When f(x) divided by (x−1)(x+2), the remainder is (x+3) When f(x) divided by (x^2 +2x+5), the remainder is (2x+1) Find the remainder if f(x) is divided by(x−1)(x^2 +2x+5)


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Question Number 116603 by ZiYangLee last updated on 05/Oct/20

$When f (x) divided by (x - 1) (x + 2),$ $the remainder is (x + 3)$ $When f (x) divided by (x^{2} + 2 x + 5),$ $the remainder is (2 x + 1)$ $Find the remainder if f (x) is divided$ $by (x - 1) (x^{2} + 2 x + 5)$
	Answered by 1549442205PVT last updated on 06/Oct/20
	$From the hypothesis we have: f(x)=p(x)(x−1)(x+2)+x+3(1) f(x)=q(x)(x^2 +2x+5)+2x+1 ⇔f(x)=q(x)([x−(−1+2i)][x−(−1−2i)] +2x+1(2) We have x^2 +2x+5=0⇔x=−1±2i (2)⇒f(−1+2i)=2(−1+2i)+1=−1+4i f(−1−2i)=2(−1−2i)+1=−1−4i We need find r(x)=ax^2 +bx+c so that f(x)=h(x)(x−1)(x^2 +2x+5)+r(x)(3) From(1)(2)we get { ((f(1)=4,f(−2)=1)),((f(−1+2i)=−1+4i,f(−1−2i)=−1−4i)) :} Put into (3) we get i)4=f(1)=a+b+c ii)−1+4i=a(−1+2i)^2 +b(−1+2i)+c ⇔−1+4i=−5a−4ai−b+2bi+c ⇔(−5a−b+c+1)+(2b−4a−4)i=0 ⇔ { ((−5a−b+c+1=0)),((2b−4a−4=0)) :} iii)−1−4i=a(−1−2i)^2 +b(−1−2i)+c ⇔−1−4i=−5a+4ai−b−2bi+c ⇔−5a−b+c+1+(4a−2b+4)i=0 ⇔ { ((−5a−b+c+1=0)),((4a−2b+4=0)) :} We obtain the system of three eqns { ((a+b+c=4(4))),((−5a−b+c+1=0(5))),((4a−2b+4=0(6))) :} Substract (4) from (5)we get 6a+2b−5=0 ,adding this equation to (6) we get 10a−1=0⇒a=1/10 ⇒b=11/10,c=28/10.Therefore we get r(x)=0.1x^2 +1.1x+2.8$
	$From the hypothesis we have :$ $f (x) = p (x) (x - 1) (x + 2) + x + 3 (1)$ $f (x) = q (x) (x^{2} + 2 x + 5) + 2 x + 1$ $\Leftrightarrow f (x) = q (x) ([x - (- 1 + 2 i)] [x - (- 1 - 2 i)]$ $+ 2 x + 1 (2)$ $We have x^{2} + 2 x + 5 = 0 \Leftrightarrow x = - 1 \pm 2 i$ $(2) \Rightarrow f (- 1 + 2 i) = 2 (- 1 + 2 i) + 1 = - 1 + 4 i$ $f (- 1 - 2 i) = 2 (- 1 - 2 i) + 1 = - 1 - 4 i$ $We need find r (x) = {ax}^{2} + bx + c so that$ $f (x) = h (x) (x - 1) (x^{2} + 2 x + 5) + r (x) (3)$ $From (1) (2) we get$ ${\begin{cases} f (1) = 4, f (- 2) = 1 \\ f (- 1 + 2 i) = - 1 + 4 i, f (- 1 - 2 i) = - 1 - 4 i \end{cases}$ $Put into (3) we get$ $i) 4 = f (1) = a + b + c$ $ii) - 1 + 4 i = a {(- 1 + 2 i)}^{2} + b (- 1 + 2 i) + c$ $\Leftrightarrow - 1 + 4 i = - 5 a - 4 ai - b + 2 bi + c$ $\Leftrightarrow (- 5 a - b + c + 1) + (2 b - 4 a - 4) i = 0$ $\Leftrightarrow {\begin{cases} - 5 a - b + c + 1 = 0 \\ 2 b - 4 a - 4 = 0 \end{cases}$ $iii) - 1 - 4 i = a {(- 1 - 2 i)}^{2} + b (- 1 - 2 i) + c$ $\Leftrightarrow - 1 - 4 i = - 5 a + 4 ai - b - 2 bi + c$ $\Leftrightarrow - 5 a - b + c + 1 + (4 a - 2 b + 4) i = 0$ $\Leftrightarrow {\begin{cases} - 5 a - b + c + 1 = 0 \\ 4 a - 2 b + 4 = 0 \end{cases}$ $We obtain the system of three eqns$ ${\begin{cases} a + b + c = 4 (4) \\ - 5 a - b + c + 1 = 0 (5) \\ 4 a - 2 b + 4 = 0 (6) \end{cases}$ $Substract (4) from (5) we get$ $6 a + 2 b - 5 = 0, adding this equation to$ $(6) we get 10 a - 1 = 0 \Rightarrow a = 1 / 10$ $\Rightarrow b = 11 / 10, c = 28 / 10 . Therefore we get$ $r (x) = 0 . {1 x}^{2} + 1 . 1 x + 2.8$
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