# `(2x^2 + x + 8)/(x^2 + 4)^2` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`(2x^2+x+8)/(x^2+4)^2`

Let`(2x^2+x+8)/(x^2+4)^2=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2`

`(2x^2+x+8)/(x^2+4)^2=((Ax+B)(x^2+4)+Cx+D)/(x^2+4)^2`

`(2x^2+x+8)/(x^2+4)^2=(Ax^3+4Ax+Bx^2+4B+Cx+D)/(x^2+4)^2`

`:.(2x^2+x+8)=Ax^3+4Ax+Bx^2+4B+Cx+D`

`2x^2+x+8=Ax^3+Bx^2+(4A+C)x+4B+D`

Equating the coefficients the like terms,

`A=0`

`B=2`

`4A+C=1`

`4B+D=8`

Plug the value of the A and B in the above equations,

`4(0)+C=1`

`C=1`

`4(2)+D=8`

`8+D=8`

`D=8-8`

`D=0`

`:.(2x^2+x+8)/(x^2+4)^2=2/(x^2+4)+x/(x^2+4)^2`

Now let's check it algebraically,

RHS=`2/(x^2+4)+x/(x^2+4)^2`

`=(2(x^2+4)+x)/(x^2+4)^2`

`=(2x^2+8+x)/(x^2+4)^2`

`=(2x^2+x+8)/(x^2+4)^2`

=LHS

Hence it is...

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`(2x^2+x+8)/(x^2+4)^2`

Let`(2x^2+x+8)/(x^2+4)^2=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2`

`(2x^2+x+8)/(x^2+4)^2=((Ax+B)(x^2+4)+Cx+D)/(x^2+4)^2`

`(2x^2+x+8)/(x^2+4)^2=(Ax^3+4Ax+Bx^2+4B+Cx+D)/(x^2+4)^2`

`:.(2x^2+x+8)=Ax^3+4Ax+Bx^2+4B+Cx+D`

`2x^2+x+8=Ax^3+Bx^2+(4A+C)x+4B+D`

Equating the coefficients the like terms,

`A=0`

`B=2`

`4A+C=1`

`4B+D=8`

Plug the value of the A and B in the above equations,

`4(0)+C=1`

`C=1`

`4(2)+D=8`

`8+D=8`

`D=8-8`

`D=0`

`:.(2x^2+x+8)/(x^2+4)^2=2/(x^2+4)+x/(x^2+4)^2`

Now let's check it algebraically,

RHS=`2/(x^2+4)+x/(x^2+4)^2`

`=(2(x^2+4)+x)/(x^2+4)^2`

`=(2x^2+8+x)/(x^2+4)^2`

`=(2x^2+x+8)/(x^2+4)^2`

=LHS

Hence it is verified.

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