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 $$A,B,C$$


ANSWER

$$C, B, A$$


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
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