Mathematics

# If $\dfrac{3x^{2}+10x+13}{(x-1)^{4}}=\dfrac{A}{(x-1)^{2}}+\dfrac{B}{(x-1)^{3}}+\dfrac{C}{(x-1)^{4}}$ then descending order of $A,B,C$

$C, B, A$

##### SOLUTION
$\dfrac{3x^{2}+10x+13}{(x-1)^{4}}=\dfrac{A}{(x-1)^{2}}+\dfrac{B}{(x-1)^{3}}+\dfrac{C}{(x-1)^{4}}$
substitute $x-1=t\Rightarrow x=(t+1)$
$\dfrac{3(t+1)^{2}+10(t+1)+13}{t^{4}}=\dfrac{3t^{2}+16t+26}{t^{4}}$
$=\dfrac{3}{t^{2}}+\dfrac{16}{t^{3}}+\dfrac{26}{t^{4}}$
$=\dfrac{3}{(x-1)^{2}}+\dfrac{16}{(x-1)^{3}}+\dfrac{26}{(x-1)^{4}}$
So, decreasing order is
$C,B,A$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
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Chapters 126
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