প্রশ্ন 1. p(x) বহুপদটোক g(x) বহুপদটোৰে হৰণ কৰা আৰু প্ৰতিটোৰে ক্ষেত্ৰত ভাগফল আৰু ভাগশেষ নির্ণয় কৰা।
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
উত্তৰঃ দিয়া আছে, p(x) p(x) = x³ – 3x² + 5x – 3 আৰু g(x) = x² – 2
\[ \begin{array}{ll} & \quad\;\, x – 3 \\[0.3em] x^2 – 2 & \overline{\smash{\big)}x^3 – 3x^2 + 5x – 3} \\[0.3em] & \underline{x^3 \quad\quad\;\, – 2x \quad\quad\;\;} \quad\quad \text{(Subtract)} \\[0.3em] & \quad\;\, -3x^2 + 7x – 3 \\[0.3em] & \quad\;\, \underline{-3x^2 \quad\quad\;\, + 6} \quad\; \text{(Subtract)} \\[0.3em] & \quad\quad\quad\quad\;\, 7x – 9 \\ \end{array} \]
(ii) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
উত্তৰঃ দিয়া আছে p(x) = X⁴ – 3x² + 4x + 5
= x⁴ + 0.x² – 3x² + 4x + 5
আৰু g(x) = x² + 1 – x
= x² – x + 1
\[ \begin{array}{ll} & \quad\;\, x^2 + x – 3 \\[0.3em] x^2 – x + 1 & \overline{\smash{\big)}x^4 + 0x^3 – 3x^2 + 4x + 5} \\[0.3em] & \underline{x^4 – x^3 + x^2 \quad\quad\quad\;\;} \quad\quad\quad \text{(Subtract)} \\[0.3em] & \quad\;\, x^3 – 4x^2 + 4x + 5 \\[0.3em] & \quad\;\, \underline{x^3 – x^2 + x \quad\quad\;\;} \quad\quad \text{(Subtract)} \\[0.3em] & \quad\quad\; -3x^2 + 3x + 5 \\[0.3em] & \quad\quad\; \underline{-3x^2 + 3x – 3} \quad\; \text{(Subtract)} \\[0.3em] & \quad\quad\quad\quad\quad\quad\; 8 \\ \end{array} \]
(iii) p(x) = x⁴ – 5x + 6 g(x) = 2 – x²
\[ \begin{array}{ll} & \quad\;\;\; -x^2 \;\;\,-2 \\[0.3em]- x^2 +2 & \overline{\smash{\big)}\; x^4 + 0x^3 + 0x^2 – 5x + 6} \\[0.3em] & \underline{\;x^4 \quad\;\;\;\;\,\phantom{0x^3} – 2x^2 \quad\;\;\;\;\;\;\;} \quad \text{(Subtract)} \\[0.3em] & \;\;\;\;\,\phantom{x^4} 0x^3 + 2x^2 – 5x + 6 \\[0.3em] & \phantom{x^4} \underline{\;\,0x^3 \quad\;\;\;\;\;\;\, + 2x^2 \quad\;\;\;\; – 4} \quad \text{(Subtract)} \\[0.3em] & \phantom{x^4 + 0x^3} \;\, -5x + 10 \\ \end{array} \]
(IV) \(P(x)= 2x^4 + 3x^3 – 2x^2 – 9x -12 ; g(x)=x^2 – 3\)
\[ \begin{array}{ll} & \quad\;\;\; 2x^2 +3x +4 \\[0.3em] x^2 – 3 & \overline{\smash{\big)}\; 2x^4 + 3x^3 -2x^2 – 9x -12} \\[0.3em] & \underline{\;2x^4 \quad\;\;\;\;\,\phantom{0x^3} – 6x^2 \quad\;\;\;\;\;\;\;} \quad \text{(Subtract)} \\[0.3em] & \quad\;\;\;\;\,\phantom{2x^4} 3x^3 + 4x^2 -9x – 12 \\[0.3em]& \phantom{x^4} \quad\;\;\;\,\underline{\;\;\,3x^3 \quad\;\;\;\;\;\,-9x\quad\;\;\;\;\,} \quad \;\;\;\text{(Subtract)} \\[0.3em] & \quad\;\;\;\;\,\phantom{3x^3} \quad\;\;\;\;\;\, 4x^2 \quad\;\;\; -12 \\ & \phantom{x^4+3x^3} \quad\;\;\;\,\underline{\;\,4x^2 \quad\;\;\;\;\,-12\quad\;\;\;\;\,} \quad \;\;\;\text{(Subtract)}\\[0.3em] & \phantom{4x^2 -12} \;\quad\;\;\;\;\; 0\\\end{array} \]
(V) \(P(x)= x^6 + 3x^2 +10 ; g(x)=x^3 + 1\)
Solution:- \[ \begin{array}{ll} & \quad\;\;\; x^3 – 1 \\[0.3em] x^3 + 1 & \overline{\smash{\big)}\; x^6 + 3x^2 +10} \\[0.3em] & \underline{\;x^6 +\; x^3 \quad\;\;\;\;\;\;\phantom{10}} \quad \text{(Subtract)} \\[0.3em] & \quad\;\;\;\;\,\ -x^3 + 3x^2 +10 \\[0.3em]& \phantom{x^4} \quad\;\;\;\,\underline{\,-x^3 \quad\;\;\;\;\;\;\, -1\quad\;\;\;\;\,} \quad \;\;\;\text{(Subtract)} \\[0.3em] & \phantom{-x^3} \;\quad\;\;\;\;\;\;\;\;\; 3x^2 + 11\\\end{array} \]
( vi) \(P(x)= 2x^5 – 5x^4 – 7x^3 – 4x^2 -10x + 11 ; g(x)=x^3 + 2x^3+2\)
\[ \begin{array}{ll} & \quad\;\;\; 2x^2 – 5x + 7 \\[0.3em] x^3 + 2x^3+2 & \overline{\smash{\big)}\; 2x^5 – 5x^4 – 7x^3 – 4x^2 -10x + 11} \\[0.3em] & \underline{\;2x^5 \phantom{-5x^4-7x^3} – 4x^2 \quad\;\;\;\;\;\;\;} \quad \text{(Subtract)} \\[0.3em] & \quad\;\;\phantom{2x^4} 5x^4 + 7x^3\phantom{x^2} -10x +11 \\[0.3em]&\quad\;\;\;\; \underline{\;\;\,5x^4 \quad\;\;\;\;\;\;\;\;\;\,-10x\quad\;\;\;\;\,} \quad \;\;\;\text{(Subtract)} \\[0.3em] & \quad\;\;\;\;\,\phantom{5x^4}\quad\;\;7x^3 \quad\;\;\;\;\;\;\;\; +11 \\ & \phantom{x^4+3x^3} \quad\;\;\;\,\underline{\;\,7x^3 \quad\;\;\;\;\,+14\quad\;\;\;\;\,} \quad \;\;\;\text{(Subtract)}\\[0.3em] & \phantom{4x^2 -12} \;\quad\;\;\;\;\phantom{7x^3} -3\\\end{array} \]
প্রশ্ন 2. দ্বিতীয় বহুপদটোক প্ৰথম বহুপদেৰে হৰণ কৰি প্ৰথম বহুপদটো দ্বিতীয় বহুপদটোৰ এটা উৎপাদক হয়নে নহয় পৰীক্ষা কৰাঃ
(i) t² – 3, 2t⁴ + 3t² – 2t² – 9t – 12
Solutrion:-
\[ \begin{array}{ll} & \quad\;\; 2t^2 + 3t + 4 \\[0.3em] t^2 – 3 & \overline{\smash{\big)}\; 2t^4 + 3t^3 – 2t^2 – 9t – 12} \\[0.3em] & \underline{2t^4 \quad\quad\;\;\;\, – 6t^2 \quad\quad\quad\quad\;\;} \quad \text{(Subtract)} \\[0.3em] & \phantom{2t^4} \; 3t^3 + 4t^2 – 9t – 12 \\[0.3em] & \phantom{2t^4} \; \underline{3t^3 \quad\quad\;\;\, – 9t \quad\quad\quad\;} \quad \text{(Subtract)} \\[0.3em] & \phantom{2t^4 + 3t^3} \; 4t^2 \quad\quad\;\;\, – 12 \\[0.3em] & \phantom{2t^4 + 3t^3} \; \underline{4t^2 \quad\quad\;\;\, – 12} \quad \text{(Subtract)} \\[0.3em] & \phantom{2t^4 + 3t^3 + 4t^2} \; 0 \\ \end{array} \]
(ii) x² + 3x + 1,3x⁴ + 5x³ – 7x² + 2x + 2
Solution:-
\[ \begin{array}{ll} & \quad\;\; 3x^2 – 4x + 2 \\[0.3em] x^2 + 3x + 1 & \overline{\smash{\big)}\; 3x^4 + 5x^3 – 7x^2 + 2x + 2} \\[0.3em] & \underline{3x^4 + 9x^3 + 3x^2 \quad\quad\quad\quad\quad\quad} \quad \text{(Subtract)} \\[0.3em] & \phantom{3x^4} -4x^3 – 10x^2 + 2x + 2 \\[0.3em] & \phantom{3x^4} \underline{-4x^3 – 12x^2 – 4x \quad\quad\quad\;\;} \quad \text{(Subtract)} \\[0.3em] & \phantom{3x^4 – 4x^3} \; 2x^2 + 6x + 2 \\[0.3em] & \phantom{3x^4 – 4x^3} \; \underline{2x^2 + 6x + 2} \quad \text{(Subtract)} \\[0.3em] & \phantom{3x^4 – 4x^3 + 2x^2} \; 0 \\ \end{array} \]
(iii) x³ – 3x + 1,x⁵ – 4x³ + x² + 3x + 1
\[ \begin{array}{ll} & \quad\;\; x^2 \quad\; – 1 \\[0.3em] x^3 – 3x + 1 & \overline{\smash{\big)}\; x^5 + 0x^4 – 4x^3 + x^2 + 3x + 1} \\[0.3em] & \quad\underline{x^5 \quad\quad\quad\quad\; – 3x^3 + x^2 \quad\quad\quad\;\;} \quad \text{(Subtract)} \\[0.3em] & \quad\phantom{x^5} 0x^4 – x^3 \quad\quad\quad\; + 3x + 1 \\[0.3em] & \quad\phantom{x^5} \underline{\quad\quad\quad\; -x^3 + 3x – 1} \quad \text{(Subtract)} \\[0.3em] & \phantom{x^5 + 0x^4} \quad\quad\quad\quad\quad\quad\; 2 \\ \end{array} \]
প্রশ্ন 3. যদি দুটা শূন্য √5/3 আৰু – √5/3, তেন্তে 3x⁴ + 6x³ – 2x² – 10x – 5 ৰ বাকী আটাইবোৰ শূন্য উলিওৱা।
উত্তৰঃ দুটা শূন্য হ’ল \(\sqrt\frac{5}{3}\) আৰু – \(\sqrt\frac{5}{3}\)
∴(\(x\) –\(\sqrt\frac{5}{3}\) ) (\(x\)+ \(\sqrt\frac{5}{3}\)) = \(x^2\)- \(\frac{5}{3}\) = \(0\)
\[ \begin{array}{ll} & \quad\;\; 3x^2 + 6x + 3 \\[0.5em] x^2 – \frac{5}{3} & \overline{\smash{\big)}\; 3x^4 + 6x^3 – 2x^2 – 10x – 5} \\[0.5em] & \underline{3x^4 \quad\quad\quad\; -\frac{5}{3}x^2 \quad\quad\quad\quad\quad} \quad \text{(Subtract)} \\[0.5em] & \phantom{3x^4} \; 6x^3 + 3x^2 – 10x – 5 \\[0.5em] & \phantom{3x^4} \; \underline{6x^3 \quad\quad\quad\; – 10x \quad\quad\quad\quad} \quad \text{(Subtract)} \\[0.5em] & \phantom{3x^4 + 6x^3} \; 3x^2 \quad\quad\; – 5 \\[0.5em] & \phantom{3x^4 + 6x^3} \; \underline{3x^2 \quad\quad\; – 5} \quad \text{(Subtract)} \\[0.5em] & \phantom{3x^4 + 6x^3 + 3x^2} \; 0 \\ \end{array} \]
\(\therefore(x^2 – \frac{5}{3})({3x^4 + 6x^3 + 3x^2})\)
\(\implies(x^2 – \frac{5}{3}){3({x^2 + 2x + 1})}\)
\(\implies(x^2 – \frac{5}{3}){3({x^2 + x+x + 1})}\)
\(\implies(x^2 – \frac{5}{3}){3({x(x+1)+1(x + 1)})}\)
\(\implies 3 (x^2 – \frac{5}{3}){(x+1)(x + 1)}\)
এতিয়া, বহুপদ ৰাশিৰ আন শূন্যবোৰ হ’ল।
x + 1 = 0, নাইবা x + 1 = 0
⇒ x= -1 ⇒ x = -1
∴ চাৰিমাত্ৰা বিশিষ্ট বহুপদ ৰাশিৰ শূন্যবোৰ হল: √5/3, -√5/3, 1 আৰু -1।
প্রশ্ন 4. x² – 3x² + x + 2 ক এটা বহুপদ g(x) ৰে হৰণ কৰাত ভাগফল x – 2 আৰু ভাগশেষ – 2x + 4 পোৱা গ’ল। g(x) উলিওৱা।
উত্তৰঃ ধৰা হল, \(x³ – 3x² + x + 2=p(x)\)
\(q(x) = x – 2\) আৰু\( r(x) = – 2x + 4\)
এই তথ্যখিনি, (বিভাজন কলৰবিধি) -ৰ লগত তুলনা কৰি পাওঁ –
\(p(x) = g(x) q(x) + r(x)\)
\(⇒ p(x) = r(x) = g(x).q(x)\)
\(\therefore x^3-3x^2+x+2 = g(x)×(x-2) + (-2x+4)\)
\(\implies (x^3-3x^2+x+2) -(-2x+4) = g(x)×(x-2)\)
\(\therefore g(x)×(x-2) = x^3-3x^2+3x-2\)
\(g(x)\) =\(\frac{x^3-3x^2+3x-2}{x-2}\)
\[ \begin{array}{ll} & \quad\;\; x^2 – x + 1 \\[0.3em] x – 2 & \overline{\smash{\big)}\; x^3 – 3x^2 + 3x – 2} \\[0.3em] & \underline{x^3 – 2x^2 \quad\quad\quad\quad} \quad \text{(Subtract)} \\[0.3em] & \phantom{x^3} \; -x^2 + 3x – 2 \\[0.3em] & \phantom{x^3} \; \underline{-x^2 + 2x \quad\quad\;} \quad \text{(Subtract)} \\[0.3em] & \phantom{x^3 – x^2} \; x – 2 \\[0.3em] & \phantom{x^3 – x^2} \; \underline{x – 2} \quad \text{(Subtract)} \\[0.3em] & \phantom{x^3 – x^2 + x} \; 0 \\ \end{array} \]
6.(i) \(3x^2− x^2 − 3x + 1\) বহুপদটোৰ এটা শূণ্য 1 । ইয়াৰ বাকী কেইটা শূণ্য নিৰ্ণয় কৰা ।
Solution:-
দিয়া আছে, প্ৰদত্ত বহুপদটো = \(3x^2− x^2 − 3x + 1\)
শূণ্য এটা হৈছে = 1
Let, p(x) = \(3x^2− x^2 − 3x + 1\)
x = 1
\(\therefore x -1 =0\)
\(\implies (x -1 ) \)হৈছে \(p(x)\) ৰ উতপাদক
\[ \begin{array}{ll} & \quad\;\, 3x^2+2x – 1 \\[0.3em] x – 1 & \overline{\smash{\big)}3x^2− x^2 − 3x + 1} \\[0.3em] & \underline{3x^3\;\; – 3x^2 \quad\quad\;\;} \quad\quad \text{(Subtract)} \\[0.3em] & \quad\;\;\;\;\;\, 2x^2 – 3x +1 \\[0.3em] & \quad\;\, \underline{\;\;2x^2 -2x} \phantom{1}\quad\;\;\; \;\;\; \text{(Subtract)} \quad\quad \\& \quad\;\;\;\;\;\,\;\;\;\;\; -x +1 \\[0.3em] & \quad\;\, \underline{\;\;\;\;\;\;\;\;\;\;-x + 1} \quad\;\;\; \;\;\; \text{(Subtract)} \\[0.3em] & \phantom{2x^2} \;\quad\;\;\;\;\; 0\\\end{array} \]