Koshe dekhi 13.2 Class 8

(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

(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

(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