Wed. Jul 17th, 2024

Koshe dekhi 13.2 Class 8

Koshe dekhi 13.2 Class 8

1. উৎপাদকে বিশ্লেষণ করি-

(i) \( 2a^{2}+ 5a + 2 \)

সমাধানঃ 

\( \begin{array}{l}2{{a}^{2}}+5a+2\\=2{{a}^{2}}+\left( {4+1} \right)a+2\\=2{{a}^{2}}+4a+a+2\\=2a\left( {a+2} \right)+1\left( {a+2} \right)\\=\left( {a+2} \right)\left( {2a+1} \right)\end{array} \)

 

(ii) \(3x^{2}+ 14x + 8 \)

সমাধানঃ 

\( \begin{array}{l}3{{x}^{2}}+14x+8\\=3{{x}^{2}}+\left( {12+2} \right)x+8\\=3{{x}^{2}}+12x+2x+8\\=3x\left( {x+4} \right)+2\left( {x+4} \right)\\=\left( {x+4} \right)\left( {3x+2} \right)\end{array} \)

 

(iii) \(2m^{2} + 7m + 6 \)

সমাধানঃ 

\( \begin{array}{l}2{{m}^{2}}+7m+6\\=2{{m}^{2}}+\left( {4+3} \right)m+6\\=2{{m}^{2}}+4m+3m+6\\=2m\left( {m+2} \right)+3\left( {m+2} \right)\\=\left( {m+2} \right)\left( {2m+3} \right)\end{array} \)

 

(iv) \(6x^{2} − x − 15 \)

সমাধানঃ 

\( \begin{array}{l}6{{x}^{2}}-x-15\\=6{{x}^{2}}-\left( {10-9} \right)x-15\\=6{{x}^{2}}-10x+9x-15\\=2x\left( {3x-5} \right)+3\left( {3x-5} \right)\\=\left( {3x-5} \right)\left( {2x+3} \right)\end{array} \)

আরও দেখুনঃ 

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ  সমাধান। 

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ  সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান। 

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

(v) \(9r^{2} + r − 8 \)

সমাধানঃ 

\( \begin{array}{l}9{{r}^{2}}+r-8\\=9{{r}^{2}}+\left( {9-8} \right)r-8\\=9{{r}^{2}}+9r-8r-8\\=9r\left( {r+1} \right)-8\left( {r+1} \right)\\=\left( {r+1} \right)\left( {9r-8} \right)\end{array} \)

 

(vi) \(6m^{2}− 11mn − 10n^{2} \)

সমাধানঃ 

\( \begin{array}{l}6{{m}^{2}}-11mn-10{{n}^{2}}\\=6{{m}^{2}}-\left( {15-4} \right)mn-10{{n}^{2}}\\=6{{m}^{2}}-15mn+4mn-10{{n}^{2}}\\=3m\left( {2m-5n} \right)+2n\left( {2m-5n} \right)\\=\left( {2m-5n} \right)\left( {3m+2n} \right)\end{array} \)

 

(vii) \(7x^{2}+ 48xy − 7y^{2} \)

সমাধানঃ 

\( \begin{array}{l}7{{x}^{2}}+48xy-7{{y}^{2}}\\=7{{x}^{2}}-\left( {49-1} \right)xy-7{{y}^{2}}\\=7{{x}^{2}}-49xy+xy-7{{y}^{2}}\\=7x\left( {x-7y} \right)+y\left( {x-7y} \right)\\=\left( {x-7y} \right)\left( {7x+y} \right)\end{array} \)

 

(viii) \( 12 + x − 6x^{2} \)

সমাধানঃ 

\( \begin{array}{l}12+x-6{{x}^{2}}\\=12+\left( {9-8} \right)x-6{{x}^{2}}\\=12+9x-8x-6{{x}^{2}}\\=3\left( {4+3x} \right)-2x\left( {4+3x} \right)\\=\left( {4+3x} \right)\left( {3-2x} \right)\end{array} \)

 

(ix) \( 6 + 5a − 6a^{2} \)

সমাধানঃ 

\( \begin{array}{l}6+5a-6{{a}^{2}}\\=6+\left( {9-4} \right)a-6{{a}^{2}}\\=6+9a-4a-6{{a}^{2}}\\=3\left( {2+3a} \right)-2a\left( {2+3a} \right)\\=\left( {2+3a} \right)\left( {3-2a} \right)\end{array} \)

 

(x) \( 6x^{2} − 13x + 6 \)

সমাধানঃ 

\( \begin{array}{l}6{{x}^{2}}-13x+6\\=6{{x}^{2}}-\left( {9+4} \right)x+6\\=6{{x}^{2}}-9x-4x+6\\=3x\left( {2x-3} \right)-2\left( {2x-3} \right)\\=\left( {2x-3} \right)\left( {3x-2} \right)\end{array} \)

 

(xi) \( 99a^{2} − 202ab + 99b^{2} \)

সমাধানঃ 

\( \begin{array}{l}99{{a}^{2}}-202ab+99{{b}^{2}}\\=99{{a}^{2}}-\left( {121+81} \right)ab+99{{b}^{2}}\\=99{{a}^{2}}-121ab-81ab+99{{b}^{2}}\\=11a\left( {9a-11b} \right)-9b\left( {9a-11b} \right)\\=\left( {9a-11b} \right)\left( {11a-9b} \right)\end{array} \)

 

(xii) \( 2a^{4} − 13a^{2} − 24 \)

সমাধানঃ 

\( \begin{array}{l}2{{a}^{4}}-13{{a}^{2}}-24\\=2{{a}^{4}}-\left( {16-3} \right){{a}^{2}}-24\\=2{{a}^{4}}-16{{a}^{2}}+3{{a}^{2}}-24\\=2{{a}^{2}}\left( {{{a}^{2}}-8} \right)+3\left( {{{a}^{2}}-8} \right)\\=\left( {{{a}^{2}}-8} \right)\left( {2{{a}^{2}}+3} \right)\end{array} \)

 

(xiii) \(8a^{4}+ 2a^{2}− 45 \)

সমাধানঃ 

\( \begin{array}{l}8{{a}^{4}}+2{{a}^{2}}-45\\=8{{a}^{4}}+\left( {20-18} \right){{a}^{2}}-45\\=8{{a}^{4}}+20{{a}^{2}}-18{{a}^{2}}-45\\=4{{a}^{2}}\left( {2{{a}^{2}}+5} \right)-9\left( {2{{a}^{2}}+5} \right)\\=\left( {2{{a}^{2}}+5} \right)\left( {4{{a}^{2}}-9} \right)\end{array} \)

আরও দেখুনঃ 

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ  সমাধান। 

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ  সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান। 

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

(xiv) \(6 (x − y)^{2} −  x + y − 15 \)

সমাধানঃ 

\(6 (x − y)^{2} −  x + y − 15 \)
\(=6{{\left( {x-y} \right)}^{2}}-\left( {x-y} \right)-15 \)

ধরি, \( \left( x-y \right) = a \)

\( \begin{array}{l}=6{{a}^{2}}-a-15\\=6{{a}^{2}}-\left( {10-9} \right)a-15\\=6{{a}^{2}}-10a+9a-15\\=2a\left( {3a-5} \right)+3\left( {3a-5} \right)\\=\left( {3a-5} \right)\left( {2a+3} \right)\end{array} \)

পুনরায় \( a\) এর পরিবর্তে \(\left(x -y \right)  \) বসিয়ে পাই –

\( \begin{array}{l}=\left\{ {3\left( {x-y} \right)-5} \right\}\left\{ {2\left( {x-y} \right)+3} \right\}\\=\left( {3x-3y-5} \right)\left( {2x-2y+3} \right)\end{array} \)

 

(xv) \(3 (a + b)^{2} − 2a − 2b − 8 \)

সমাধানঃ 

\(3 (a + b)^{2} − 2a − 2b − 8 \)
\( =3 (a + b)^{2} − 2\left (a + b \right ) − 8 \)

ধরি, \( \left( a+b \right) = x \)

\( \begin{array}{l}=3{{x}^{2}}-2x-8\\=3{{x}^{2}}-\left( {6-4} \right)x-8\\=3{{x}^{2}}-6x+4x-8\\=3x\left( {x-2} \right)+4\left( {x-2} \right)\\=\left( {x-2} \right)\left( {3x+4} \right)\end{array} \)

পুনরায় \( x\) এর পরিবর্তে \(\left(a +b \right)  \) বসিয়ে পাই –

\( \begin{array}{l}=\left( {a+b-2} \right)\left\{ {3\left( {a+b} \right)+4} \right\}\\=\left( {a+b-2} \right)\left( {3a+3b+4} \right)\end{array} \)

 

(xvi) \(6 (a+b)^{2} + 5(a^{2} – b^{2}) − 6 (a − b)^{2} \)

সমাধানঃ 

\(6 (a+b)^{2} + 5(a^{2} – b^{2}) − 6 (a − b)^{2} \)
\( = 6 (a+b)^{2} + 5(a+b)(a-b) − 6 (a − b)^{2} \)

ধরি, \( \left( a+b \right) = x \) এবং  \( \left( a-b \right) = y \)

\( \begin{array}{l}=6{{x}^{2}}+5xy-6{{y}^{2}}\\=6{{x}^{2}}+\left( {9-4} \right)xy-6{{y}^{2}}\\=6{{x}^{2}}+9xy-4xy-6{{y}^{2}}\\=3x\left( {2x+3y} \right)-2y\left( {2x+3y} \right)\\=\left( {2x+3y} \right)\left( {3x-2y} \right)\end{array} \)

পুনরায় \( x\) এর পরিবর্তে \(\left(a +b \right)  \) এবং  \( y\) এর পরিবর্তে \(\left(a-b \right)  \) বসিয়ে পাই –

\( \begin{array}{l}=\left\{ {2\left( {a+b} \right)+3\left( {a-b} \right)} \right\}\left\{ {3\left( {a+b} \right)-2\left( {a-b} \right)} \right\}\\=\left( {2a+2b+3a-3b} \right)\left( {3a+3b-2a+2b} \right)\\=\left( {5a-b} \right)\left( {a+5b} \right)\end{array} \)

 

2. নীচের বীজগাণিতিক সংখ্যামালাগুলি দুটি বর্গের অন্তররূপে প্রকাশ করে উৎপাদকে বিশ্লেষণ করি—
(i) \(x^{2}− 2x − 3 \)

সমাধানঃ 

\( \begin{array}{l}{{x}^{2}}-2x-3\\=\left[ {{{{\left( x \right)}}^{2}}-2.\left( x \right).\left( 1 \right)+{{{\left( 1 \right)}}^{2}}} \right]-3-{{\left( 1 \right)}^{2}}\end{array} \)

[যেহেতু,  \( a^2 -2ab +b^2 = (a-b)^2 \) ]

\( \begin{array}{l}={{\left( {x-1} \right)}^{2}}-3-1\\={{\left( {x-1} \right)}^{2}}-4\\={{\left( {x-1} \right)}^{2}}-{{\left( 2 \right)}^{2}}\end{array} \)

[যেহেতু,  \( (a^2 -b^2) = (a+b)(a-b) \) ]

\( \begin{array}{l}=\left\{ {\left( {x-1} \right)+2} \right\}\left\{ {\left( {x-1} \right)-2} \right\}\\=\left( {x-1+2} \right)\left( {x-1-2} \right)\\=\left( {x+1} \right)\left( {x-3} \right)\end{array} \)

আরও দেখুনঃ 

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ  সমাধান। 

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ  সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান। 

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

(ii) \(x^{2}+ 5x + 6 \)

সমাধানঃ 

\( \begin{array}{l}{{x}^{2}}+5x+6\\=\left[ {{{{\left( x \right)}}^{2}}+2.\left( x \right).\left( {\frac{5}{2}} \right)+{{{\left( {\frac{5}{2}} \right)}}^{2}}} \right]+6-{{\left( {\frac{5}{2}} \right)}^{2}}\end{array} \)

[যেহেতু,  \( a^2 +2ab +b^2 = (a+b)^2 \) ]

\( \begin{array}{l}={{\left( {x+\frac{5}{2}} \right)}^{2}}+6-\frac{{25}}{4}\\={{\left( {x+\frac{5}{2}} \right)}^{2}}+\frac{{24-25}}{4}\\={{\left( {x+\frac{5}{2}} \right)}^{2}}-\frac{1}{4}\\={{\left( {x+\frac{5}{2}} \right)}^{2}}-{{\left( {\frac{1}{2}} \right)}^{2}}\end{array} \)

[যেহেতু,  \( (a^2 -b^2) = (a+b)(a-b) \) ]

\( \begin{array}{l}=\left\{ {\left( {x+\frac{5}{2}} \right)+\frac{1}{2}} \right\}\left\{ {\left( {x+\frac{5}{2}} \right)-\frac{1}{2}} \right\}\\=\left( {x+\frac{5}{2}+\frac{1}{2}} \right)\left( {x+\frac{5}{2}-\frac{1}{2}} \right)\\=\left( {x+\frac{{5+1}}{2}} \right)\left( {x+\frac{{5-1}}{2}} \right)\\=\left( {x+\frac{6}{2}} \right)\left( {x+\frac{4}{2}} \right)\\=\left( {x+3} \right)\left( {x+2} \right)\end{array} \)

 

(iii) \(3x^{2}− 7x − 6 \)

সমাধানঃ 

\( \begin{array}{l}3{{x}^{2}}-7x-6\\=3\left( {{{x}^{2}}-\frac{7}{3}x-2} \right)\\=3\left[ {\left\{ {{{{\left( x \right)}}^{2}}-2.\left( x \right).\left( {\frac{7}{6}} \right)+{{{\left( {\frac{7}{6}} \right)}}^{2}}} \right\}-2-{{{\left( {\frac{7}{6}} \right)}}^{2}}} \right]\end{array} \)

[যেহেতু,  \( a^2 -2ab +b^2 = (a-b)^2 \) ]

\( \begin{array}{l}=3\left[ {{{{\left( {x-\frac{7}{6}} \right)}}^{2}}-\left( {2+\frac{{49}}{{36}}} \right)} \right]\\=3\left[ {{{{\left( {x-\frac{7}{6}} \right)}}^{2}}-\left( {\frac{{72+49}}{{36}}} \right)} \right]\\=3\left[ {{{{\left( {x-\frac{7}{6}} \right)}}^{2}}-\left( {\frac{{121}}{{36}}} \right)} \right]\\=3\left[ {{{{\left( {x-\frac{7}{6}} \right)}}^{2}}-{{{\left( {\frac{{11}}{6}} \right)}}^{2}}} \right]\end{array} \)

[যেহেতু,  \( (a^2 -b^2) = (a+b)(a-b) \) ]

\( \begin{array}{l}=3\left\{ {\left( {x-\frac{7}{6}} \right)+\frac{{11}}{6}} \right\}\left\{ {\left( {x-\frac{7}{6}} \right)-\frac{{11}}{6}} \right\}\\=3\left( {x-\frac{7}{6}+\frac{{11}}{6}} \right)\left( {x-\frac{7}{6}-\frac{{11}}{6}} \right)\\=3\left\{ {x-\left( {\frac{7}{6}-\frac{{11}}{6}} \right)} \right\}\left\{ {x-\left( {\frac{7}{6}+\frac{{11}}{6}} \right)} \right\}\\=3\left( {x-\frac{{7-11}}{6}} \right)\left( {x-\frac{{7+11}}{6}} \right)\\=3\left( {x+\frac{4}{6}} \right)\left( {x-\frac{{18}}{6}} \right)\\=3\left( {x+\frac{2}{3}} \right)\left( {x-3} \right)\\=\left( {3x+2} \right)\left( {x-3} \right)\end{array} \)

 

(iv) \(3a^{2}−2a − 5 \)

সমাধানঃ 

\( \begin{array}{l}3{{a}^{2}}-2a-5\\=3\left( {{{a}^{2}}-\frac{2}{3}a-\frac{5}{3}} \right)\\=3\left[ {\left\{ {{{{\left( a \right)}}^{2}}-2.\left( a \right).\left( {\frac{1}{3}} \right)+{{{\left( {\frac{1}{3}} \right)}}^{2}}} \right\}-\frac{5}{3}-{{{\left( {\frac{1}{3}} \right)}}^{2}}} \right]\end{array} \)

[যেহেতু,  \( a^2 -2ab +b^2 = (a-b)^2 \) ]

\( \begin{array}{l}=3\left[ {{{{\left( {a-\frac{1}{3}} \right)}}^{2}}-\left( {\frac{5}{3}+\frac{1}{9}} \right)} \right]\\=3\left[ {{{{\left( {a-\frac{1}{3}} \right)}}^{2}}-\left( {\frac{{15+1}}{9}} \right)} \right]\\=3\left[ {{{{\left( {a-\frac{1}{3}} \right)}}^{2}}-\left( {\frac{{16}}{9}} \right)} \right]\\=3\left[ {{{{\left( {a-\frac{1}{3}} \right)}}^{2}}-{{{\left( {\frac{4}{3}} \right)}}^{2}}} \right]\end{array} \)

[যেহেতু,  \( (a^2 -b^2) = (a+b)(a-b) \) ]

\( \begin{array}{l}=3\left\{ {\left( {a-\frac{1}{3}} \right)+\frac{4}{3}} \right\}\left\{ {\left( {a-\frac{1}{3}} \right)-\frac{4}{3}} \right\}\\=3\left( {a-\frac{1}{3}+\frac{4}{3}} \right)\left( {a-\frac{1}{3}-\frac{4}{3}} \right)\\=3\left\{ {a-\left( {\frac{1}{3}-\frac{4}{3}} \right)} \right\}\left\{ {a-\left( {\frac{1}{3}+\frac{4}{3}} \right)} \right\}\\=3\left( {a-\frac{{1-4}}{3}} \right)\left( {a-\frac{{1+4}}{3}} \right)\\=3\left( {a+\frac{3}{3}} \right)\left( {a-\frac{5}{3}} \right)\\=3\left( {a+1} \right)\left( {a-\frac{5}{3}} \right)\\=\left( {a+1} \right)\left( {3a-5} \right)\end{array} \)

আরও দেখুনঃ 

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ  সমাধান। 

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ  সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান। 

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

3. উৎপাদকে বিশ্লেষণ করি—

(i) \( ax^{2} + (a^{2}+ 1) x + a \)

সমাধানঃ 

\( \begin{array}{l}a x^{2}+\left(a^{2}+1\right) x+a \\ =a x^{2}+a^{2} x+x+a \\ =a x(x+a)+1(x+a) \\ =(x+a)(a x+1)\end{array} \)

 

(ii) \(x^{2}+ 2ax + (a + b)(a − b) \)

সমাধানঃ 

\( \begin{array}{l}x^{2}+2 a x+(a+b)(a-b) \\ =x^{2}+\{(a+b)+(a-b)\} x+(a+b)(a-b) \\ =x^{2}+(a+b) x+(a-b) x+(a+b)(a-b) \\ =x(x+a+b)+(a-b)(x+a+b) \\ =(x+a+b)(x+a-b)\end{array} \)

 

(iii) \(ax^{2} − (a^{2}+ 1)x + a \)

সমাধানঃ 

\( \begin{array}{l}a x^{2}-\left(a^{2}+1\right) x+a \\ =a x^{2}-a^{2} x-x+a \\ =a x(x-a)-1(x-a) \\ =(x-a)(a x-1)\end{array} \)

 

(iv) \(ax^{2} + (a^{2} − 1) x − a \)

সমাধানঃ 

\( \begin{array}{l}a x^{2}+\left(a^{2}-1\right) x-a \\ =a x 2+a^{2} x-x-a \\ =a x(x+a)-1(x+a) \\ =(x+a)(a x-1) \\ =(x+a)(a x-1)\end{array} \)

 

(v) \(ax^{2} − (a^{2} − 2) x − 2a \)

সমাধানঃ 

\( \begin{array}{l}a x^{2}-\left(a^{2}-2\right) x-2 a \\ =a x^{2}-a^{2} x+2 x-2 a \\ =a x(x-a)+2(x-a) \\ =(x-a)(a x+2)\end{array} \)

 

(vi) \( a^{2}+1- \frac{6}{a^{2}} \)

সমাধানঃ 

\( \begin{array}{l}{{a}^{2}}+1-\frac{6}{{{{a}^{2}}}}\\=\frac{{{{a}^{4}}+{{a}^{2}}-6}}{{{{a}^{2}}}}\\=\frac{{{{a}^{4}}+\left( {3-2} \right){{a}^{2}}-6}}{{{{a}^{2}}}}\\=\frac{{{{a}^{4}}+3{{a}^{2}}-2{{a}^{2}}-6}}{{{{a}^{2}}}}\\=\frac{{{{a}^{2}}\left( {{{a}^{2}}+3} \right)-2\left( {{{a}^{2}}+3} \right)}}{{{{a}^{2}}}}\\=\frac{{\left( {{{a}^{2}}+3} \right)\left( {{{a}^{2}}-2} \right)}}{{{{a}^{2}}}}\end{array} \)

 

Koshe Dekhi 13.2 Class 8

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আরও দেখুনঃ 

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ  সমাধান। 

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ  সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান। 

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান। 

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

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