2}}-24pq+16{{q}^{2}}+3ap-4aq\]

\[={{\left( 3p \right)}^{2}}-2\times 3p\times 4q+{{\left( 4q \right)}^{2}}+a\left( 3p-4q \right)\]

\[={{\left( 3p-4q \right)}^{2}}+a\left( 3p-4q \right)\]

\[=\left( 3p-4q \right)\left( 3p-4q+a \right)\]

 

4. 4x4 + 81                                     

উত্তর-

\[4{{x}^{4}}+81\]

\[={{\left( 2{{x}^{2}} \right)}^{2}}+{{\left( 9 \right)}^{2}}\]

\[={{\left( 2{{x}^{2}}+9 \right)}^{2}}-2\times 2{{x}^{2}}\times 9\]

\[={{\left( 2{{x}^{2}}+9 \right)}^{2}}-36{{x}^{2}}\]

\[={{\left( 2{{x}^{2}}+9 \right)}^{2}}-{{\left( 6x \right)}^{2}}\]

\[=\left( 2{{x}^{2}}+9+6x \right)\left( 2{{x}^{2}}+9-6x \right)\]

\[=\left( 2{{x}^{2}}+6x+9 \right)\left( 2{{x}^{2}}-6x+9 \right)\]

 

5. x4 – 7x2 + 1                               

উত্তর-

\[{{x}^{4}}-7{{x}^{2}}+1\]

\[={{\left( {{x}^{2}} \right)}^{2}}+{{\left( 1 \right)}^{2}}-7{{x}^{2}}\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}-2{{x}^{2}}-7{{x}^{2}}\,\,\left[ \because {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \right]\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}-9{{x}^{2}}\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}-{{\left( 3x \right)}^{2}}\]

\[=\left( {{x}^{2}}+1+3x \right)\left( {{x}^{2}}+1-3x \right)\]

\[=\left( {{x}^{2}}+3x+1 \right)\left( {{x}^{2}}-3x+1 \right)\]

 

6. p4 – 11 p2q2 + q4

উত্তর-

\[{{p}^{4}}-11{{p}^{2}}{{q}^{2}}+{{q}^{4}}\]

\[={{\left( {{p}^{2}} \right)}^{2}}+{{\left( {{q}^{2}} \right)}^{2}}-11{{p}^{2}}{{q}^{2}}\]

\[={{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}}+2{{p}^{2}}{{q}^{2}}-11{{p}^{2}}{{q}^{2}}\,\,\left[ \because {{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab \right]\]

\[={{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}}-9{{p}^{2}}{{q}^{2}}\]

\[={{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}}-{{\left( 3pq \right)}^{2}}\]

\[=\left( {{p}^{2}}-{{q}^{2}}+3pq \right)\left( {{p}^{2}}-{{q}^{2}}-3pq \right)\]

\[=\left( {{p}^{2}}+3pq-{{q}^{2}} \right)\left( {{p}^{2}}-3pq-{{q}^{2}} \right)\]

 

7. a2 + b2 – c2 – 2ab                     

উত্তর-

\[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-2ab\]

\[=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)-{{c}^{2}}\]

\[={{\left( a-b \right)}^{2}}-{{c}^{2}}\]

\[=\left( a-b+c \right)\left( a-b-c \right)\]

 

8. 3a(3a + 2c) – 4b(b + c)          

উত্তর-

\[3a\left( 3a+2c \right)-4b\left( b+c \right)\]

\[=9{{a}^{2}}+6ac-4{{b}^{2}}-4bc\]

\[={{\left( 3a \right)}^{2}}-{{\left( 2b \right)}^{2}}+6ac-4bc\]

\[=\left( 3a+2b \right)\left( 3a-2b \right)+2c\left( 3a-2b \right)\]

\[=\left( 3a-2b \right)\left( 3a+2b+2c \right)\]

 

9. a2 – 6ab + 12bc – 4c2

উত্তর-

\[{{a}^{2}}-6ab+12bc-4{{c}^{2}}\]

\[={{\left( a \right)}^{2}}-{{\left( 2c \right)}^{2}}-6ab+12bc\]

\[=\left( a+2c \right)\left( a-2c \right)-6b\left( a-2c \right)\]

\[=\left( a-2c \right)\left( a+2c-6b \right)\]

\[=\left( a-2c \right)\left( a-6b+2c \right)\]

 

10. 3a2 + 4ab + b2 – 2ac – 8c2     

উত্তর-

\[3{{a}^{2}}+4ab+{{b}^{2}}-2ac-{{c}^{2}}\]

\[=4{{a}^{2}}+4ab+{{b}^{2}}-{{a}^{2}}-2ac-{{c}^{2}}\]

\[={{\left( 2a \right)}^{2}}+2\times 2a\times b+{{\left( b \right)}^{2}}-\left( {{a}^{2}}+2ac+{{c}^{2}} \right)\]

\[={{\left( 2a+b \right)}^{2}}-{{\left( a+c \right)}^{2}}\]

\[=\left\{ \left( 2a+b \right)+\left( a+c \right) \right\}\left\{ \left( 2a+b \right)-\left( a+c \right) \right\}\]

\[=\left( 2a+b+a+c \right)\left( 2a+b-a-c \right)\]

\[=\left( 3a+b+c \right)\left( a+b-c \right)\]

 

11. x2 – y2 – 6ax + 2ay + a2        

উত্তর-

\[{{x}^{2}}-{{y}^{2}}-6ax+2ay+8{{a}^{2}}\]

\[={{x}^{2}}-6ax+9{{a}^{2}}-{{y}^{2}}+2ay-{{a}^{2}}\]

\[={{\left( x \right)}^{2}}-2\times x\times 3a+{{\left( 3a \right)}^{2}}-\left( {{y}^{2}}-2ay+{{a}^{2}} \right)\]

\[={{\left( x-3a \right)}^{2}}-{{\left( y-a \right)}^{2}}\]

\[=\left\{ \left( x-3a \right)+\left( y-a \right) \right\}\left\{ \left( x-3a \right)-\left( y-a \right) \right\}\]

\[=\left( x-3a+y-a \right)\left( x-3a-y+a \right)\]

\[=\left( x+y-4a \right)\left( x-y-2a \right)\]

 

12. a2 – 9b2 + 4c2 – 25d2 – 4ac + 30bd

উত্তর-

\[{{a}^{2}}-9{{b}^{2}}+4{{c}^{2}}-25{{d}^{2}}-4ac+30bd\]

\[={{a}^{2}}-4ac+4{{c}^{2}}-9{{b}^{2}}+30bd-25{{d}^{2}}\]

\[=\left\{ {{a}^{2}}-4ac+4{{c}^{2}} \right\}-\left\{ 9{{b}^{2}}-30bd+25{{d}^{2}} \right\}\]

\[=\left\{ {{a}^{2}}-2\times a\times 2c+{{\left( 2c \right)}^{2}} \right\}-\left\{ {{\left( 3b \right)}^{2}}-2\times 3b\times 5d+{{\left( 5d \right)}^{2}} \right\}\]

\[={{\left( a-2c \right)}^{2}}-{{\left( 3b-5d \right)}^{2}}\]

\[=\left\{ \left( a-2c \right)+\left( 3b-5d \right) \right\}\left\{ \left( a-2c \right)-\left( 3b-5d \right) \right\}\]

\[=\left( a-2c+3b-5d \right)\left( a-2c-3b+5d \right)\]

\[=\left( a+3b-2c-5d \right)\left( a-3b-2c+5d \right)\]

 

13. 3a2 – b2 – c2 + 2ab – 2bc + 2ca         

উত্তর-

\[3{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+2ab-2bc+2ca\]

\[=3{{a}^{2}}-ab-ac+3ab-{{b}^{2}}-bc+3ac-bc-{{c}^{2}}\]

\[=a\left( 3a-b-c \right)+b\left( 3a-b-c \right)+c\left( 3a-b-c \right)\]

\[=\left( 3a-b-c \right)\left( a+b+c \right)\]

 

14. x2 – 2x – 22499       

উত্তর-

\[{{x}^{2}}-2x-22499\]

\[={{x}^{2}}-\left( 151-149 \right)x-22499\]

\[={{x}^{2}}-151x+149x-22499\]

\[=x\left( x-151 \right)+149\left( x-151 \right)\]

\[=\left( x-151 \right)\left( x+149 \right)\]

 

15. (x2 – y2)(a2 – b2) + 4abxy

উত্তর-

\[\left( {{x}^{2}}-{{y}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)+4abxy\]

\[={{a}^{2}}{{x}^{2}}-{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{y}^{2}}+4abxy\]

\[={{a}^{2}}{{x}^{2}}+2abxy+{{b}^{2}}{{y}^{2}}-{{b}^{2}}{{x}^{2}}+2abxy-{{a}^{2}}{{y}^{2}}\]

\[={{\left( ax \right)}^{2}}+2\times ax\times by+{{\left( by \right)}^{2}}-\left\{ {{\left( bx \right)}^{2}}-2\times bx\times ay+{{\left( ay \right)}^{2}} \right\}\]

\[={{\left( ax+by \right)}^{2}}-{{\left( bx-ay \right)}^{2}}\]

\[=\left\{ \left( ax+by \right)+\left( bx-ay \right) \right\}\left\{ \left( ax+by \right)-\left( bx-ay \right) \right\}\]

\[=\left( ax+by+bx-ay \right)\left( ax+by-bx+ay \right)\]

\[=\left( ax+bx-ay+by \right)\left( ax-bx+ay+by \right)\]

কষে দেখি – 8.3

নীচের বীজগাণিতিক সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি –

 

1. t9 – 512                                      

উত্তর-

\[{{t}^{9}}-512\]

\[={{\left( {{t}^{3}} \right)}^{3}}-{{\left( 8 \right)}^{3}}\]

\[=\left( {{t}^{3}}-8 \right)\left\{ {{\left( {{t}^{3}} \right)}^{2}}+{{t}^{3}}\times 8+{{\left( 8 \right)}^{2}} \right\}\]

\[\left[ \because {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \right]\]

\[=\left\{ {{\left( t \right)}^{3}}-{{\left( 2 \right)}^{3}} \right\}\left( {{t}^{6}}+8{{t}^{3}}+64 \right)\]

\[=\left( t-2 \right)\left( {{t}^{2}}+2t+4 \right)\left( {{t}^{6}}+8{{t}^{3}}+64 \right)\]

 

2. 729p6 – q6                                 

উত্তর-

\[729{{p}^{6}}-{{q}^{6}}\]

\[={{\left( 27{{p}^{3}} \right)}^{2}}-{{\left( {{q}^{3}} \right)}^{2}}\]

\[=\left( 27{{p}^{3}}+{{q}^{3}} \right)\left( 27{{p}^{3}}-{{q}^{3}} \right)\]

\[=\left\{ {{\left( 3p \right)}^{3}}+{{\left( q \right)}^{3}} \right\}\left\{ {{\left( 3p \right)}^{3}}-{{\left( q \right)}^{3}} \right\}\]

\[=\left( 3p+q \right)\left\{ {{\left( 3p \right)}^{2}}-3p\times q+{{\left( q \right)}^{2}} \right\}\left( 3p-q \right)\left\{ {{\left( 3p \right)}^{2}}+3p\times q+{{\left( q \right)}^{2}} \right\}\]

\[=\left( 3p+q \right)\left( 9{{p}^{2}}-3pq+{{q}^{2}} \right)\left( 3p-q \right)\left( 9{{p}^{2}}+3pq+{{q}^{2}} \right)\]

\[=\left( 3p+q \right)\left( 3p-q \right)\left( 9{{p}^{2}}-3pq+{{q}^{2}} \right)\left( 9{{p}^{2}}+3pq+{{q}^{2}} \right)\]

 

3. 8(p – 3)3 + 343

উত্তর-

\[8{{\left( p-3 \right)}^{3}}+343\]

\[={{\left\{ 2\left( p-3 \right) \right\}}^{3}}+{{\left( 7 \right)}^{3}}\]

\[=\left\{ 2\left( p-3 \right)+7 \right\}\left[ {{\left\{ 2\left( p-3 \right) \right\}}^{2}}-2\left( p-3 \right)\times 7+{{\left( 7 \right)}^{2}} \right]\]

\[=\left( 2p-6+7 \right)\left\{ 4{{\left( p-3 \right)}^{2}}-14\left( p-3 \right)+49 \right\}\]

\[=\left( 2p+1 \right)\left\{ 4\left( {{p}^{2}}-6p+9 \right)-14p+42+49 \right\}\]

\[=\left( 2p+1 \right)\left( 4{{p}^{2}}-24p+36-14p+91 \right)\]

\[=\left( 2p+1 \right)\left( 4{{p}^{2}}-38p+127 \right)\]

 

4.      \frac1{8a^3}+\frac8{b^3}                       

উত্তর-

\[\frac{1}{8{{a}^{3}}}+\frac{8}{{{b}^{3}}}\]

\[={{\left( \frac{1}{2a} \right)}^{3}}+{{\left( \frac{2}{b} \right)}^{3}}\]

\[=\left( \frac{1}{2a}+\frac{2}{b} \right)\left\{ {{\left( \frac{1}{2a} \right)}^{2}}-\frac{1}{2a}\times \frac{2}{b}+{{\left( \frac{2}{b} \right)}^{2}} \right\}\]

\[=\left( \frac{1}{2a}+\frac{2}{b} \right)\left( \frac{1}{4{{a}^{2}}}-\frac{1}{ab}+\frac{4}{{{b}^{2}}} \right)\]

 

5. (2a3 – b3)3 – b9

উত্তর-

\[{{\left( 2{{a}^{3}}-{{b}^{3}} \right)}^{3}}-{{b}^{9}}\]

\[={{\left( 2{{a}^{3}}-{{b}^{3}} \right)}^{3}}-{{\left( {{b}^{3}} \right)}^{3}}\]

\[=\left( 2{{a}^{3}}-{{b}^{3}}-{{b}^{3}} \right)\left\{ {{\left( 2{{a}^{3}}-{{b}^{3}} \right)}^{2}}+\left( 2{{a}^{3}}-{{b}^{3}} \right){{b}^{3}}+{{\left( {{b}^{3}} \right)}^{2}} \right\}\]

\[=\left( 2{{a}^{3}}-2{{b}^{3}} \right)\left\{ 4{{a}^{6}}-4{{a}^{3}}{{b}^{3}}+{{b}^{6}}+2{{a}^{3}}{{b}^{3}}-{{b}^{6}}+{{b}^{6}} \right\}\]

\[=2\left( {{a}^{3}}-{{b}^{3}} \right)\left( 4{{a}^{6}}-2{{a}^{3}}{{b}^{3}}+{{b}^{6}} \right)\]

\[=2\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\left( 4{{a}^{6}}-2{{a}^{3}}{{b}^{3}}+{{b}^{6}} \right)\]

 

6. AR3 – Ar3 + AR2h – Ar2h        

উত্তর-

\[A{{R}^{3}}-A{{r}^{3}}+A{{R}^{2}}h-A{{r}^{2}}h\]

\[=A\left( {{R}^{3}}-{{r}^{3}} \right)+Ah\left( {{R}^{2}}-{{r}^{2}} \right)\]

\[=A\left( R-r \right)\left( {{R}^{2}}-Rr+{{r}^{2}} \right)+Ah\left( R+r \right)\left( R-r \right)\]

\[=A\left( R-r \right)\left\{ \left( {{R}^{2}}-Rr+{{r}^{2}} \right)+h\left( R+r \right) \right\}\]

\[=A\left( R-r \right)\left( {{R}^{2}}-Rr+{{r}^{2}}+Rh+rh \right)\]

 

7. a3 + 3a2b + 3ab2 + b3 – 8

উত্তর-

\[{{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8\]

\[={{\left( a+b \right)}^{3}}-{{\left( 2 \right)}^{3}}\]

\[=\left( a+b-2 \right)\left\{ {{\left( a+b \right)}^{2}}+\left( a+b \right)\times 2+{{\left( 2 \right)}^{2}} \right\}\]

\[=\left( a+b-2 \right)\left( {{a}^{2}}+2ab+{{b}^{2}}+2a+2b+4 \right)\]

 

8. 32x4 – 500x                              

উত্তর-

\[32{{x}^{4}}-500x\]

\[=4x\left( 8{{x}^{3}}-125 \right)\]

\[=4x\left\{ {{\left( 2x \right)}^{3}}-{{\left( 5 \right)}^{3}} \right\}\]

\[=4x\left( 2x-5 \right)\left\{ {{\left( 2x \right)}^{2}}+2x\times 5+{{\left( 5 \right)}^{2}} \right\}\]

\[=4x\left( 2x-5 \right)\left( 4{{x}^{2}}+10x+25 \right)\]

 

9. 8a3 – b3 – 4ax + 2bx               

উত্তর-

\[8{{a}^{3}}-{{b}^{3}}-4ax+2bx\]

\[={{\left( 2a \right)}^{3}}-{{\left( b \right)}^{3}}-2x\left( 2a-b \right)\]

\[=\left( 2a-b \right)\left\{ {{\left( 2a \right)}^{2}}+2a\times b+{{b}^{2}} \right\}-2x\left( 2a-b \right)\]

\[=\left( 2a-b \right)\left( 4{{a}^{2}}+2ab+{{b}^{2}}-2x \right)\]

 

10. x3 – 6x2 + 12x – 35

উত্তর-

মনেকরি, f\left( x \right)={{x}^{3}}-6{{x}^{2}}+12x-35

দেখছি, f\left( 5 \right)={{\left( 5 \right)}^{3}}-6{{\left( 5 \right)}^{2}}+12\left( 5 \right)-35=125-150+60-35=0

\left( x-5 \right),\,\,f\left( x \right) -এর একটি উৎপাদক।

\[\therefore \,\,{{x}^{3}}-6{{x}^{2}}+12x-35\]

\[={{x}^{3}}-5{{x}^{2}}-{{x}^{2}}+5x+7x-35\]

\[={{x}^{2}}\left( x-5 \right)-x\left( x-5 \right)+7\left( x-5 \right)\]

\[=\left( x-5 \right)\left( {{x}^{2}}-x+7 \right)\]

কষে দেখি – 8.4

নীচের বীজগাণিতিক সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি

 

1. x3 + y3 – 12xy + 64                                               

উত্তর-

\[{{x}^{3}}+{{y}^{3}}-12xy+64\]

\[={{x}^{3}}+{{y}^{3}}+{{\left( 4 \right)}^{3}}-3\times x\times y\times 4\]

\[=\left( x+y+4 \right)\left( {{x}^{2}}+{{y}^{2}}+{{4}^{2}}-xy-y\times 4-4\times x \right)\]

\[\left[ \because {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right) \right]\]

\[=\left( x+y+4 \right)\left( {{x}^{2}}+{{y}^{2}}+16-xy-4y-4x \right)\]

 

2. 8x3 – y3 + 1 + 6xy                                                 

উত্তর-

\[8{{x}^{3}}-{{y}^{3}}+1+6xy\]

\[={{\left( 2x \right)}^{3}}+{{\left( -y \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\times 2x\times \left( -y \right)\times 1\]

\[=\left\{ 2x+\left( -y \right)+1 \right\}\left\{ {{\left( 2x \right)}^{2}}+{{\left( -y \right)}^{2}}+{{\left( 1 \right)}^{2}}-\left( 2x \right)\left( -y \right)-\left( -y \right)\left( 1 \right)-\left( 1 \right)\left( 2x \right) \right\}\]

\[\left[ \because {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right) \right]\]

\[=\left( 2x-y+1 \right)\left( 4{{x}^{2}}+{{y}^{2}}+1+2xy+y-2x \right)\]

 

3. 8a3 – 27b3 – 1 – 18ab                                          

উত্তর-

\[8{{a}^{3}}-27{{b}^{3}}-1-18ab\]

\[={{\left( 2a \right)}^{3}}+{{\left( -3b \right)}^{3}}+{{\left( -1 \right)}^{3}}-3\left( 2a \right)\left( -3b \right)\left( -1 \right)\]

\[=\left\{ 2a+\left( -3b \right)+\left( -1 \right) \right\}\left\{ {{\left( 2a \right)}^{2}}+{{\left( -3b \right)}^{2}}+{{\left( -1 \right)}^{2}}-\left( 2a \right)\left( -3b \right)-\left( -3b \right)\left( -1 \right)-\left( -1 \right)\left( 2a \right) \right\}\]

\[\left[ \because {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right) \right]\]

\[=\left( 2a-3b-1 \right)\left( 4{{a}^{2}}+9{{b}^{2}}+1+6ab-3b+2a \right)\]

 

4. 1 + 8x3 + 18xy – 27y3

উত্তর-

\[1+8{{x}^{3}}+18xy-27{{y}^{3}}\]

\[={{\left( 1 \right)}^{3}}+{{\left( 2x \right)}^{3}}+{{\left( -3y \right)}^{3}}-3\left( 1 \right)\left( 2x \right)\left( -3y \right)\]

\[=\left\{ 1+2x+\left( -3y \right) \right\}\left\{ {{\left( 1 \right)}^{2}}+{{\left( 2x \right)}^{2}}+{{\left( -3y \right)}^{2}}-\left( 1 \right)\left( 2x \right)-\left( 2x \right)\left( -3y \right)-\left( -3y \right)\left( 1 \right) \right\}\]

\[=\left( 1+2x-3y \right)\left( 1+4{{x}^{2}}+9{{y}^{2}}-2x+6xy+3y \right)\]

 

5. (3a – 2b)3 + (2b – 5c)3 + (5c – 3a)3   

উত্তর-

আমরা জানি,

\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)+3abc\]

ধরি, x=\left( 3a-2b \right);\,\,y=\left( 2b-5c \right);\,\,z=\left( 5c-3a \right)

\[\therefore x+y+z=3a-2b+2b-5c+5c-3a=0\]

এখন,

\[{{\left( 3a-2b \right)}^{3}}+{{\left( 2b-5c \right)}^{3}}+{{\left( 5c-3a \right)}^{3}}\]

\[={{x}^{3}}+{{y}^{3}}+{{z}^{3}}\]

\[=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=\left( 0 \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=3xyz\]

\[=3\left( 3a-2b \right)\left( 2b-5c \right)\left( 5c-3a \right)\]

 

6. (2x – y)3 – (x + y)3 + (2y – x)3             

উত্তর-

ধরি, x=\left( 2x-y \right);\,\,y=-\left( x+y \right);\,\,z=\left( 2y-x \right)

\[\therefore x+y+z=2x-y-x-y+2y-x=0\]

এখন,

\[{{\left( 2x-y \right)}^{3}}-{{\left( x+y \right)}^{3}}+{{\left( 2y-x \right)}^{3}}\]

\[={{x}^{3}}+{{y}^{3}}+{{z}^{3}}\]

\[=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=\left( 0 \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=3xyz\]

\[=-3\left( 2x-y \right)\left( x+y \right)\left( 2y-x \right)\]

\[=3\left( 2x-y \right)\left( x+y \right)\left( x-2y \right)\]

 

7. a6 + 32a3 – 64            

উত্তর-

\[{{a}^{6}}+32{{a}^{3}}-64\]

\[={{a}^{6}}+8{{a}^{3}}-64+24{{a}^{3}}\]

\[={{\left( {{a}^{2}} \right)}^{3}}+{{\left( 2a \right)}^{3}}+{{\left( -4 \right)}^{3}}-3\left( {{a}^{2}} \right)\left( 2a \right)\left( -4 \right)\]

\[=\left( {{a}^{2}}+2a-4 \right)\left\{ {{\left( {{a}^{2}} \right)}^{2}}+{{\left( 2a \right)}^{2}}+{{\left( -4 \right)}^{2}}-\left( {{a}^{2}} \right)\left( 2a \right)-\left( 2a \right)\left( -4 \right)-\left( -4 \right)\left( {{a}^{2}} \right) \right\}\]

\[=\left( {{a}^{2}}+2a-4 \right)\left( {{a}^{4}}+4{{a}^{2}}+16-2{{a}^{3}}+8a+4{{a}^{2}} \right)\]

\[=\left( {{a}^{2}}+2a-4 \right)\left( {{a}^{4}}-2{{a}^{3}}+8{{a}^{2}}+8a+16 \right)\]

 

8. a6 – 18a3 + 125

উত্তর-

\[{{a}^{6}}-18{{a}^{3}}+125\]

\[={{a}^{6}}+27{{a}^{3}}+125-45{{a}^{3}}\]

\[={{\left( {{a}^{2}} \right)}^{3}}+{{\left( 3a \right)}^{3}}+{{\left( 5 \right)}^{3}}-3\left( {{a}^{2}} \right)\left( 3a \right)\left( 5 \right)\]

\[=\left( {{a}^{2}}+3a+5 \right)\left\{ {{\left( {{a}^{2}} \right)}^{2}}+{{\left( 3a \right)}^{2}}+{{\left( 5 \right)}^{2}}-\left( {{a}^{2}} \right)\left( 3a \right)-\left( 3a \right)\left( 5 \right)-\left( 5 \right)\left( {{a}^{2}} \right) \right\}\]

\[=\left( {{a}^{2}}+3a+5 \right)\left( {{a}^{4}}+9{{a}^{2}}+25-3{{a}^{3}}-15a-5{{a}^{2}} \right)\]

\[=\left( {{a}^{2}}+3a+5 \right)\left( {{a}^{4}}-3{{a}^{3}}+4{{a}^{2}}-15a+25 \right)\]

 

9. p3(q – r)3 + q3(r – p)3 + r3(p – q)3

উত্তর-

\[{{p}^{3}}{{\left( q-r \right)}^{3}}+{{q}^{3}}{{\left( r-p \right)}^{3}}+{{r}^{3}}{{\left( p-q \right)}^{3}}\]

\[={{\left\{ p\left( q-r \right) \right\}}^{3}}+{{\left\{ q\left( r-p \right) \right\}}^{3}}+{{\left\{ r\left( p-q \right) \right\}}^{3}}………\left( i \right)\]

ধরি, x=p\left( q-r \right);\,\,y=q\left( r-p \right);\,\,z=r\left( p-q \right)

\[\therefore \,\,x+y+z=p\left( q-r \right)+q\left( r-p \right)+r\left( p-q \right)\]

\[=pq-pr+qr-pq+pr-qr\]

\[=0\]

এখন, (i) নং থেকে পাই,

\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}\]

\[=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=\left( 0 \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)+3xyz\]

\[=3xyz\]

\[=3p\left( q-r \right)q\left( r-p \right)r\left( p-q \right)\]

\[=3pqr\left( q-r \right)\left( r-p \right)\left( p-q \right)\]

 

10. p^3+\frac1{p^3}+\frac{26}{27}

উত্তর-

\[{{p}^{3}}+\frac{1}{{{p}^{3}}}+\frac{26}{27}\]

\[={{p}^{3}}+\frac{1}{{{p}^{3}}}+1-\frac{1}{27}\]

\[={{\left( p \right)}^{3}}+{{\left( \frac{1}{p} \right)}^{3}}+{{\left( -\frac{1}{3} \right)}^{3}}-3\left( p \right)\left( \frac{1}{p} \right)\left( -\frac{1}{3} \right)\]

\[=\left( p+\frac{1}{p}-\frac{1}{3} \right)\left\{ {{\left( p \right)}^{2}}+{{\left( \frac{1}{p} \right)}^{2}}+{{\left( -\frac{1}{3} \right)}^{2}}-\left( p \right)\left( \frac{1}{p} \right)-\left( \frac{1}{p} \right)\left( -\frac{1}{3} \right)-\left( -\frac{1}{3} \right)\left( p \right) \right\}\]

\[=\left( p+\frac{1}{p}-\frac{1}{3} \right)\left( {{p}^{2}}+\frac{1}{{{p}^{2}}}+\frac{1}{9}-1+\frac{1}{3p}+\frac{p}{3} \right)\]

\[=\left( p+\frac{1}{p}-\frac{1}{3} \right)\left( {{p}^{2}}+\frac{1}{{{p}^{2}}}+\frac{1}{3p}+\frac{p}{3}-\frac{8}{9} \right)\]

কষে দেখি – 8.5

1. নীচের বীজগাণিতিক সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি

 

(i) (a +b)2 – 5a – 5b + 6                                          

উত্তর-

\[{{\left( a+b \right)}^{2}}-5a-5b+6\]

\[={{\left( a+b \right)}^{2}}-5\left( a+b \right)+6\]

ধরি, x=a+b

\[={{x}^{2}}-5x+6\]

\[={{x}^{2}}-\left( 3+2 \right)x+6\]

\[={{x}^{2}}-3x-2x+6\]

\[=x\left( x-3 \right)-2\left( x-3 \right)\]

\[=\left( x-3 \right)\left( x-2 \right)\]

\[=\left( a+b-3 \right)\left( a+b-2 \right)\,\,\left[ \because \,\,x=a+b \right]\]

 

(ii) (x + 1)(x + 2)(3x – 1)(3x – 4) + 12                

উত্তর-

\[\left( x+1 \right)\left( x+2 \right)\left( 3x-1 \right)\left( 3x-4 \right)+12\]

\[=\left\{ \left( x+1 \right)\left( 3x-1 \right) \right\}\left\{ \left( x+2 \right)\left( 3x-4 \right) \right\}+12\]

\[=\left\{ 3{{x}^{2}}-x+3x-1 \right\}\left\{ 3{{x}^{2}}-4x+6x-8 \right\}+12\]

\[=\left( 3{{x}^{2}}+2x-1 \right)\left( 3{{x}^{2}}+2x-8 \right)+12\]

ধরি, z=3{{x}^{2}}+2x

\[=\left( z-1 \right)\left( z-8 \right)+12\]

\[={{z}^{2}}-8z-z+8+12\]

\[={{z}^{2}}-9z+20\]

\[={{z}^{2}}-\left( 5+4 \right)z+20\]

\[={{z}^{2}}-5z-4z+20\]

\[=z\left( z-5 \right)-4\left( z-5 \right)\]

\[=\left( z-5 \right)\left( z-4 \right)\]

\[=\left( 3{{x}^{2}}+2x-5 \right)\left( 3{{x}^{2}}+2x-4 \right)\,\left[ \because z=3{{x}^{2}}+2x \right]\]

\[=\left\{ 3{{x}^{2}}+\left( 5-3 \right)x-5 \right\}\left( 3{{x}^{2}}+2x-4 \right)\]

\[=\left\{ 3{{x}^{2}}+5x-3x-5 \right\}\left( 3{{x}^{2}}+2x-4 \right)\]

\[=\left\{ x\left( 3x+5 \right)-1\left( 3x+5 \right) \right\}\left( 3{{x}^{2}}+2x-4 \right)\]

\[=\left( 3x+5 \right)\left( x-1 \right)\left( 3{{x}^{2}}+2x-4 \right)\]

 

(iii) x(x2 – 1)(x + 2) – 8                                          

উত্তর-

\[x\left( {{x}^{2}}-1 \right)\left( x+2 \right)-8\]

\[=x\left( x+1 \right)\left( x-1 \right)\left( x+2 \right)-8\]

\[=\left( {{x}^{2}}+x \right)\left( {{x}^{2}}+2x-x-2 \right)-8\]

\[=\left( {{x}^{2}}+x \right)\left( {{x}^{2}}+x-2 \right)-8\]

ধরি, z={{x}^{2}}+x

\[=z\left( z-2 \right)-8\]

\[={{z}^{2}}-2z-8\]

\[={{z}^{2}}-\left( 4-2 \right)z-8\]

\[={{z}^{2}}-4z+2z-8\]

\[=z\left( z-4 \right)+2\left( z-4 \right)\]

\[=\left( z-4 \right)\left( z-2 \right)\]

\[=\left( {{x}^{2}}+x-4 \right)\left( {{x}^{2}}+x-2 \right)\,\,\left[ \because \,z={{x}^{2}}+x \right]\]

 

(iv) 7(a2 + b2)2 – 15(a4 – b4) + 8(a2 – b2)2           

উত্তর-

\[7{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-15\left( {{a}^{4}}-{{b}^{4}} \right)+8{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}\]

\[=7{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-15\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)+8{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}\]

ধরি, x=\left( {{a}^{2}}+{{b}^{2}} \right);\,\,y=\left( {{a}^{2}}-{{b}^{2}} \right)

\[=7{{x}^{2}}-15xy+8{{y}^{2}}\]

\[=7{{x}^{2}}-\left( 7+8 \right)xy+8{{y}^{2}}\]

\[=7{{x}^{2}}-7xy-8xy+8{{y}^{2}}\]

\[=7x\left( x-y \right)-8y\left( x-y \right)\]

\[=\left( x-y \right)\left( 7x-8y \right)\]

\[=\left\{ \left( {{a}^{2}}+{{b}^{2}} \right)-\left( {{a}^{2}}-{{b}^{2}} \right) \right\}\left\{ 7\left( {{a}^{2}}+{{b}^{2}} \right)-8\left( {{a}^{2}}-{{b}^{2}} \right) \right\}\]

\[\left[ \because \,\,x=\left( {{a}^{2}}+{{b}^{2}} \right);\,\,y=\left( {{a}^{2}}-{{b}^{2}} \right) \right]\]

\[=\left\{ {{a}^{2}}+{{b}^{2}}-{{a}^{2}}+{{b}^{2}} \right\}\left\{ 7{{a}^{2}}+7{{b}^{2}}-8{{a}^{2}}+8{{b}^{2}} \right\}\]

\[=2{{b}^{2}}\left( 15{{b}^{2}}-{{a}^{2}} \right)\]

 

(v) (x2 – 1)2 + 8x(x2 + 1) + 19x2

উত্তর-

\[{{\left( {{x}^{2}}-1 \right)}^{2}}+8x\left( {{x}^{2}}+1 \right)+19{{x}^{2}}\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}-4{{x}^{2}}+8x\left( {{x}^{2}}+1 \right)+19{{x}^{2}}\]

\[\left[ \because {{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab \right]\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}+8x\left( {{x}^{2}}+1 \right)+15{{x}^{2}}\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}+\left( 5+3 \right)x\left( {{x}^{2}}+1 \right)+15{{x}^{2}}\]

\[={{\left( {{x}^{2}}+1 \right)}^{2}}+5x\left( {{x}^{2}}+1 \right)+3x\left( {{x}^{2}}+1 \right)+15{{x}^{2}}\]

\[=\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+1+5x \right)+3x\left( {{x}^{2}}+1+5x \right)\]

\[=\left( {{x}^{2}}+1+5x \right)\left( {{x}^{2}}+1+3x \right)\]

\[=\left( {{x}^{2}}+3x+1 \right)\left( {{x}^{2}}+5x+1 \right)\]

 

(vi) (a – 1)x2 – x – (a – 2)           

উত্তর-

\[\left( a-1 \right){{x}^{2}}-x-\left( a-2 \right)\]

\[=\left( a-1 \right){{x}^{2}}-\left\{ \left( a-1 \right)-\left( a-2 \right) \right\}x-\left( a-2 \right)\]

\[=\left( a-1 \right){{x}^{2}}-\left( a-1 \right)x+\left( a-2 \right)x-\left( a-2 \right)\]

\[=\left( a-1 \right)x\left\{ x-1 \right\}+\left( a-2 \right)\left\{ x-1 \right\}\]

\[=\left\{ x-1 \right\}\left\{ \left( a-1 \right)x+\left( a-2 \right) \right\}\]

\[=\left( x-1 \right)\left( ax-x+a-2 \right)\]

 

(vii) (a – 1)x2 + a2xy + (a + 1)y2             

উত্তর-

\[\left( a-1 \right){{x}^{2}}+{{a}^{2}}xy+\left( a+1 \right){{y}^{2}}\]

\[=\left( a-1 \right){{x}^{2}}+\left( {{a}^{2}}-1+1 \right)xy+\left( a+1 \right){{y}^{2}}\]

\[=\left( a-1 \right){{x}^{2}}+\left( {{a}^{2}}-1 \right)xy+xy+\left( a+1 \right){{y}^{2}}\]

\[=\left( a-1 \right){{x}^{2}}+\left( a+1 \right)\left( a-1 \right)xy+xy+\left( a+1 \right){{y}^{2}}\]

\[=\left( a-1 \right)x\left\{ x+\left( a+1 \right)y \right\}+y\left\{ x+\left( a+1 \right)y \right\}\]

\[=\left\{ x+\left( a+1 \right)y \right\}\left\{ \left( a-1 \right)x+y \right\}\]

\[=\left( x+ay+y \right)\left( ax-x+y \right)\]

 

(viii) x2 – qx – p2 + 5pq – 6q2

উত্তর-

\[{{x}^{2}}-qx-{{p}^{2}}+5pq-6{{q}^{2}}\]

\[={{x}^{2}}-qx-\left\{ {{p}^{2}}-5pq+6{{q}^{2}} \right\}\]

\[={{x}^{2}}-qx-\left\{ {{p}^{2}}-\left( 3+2 \right)pq+6{{q}^{2}} \right\}\]

\[={{x}^{2}}-qx-\left\{ {{p}^{2}}-3pq-2pq+6{{q}^{2}} \right\}\]

\[={{x}^{2}}-qx-\left\{ p\left( p-3q \right)-2q\left( p-3q \right) \right\}\]

\[={{x}^{2}}-qx-\left( p-3q \right)\left( p-2q \right)\]

\[={{x}^{2}}-\left\{ \left( p-2q \right)-\left( p-3q \right) \right\}x-\left( p-3q \right)\left( p-2q \right)\]

\[={{x}^{2}}-\left( p-2q \right)x+\left( p-3q \right)x-\left( p-3q \right)\left( p-2q \right)\]

\[=x\left\{ x-\left( p-2q \right) \right\}+\left( p-3q \right)\left\{ x-\left( p-2q \right) \right\}\]

\[=\left\{ x-\left( p-2q \right) \right\}\left\{ x+\left( p-3q \right) \right\}\]

\[=\left( x-p+2q \right)\left( x+p-3q \right)\]

 

(ix) 2\left(a^2+\frac1{a^2}\right)-\left(a-\frac1a\right)-7

উত্তর-

\[2\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)-\left( a-\frac{1}{a} \right)-7\]

\[=2\left\{ {{\left( a-\frac{1}{a} \right)}^{2}}-2\times a\times \frac{1}{a} \right\}-\left( a-\frac{1}{a} \right)-7\]

\[=2{{\left( a-\frac{1}{a} \right)}^{2}}-4-\left( a-\frac{1}{a} \right)-7\]

\[=2{{\left( a-\frac{1}{a} \right)}^{2}}-\left( a-\frac{1}{a} \right)-3\]

\[=2{{\left( a-\frac{1}{a} \right)}^{2}}-\left( 3-2 \right)\left( a-\frac{1}{a} \right)-3\]

\[=2{{\left( a-\frac{1}{a} \right)}^{2}}-3\left( a-\frac{1}{a} \right)+2\left( a-\frac{1}{a} \right)-3\]

\[=\left( a-\frac{1}{a} \right)\left\{ 2\left( a-\frac{1}{a} \right)-3 \right\}+1\left\{ 2\left( a-\frac{1}{a} \right)-3 \right\}\]

\[=\left\{ 2\left( a-\frac{1}{a} \right)-3 \right\}\left\{ \left( a-\frac{1}{a} \right)+1 \right\}\]

\[=\left( 2a-\frac{2}{a}-3 \right)\left( a-\frac{1}{a}+1 \right)\]

\[=\left( 2a-4+1-\frac{2}{a} \right)\left( a-\frac{1}{a}+1 \right)\]

\[=\left( 2a-4+\frac{a}{a}-\frac{2}{a} \right)\left( a-\frac{1}{a}+1 \right)\]

\[=\left\{ 2\left( a-2 \right)+\frac{1}{a}\left( a-2 \right) \right\}\left( a-\frac{1}{a}+1 \right)\]

\[=\left( a-2 \right)\left( 2+\frac{1}{a} \right)\left( a-\frac{1}{a}+1 \right)\]

 

(x) (x2 – x)y2 + y – (x2 + x)

উত্তর-

\[\left( {{x}^{2}}-x \right){{y}^{2}}+y-\left( {{x}^{2}}+x \right)\]

\[={{x}^{2}}{{y}^{2}}-x{{y}^{2}}+y-{{x}^{2}}-x\]

\[={{x}^{2}}{{y}^{2}}-{{x}^{2}}y-xy-x{{y}^{2}}+xy+y+{{x}^{2}}y-{{x}^{2}}-x\]

\[=xy\left( xy-x-1 \right)-y\left( xy-x-1 \right)+x\left( xy-x-1 \right)\]

\[=\left( xy-x-1 \right)\left( xy-y+x \right)\]

 

2. বহু বিকল্পীয় প্রশ্ন (M.C.Q):

(i) a2 – b2 = 11 × 9 এবং a ও b ধনাত্মক পূর্ণসংখ্যা (a>b) হলে,

(a) a=11, b=9  (b) a=33, b=3  (c) a=10, b=1  (d) a=100, b=1

উত্তর-

\[{{a}^{2}}-{{b}^{2}}=11\times 9\]

\[\Rightarrow \left( a+b \right)\left( a-b \right)=11\times 9\]

\[\therefore \left( a+b \right)=11;\,\,\left( a-b \right)=9\]

\[\Rightarrow 2a=20\Rightarrow a=10;\,\,b=11-10=1\]

সঠিক উত্তরটি হল –  (c) a=10, b=1 

(ii) যদি \frac ab+\frac ba=1 হয়, তাহলে a3 + b3 –এর মান

(a) 1  (b) a   (c) b   (d) 0 

উত্তর-

\[\frac{a}{b}+\frac{b}{a}=1\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=ab\]

\[\Rightarrow {{a}^{2}}-ab+{{b}^{2}}=0\]

\[\therefore \,\,{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)=0\]

সঠিক উত্তরটি হল –  (d) 0 

(iii) 253 – 753 + 503 + 3×25×75×50 –এর মান

(a) 150  (b) 0  (c) 25   (d) 50

উত্তর-

\[{{25}^{3}}-{{75}^{3}}+{{50}^{3}}+3\times 25\times 75\times 50\]

\[={{\left( 25 \right)}^{3}}+{{\left( -75 \right)}^{3}}+{{\left( 50 \right)}^{3}}-3\left( 25 \right)\left( -75 \right)\left( 50 \right)\]

\[=\left( 25-75+50 \right)\left\{ {{\left( 25 \right)}^{2}}+{{\left( -75 \right)}^{2}}+{{\left( 50 \right)}^{2}}-\left( 25 \right)\left( -75 \right)-\left( -75 \right)\left( 50 \right)-\left( 50 \right)\left( 25 \right) \right\}\]

\[=0\]

  ∴ সঠিক উত্তরটি হল –  (b) 0 

(iv) a + b +c = 0 হলে, -এর মান  \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}

(a) 0  (b) 1   (c) –1    (d) 3

উত্তর-

\[\frac{{{a}^{2}}}{bc}+\frac{{{b}^{2}}}{ca}+\frac{{{c}^{2}}}{ab}\]

\[=\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}\]

\[=\frac{\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)+3abc}{abc}\]

\[=\frac{0+3abc}{abc}\,\,\left[ \because \,\,a+b+c=0 \right]\]

\[=\frac{3abc}{abc}=3\]

সঠিক উত্তরটি হল –  (d) 3

(v) x2 – px + 12 = (x – 3)(x – a) একটি অভেদ হলে, a ও p এর মান যথাক্রমে

(a) a=4, p=7  (b) a=7, p=4  (c) a=4, p=–7  (d) a=–4, p=7

উত্তর-

\[{{x}^{2}}-px+12=\left( x-3 \right)\left( x-a \right)\]

\[\Rightarrow {{x}^{2}}-px+12={{x}^{2}}-\left( a+3 \right)x+3a\]

\[\therefore \,\,3a=12;\,\,a+3=p\]

\[\Rightarrow a=4;\,\,p=4+3=7\]

সঠিক উত্তরটি হল –  (a) a=4, p=7 

 

3. সংক্ষিপ্ত উত্তর ভিত্তিক প্রশ্ন

(i) \frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}-এর সরলতম মান লিখি।

উত্তর-

এখানে, \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{c}^{2}}-{{a}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)=0 এবং \left( b-c \right)+\left( c-a \right)+\left( a-b \right)=0

\[\because x+y+z=0\Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz\]

\[\therefore \frac{{{\left( {{b}^{2}}-{{c}^{2}} \right)}^{3}}+{{\left( {{c}^{2}}-{{a}^{2}} \right)}^{3}}+{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{3}}}{{{\left( b-c \right)}^{3}}+{{\left( c-a \right)}^{3}}+{{\left( a-b \right)}^{3}}}\]

\[=\frac{3\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)}{3\left( b-c \right)\left( c-a \right)\left( a-b \right)}\]

\[=\frac{\left( b+c \right)\left( b-c \right)\left( c+a \right)\left( c-a \right)\left( a+b \right)\left( a-b \right)}{\left( b-c \right)\left( c-a \right)\left( a-b \right)}\]

\[=\left( b+c \right)\left( c+a \right)\left( a+b \right)\]

(ii) a3 + b3 + c3 – 3abc = 0 এবং a + b +c ≠ 0 হলে, a, b ও c –এর মধ্যে সম্পর্ক লিখি।

উত্তর-

\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\]

\[\Rightarrow \left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)=0\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\,\,\left[ \because a+b+c\ne 0 \right]\]

\[\Rightarrow 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0\]

\[\Rightarrow \left( {{a}^{2}}-2ab+{{b}^{2}} \right)+\left( {{b}^{2}}-2bc+{{c}^{2}} \right)+\left( {{c}^{2}}-2ca+{{a}^{2}} \right)=0\]

\[\Rightarrow {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}}=0\]

\[\Rightarrow a-b=0;\,\,b-c=0;\,\,c-a=0\]

\[\therefore \,\,a=b=c\]

(iii) a2 – b2 = 224 এবং a ও b (a<b) ঋনাত্মক পূর্নসংখ্যা হলে, a ও b –এর মান লিখি।

উত্তর-

\[{{a}^{2}}-{{b}^{2}}=224\]

\[\Rightarrow {{a}^{2}}-{{b}^{2}}=225-1\]

\[\Rightarrow {{a}^{2}}-{{b}^{2}}={{\left( 15 \right)}^{2}}-{{\left( 1 \right)}^{2}}\]

\[{{a}^{2}}-{{b}^{2}}={{\left( -15 \right)}^{2}}-{{\left( 1 \right)}^{2}}\]

\[\therefore \,\,a=-15,\,\,b=1\]

(iv) 3x = a + b + c হলে, (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a)(x – b)(x – c) –এর মান কত লিখি।

উত্তর-

\[3x=a+b+c\]

\[\Rightarrow 3x-a-b-c=0\]

\[\Rightarrow \left( x-a \right)+\left( x-b \right)+\left( x-c \right)=0\]

\[\therefore {{\left( x-a \right)}^{3}}+{{\left( x-b \right)}^{3}}+{{\left( x-c \right)}^{3}}-3\left( x-a \right)\left( x-b \right)\left( x-c \right)=0\]

(v) 2x2 + px + 6 = (2x – a)(x – 2) একটি অভেদ হলে, a ও p –এর মান কত লিখি।

উত্তর-

\[2{{x}^{2}}+px+6=\left( 2x-a \right)\left( x-2 \right)\]

\[\Rightarrow 2{{x}^{2}}+px+6=2{{x}^{2}}-\left( 4+a \right)x+2a\]

\[\therefore \,2a=6;\,\,p=-\left( 4+a \right)\]

\[\Rightarrow a=3;\,\,p=-\left( 4+3 \right)=-7\]

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