⋓si⋒g ChineseRemainderTheorm ∂etermine polynomial p(x) such that p(x)≡8(mod x+1) p(x)≡−24(mod x+3) p(x)≡6(mod x) p(x)≡0(mod x+2)


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Question Number 67697 by Rasheed.Sindhi last updated on 30/Aug/19

$⋓ si ⋒ g ChineseRemainderTheorm$ $\partial etermine polynomial p (x) such that$ $p (x) \equiv 8 (\mod x + 1)$ $p (x) \equiv - 24 (\mod x + 3)$ $p (x) \equiv 6 (\mod x)$ $p (x) \equiv 0 (\mod x + 2)$
	Answered by Rasheed.Sindhi last updated on 01/Sep/19
	${ ((p(x)≡8(mod x+1))),((p(x)≡−24(mod x+3))),((p(x)≡6(mod x))),((p(x)≡0(mod x+2))) :} Let m_1 =x+1,m_2 =x+3,m_3 =x,m_4 =x+2 Note that (m_i ,m_j )=1 for ∀ i≠j (pairwise coprime) Let M=m_1 m_2 m_3 m_4 =(x+1)(x+3)(x)(x+2) M_1 =(M/m_1 )=m_2 m_3 m_4 =(x+3)(x)(x+2) M_2 =(M/m_2 )=m_1 m_3 m_4 =(x+1)(x)(x+2) M_3 =(M/m_3 )=m_1 m_2 m_4 =(x+1)(x+3)(x+2) M_4 =(M/m_4 )=m_1 m_2 m_3 =(x+1)(x+3)(x) Note that (M_i ,m_i )=1 ∴ M_i y≡1(mod m_i ) (x+3)(x)(x+2)H_1 (x)≡1(mod x+1) (x^3 +5x^2 +6x)H_1 (x)≡1(mod x+1) Let H_1 (x)=h_1 (a constant) (((−1)),h_1 ,( 5h_1 ),( 6h_1 ),( −1)),(,,(−h_1 ),(−4h_1 ),( −2h_1 )),(,h_1 ,( 4h_1 ),( 2h_1 ),( −2h_1 −1)) ) −2h_1 −1=0⇒h_1 =−1/2 (x+1)(x)(x+2)H_2 (x)≡1(mod x+3) (x^3 +3x^2 +2x)H_2 (x)≡1(mod x+3) If H_2 (x)=h_2 (constant) h_2 =−1/6 (By similar process) (x+1)(x+3)(x+2)H_3 (x)≡1(mod x) (x^3 +6x^2 +11x+6)H_3 (x)≡1(mod x) If H_3 (x)=h_3 (constant) (((0)),h_3 ,(6h_3 ),(11h_3 ),(6h_3 −1)),(,,0,0,0),(,h_3 ,(6h_3 ),(11h_3 ),(6h_3 −1)) ) 6h_3 −1=0⇒h_3 =1/6 (x+1)(x+3)(x)H_4 (x)≡1(mod x+2) (x^3 +4x^2 +3x)H_4 (x)≡1(mod x+2) If H_4 (x)=h_4 (((−2)),h_4 ,( 4h_4 ),( 3h_4 ),(−1)),(,,(−2h4),(−4h_4 ),( 2h_4 )),(,h_4 ,( 2h_4 ),(−h_4 ),(2h_4 −1)) ) 2h_4 −1=0⇒h_4 =1/2 x=8(x^3 +5x^2 +6x)(−1/2) +(−24)(x^3 +3x^2 +2x)(−1/6) +6(x^3 +6x^2 +11x+6)(1/6) +(0)(x^3 +4x^2 +3x)(1/2) x=−4x^3 −20x^2 −24x +4x^3 +12x^2 + 8x + x^3 + 6x^2 +11x+6 x=x^3 −2x^2 −5x+6$
	${\begin{cases} p (x) \equiv 8 (\mod x + 1) \\ p (x) \equiv - 24 (\mod x + 3) \\ p (x) \equiv 6 (\mod x) \\ p (x) \equiv 0 (\mod x + 2) \end{cases}$ $Let m_{1} = x + 1, m_{2} = x + 3, m_{3} = x, m_{4} = x + 2$ $Note that (m_{i}, m_{j}) = 1 for \forall i \neq j$ $(pairwise coprime)$ $Let M = m_{1} m_{2} m_{3} m_{4} = (x + 1) (x + 3) (x) (x + 2)$ $M_{1} = \frac{M}{m_{1}} = m_{2} m_{3} m_{4} = (x + 3) (x) (x + 2)$ $M_{2} = \frac{M}{m_{2}} = m_{1} m_{3} m_{4} = (x + 1) (x) (x + 2)$ $M_{3} = \frac{M}{m_{3}} = m_{1} m_{2} m_{4} = (x + 1) (x + 3) (x + 2)$ $M_{4} = \frac{M}{m_{4}} = m_{1} m_{2} m_{3} = (x + 1) (x + 3) (x)$ $Note that (M_{i}, m_{i}) = 1$ $∴ M_{i} y \equiv 1 (\mod m_{i})$ $(x + 3) (x) (x + 2) H_{1} (x) \equiv 1 (\mod x + 1)$ $(x^{3} + {5 x}^{2} + 6 x) H_{1} (x) \equiv 1 (\mod x + 1)$ $Let H_{1} (x) = h_{1} (a constant)$ $(\begin{matrix} - 1) & h_{1} & {5 h}_{1} & {6 h}_{1} & - 1 \\ - h_{1} & - {4 h}_{1} & - {2 h}_{1} \\ h_{1} & {4 h}_{1} & {2 h}_{1} & - {2 h}_{1} - 1 \end{matrix})$ $- {2 h}_{1} - 1 = 0 \Rightarrow h_{1} = - 1 / 2$ $(x + 1) (x) (x + 2) H_{2} (x) \equiv 1 (\mod x + 3)$ $(x^{3} + {3 x}^{2} + 2 x) H_{2} (x) \equiv 1 (\mod x + 3)$ $If H_{2} (x) = h_{2} (constant)$ $h_{2} = - 1 / 6 (By similar process)$ $(x + 1) (x + 3) (x + 2) H_{3} (x) \equiv 1 (\mod x)$ $(x^{3} + {6 x}^{2} + 11 x + 6) H_{3} (x) \equiv 1 (\mod x)$ $If H_{3} (x) = h_{3} (constant)$ $(\begin{matrix} 0) & h_{3} & {6 h}_{3} & {11 h}_{3} & {6 h}_{3} - 1 \\ 0 & 0 & 0 \\ h_{3} & {6 h}_{3} & {11 h}_{3} & {6 h}_{3} - 1 \end{matrix})$ ${6 h}_{3} - 1 = 0 \Rightarrow h_{3} = 1 / 6$ $(x + 1) (x + 3) (x) H_{4} (x) \equiv 1 (\mod x + 2)$ $(x^{3} + {4 x}^{2} + 3 x) H_{4} (x) \equiv 1 (\mod x + 2)$ $If H_{4} (x) = h_{4}$ $(\begin{matrix} - 2) & h_{4} & {4 h}_{4} & {3 h}_{4} & - 1 \\ - 2 h 4 & - {4 h}_{4} & {2 h}_{4} \\ h_{4} & {2 h}_{4} & - h_{4} & {2 h}_{4} - 1 \end{matrix})$ ${2 h}_{4} - 1 = 0 \Rightarrow h_{4} = 1 / 2$ $x = 8 (x^{3} + {5 x}^{2} + 6 x) (- 1 / 2)$ $+ (- 24) (x^{3} + {3 x}^{2} + 2 x) (- 1 / 6)$ $+ 6 (x^{3} + {6 x}^{2} + 11 x + 6) (1 / 6)$ $+ (0) (x^{3} + {4 x}^{2} + 3 x) (1 / 2)$ $x = - {4 x}^{3} - {20 x}^{2} - 24 x$ $+ {4 x}^{3} + {12 x}^{2} + 8 x$ $+ x^{3} + {6 x}^{2} + 11 x + 6$ $x = x^{3} - {2 x}^{2} - 5 x + 6$
	Commented by Prithwish sen last updated on 02/Sep/19

	$excelent .$
	Commented by mr W last updated on 03/Sep/19

	$n i c e c r e a t i o n!$
	Commented by Rasheed.Sindhi last updated on 04/Sep/19

	$Thank you both sirs!$
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