Decompose the function P(x) = ((x^4 +2x^3 +6x^2 +20x+6)/(x^3 +x^2 +x)) in partial fractions.


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Question Number 137252 by mey3nipaba last updated on 31/Mar/21
$Decompose the function P(x) = ((x^4 +2x^3 +6x^2 +20x+6)/(x^3 +x^2 +x)) in partial fractions.$
$Decompose the function P (x) = \frac{x^{4} + {2 x}^{3} + {6 x}^{2} + 20 x + 6}{x^{3} + x^{2} + x}$ $in partial fractions .$
	Answered by bemath last updated on 31/Mar/21
	$⇔ x+1 + ((4x^2 +19x+6)/(x(x^2 +x+1))) = ⇔x+1 + (A/x) + ((Bx+C)/(x^2 +x+1)) ⇔ 4x^2 +19x+6 = A(x^2 +x+1)+(Bx+C)x ⇔ A = [((4x^2 +19x+6)/(x^2 +x+1)) ]_(x=0) = 6 put x=1⇒29 = 18+B+C ⇒ B+C = 11 put x=−1⇒−9= 6 −(−B+C) ⇒ −15 = B−C we get { ((B=−2)),((C=13)) :} ⇔ x+1 +(6/x)+ ((13−2x)/(x^2 +x+1))$
	$\Leftrightarrow x + 1 + \frac{4 x^{2} + 19 x + 6}{x (x^{2} + x + 1)} =$ $\Leftrightarrow x + 1 + \frac{A}{x} + \frac{B x + C}{x^{2} + x + 1}$ $\Leftrightarrow 4 x^{2} + 19 x + 6 = A (x^{2} + x + 1) + (B x + C) x$ $\Leftrightarrow A = {[\frac{{4 x}^{2} + 19 x + 6}{x^{2} + x + 1}]}_{x = 0} = 6$ $put x = 1 \Rightarrow 29 = 18 + B + C$ $\Rightarrow B + C = 11$ $put x = - 1 \Rightarrow - 9 = 6 - (- B + C)$ $\Rightarrow - 15 = B - C$ $we get {\begin{cases} B = - 2 \\ C = 13 \end{cases}$ $\Leftrightarrow x + 1 + \frac{6}{x} + \frac{13 - 2 x}{x^{2} + x + 1}$
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