If 5x2+2x3+3=A1x+A2x+A3x2+1, then A1,A2,A3=

5x2+2x3+x=A1x+A2x+A3x2+15x2+2=A1(x2+1)+(A2x+A3)xx=0 ⇒ 2=A1co.ef x2 ⇒ 5=A1+A2 ⇒ A2=3co.eff x⇒A3=0(A1,A2,A3)=(2,3,0)

If 5x2+2x3+3=A1x+A2x+A3x2+1, then A1,A2,A3=

5x2+2x3+x=A1x+A2x+A3x2+15x2+2=A1(x2+1)+(A2x+A3)xx=0 ⇒ 2=A1co.ef x2 ⇒ 5=A1+A2 ⇒ A2=3co.eff x⇒A3=0(A1,A2,A3)=(2,3,0)

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$If \frac{5 x^{2} + 2}{x^{3} + 3} = \frac{A_{1}}{x} + \frac{A_{2} x + A_{3}}{x^{2} + 1}, then (A_{1}, A_{2}, A_{3}) =$

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$(0, 2, 3)$

$(3, 0, 2)$

$(2, 3, 0)$

$(2, 0, 3)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \frac{5 x^{2} + 2}{x^{3} + x} & = & \frac{A_{1}}{x} + \frac{A_{2} x + A_{3}}{x^{2} + 1} \\ 5 x^{2} + 2 & = & A_{1} (x^{2} + 1) + (A_{2} x + A_{3}) x \\ x & = & 0 \Rightarrow 2 = A_{1} \\ co . ef x^{2} & \Rightarrow & 5 = A_{1} + A_{2} \Rightarrow A_{2} = 3 \\ co . eff x & \Rightarrow & A_{3} = 0 \\ (A_{1}, A_{2}, A_{3}) & = & (2, 3, 0) \end{array}$