Question

I wonder how to do this?

Answer 1

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Answer 2

Answer:

A+B+C+D= 1

Step-by-step explanation:

If

$\frac{3x^{3}-2x^{2}+2x-1}{x^{4}+x^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{Cx+D}{x^{2}+1}$

Find the value of A+B+C+D

$\frac{3x^{3}-2x^{2}+2x-1}{x^{4}+x^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{Cx+D}{x^{2}+1}\\\\\Rightarrow\frac{3x^{3}-2x^{2}+2x-1}{x^{4}+x^{2}}=\frac{A\times(x^{3}+x)+B\times(x^{2}+1)+(Cx+D)\times{x^{2}}}{x^{2}(x^{2}+1)}\\\\\Rightarrow\frac{3x^{3}-2x^{2}+2x-1}{x^{4}+x^{2}}=\frac{Ax^{3}+Ax+Bx^{2}+B+Cx^{3}+Dx^{2}}{x^{4}+x^{2}}\\\\\Rightarrow{3x^{3}-2x^{2}+2x-1}={Ax^{3}+Ax+Bx^{2}+B+Cx^{3}+Dx^{2}}\\\\\Rightarrow{3x^{3}-2x^{2}+2x-1}={Ax^{3}+Cx^{3}+Bx^{2}+Dx^{2}}+Ax+B$

$\\\\\Rightarrow{3x^{3}-2x^{2}+2x-1}=(A+C)x^{3}+(B+D)x^{2}+Ax+B$

By comparing constant term of both side we get

B=-1

By comparing co-efficient of $x$ on both side we get

A=2

By comparing co-efficient of $x^{2}$ on both side we get  

$B+D=-2\\\\\Rightarrow{D}=-2-B\\\\\Rightarrow{D}=-2-(-1)\\\\\Rightarrow{D}=-2+1=-1$

By comparing co-efficient of [tex[x^{4}[/tex] on both side we get  

$A+C=3\\\\\Rightarrow{C}=3-A\\\\\Rightarrow{C}=3-2\\\\\Rightarrow{C}=1$

Therefore

$A+B+C+D= 2+(-1)+1+(-1)=2-1+1-1=1$