# `(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x)` Write the partial fraction decomposition of the improper rational expression.

A bit more extension for my above solution

Now in the above expression we need to simplify the

`(2x-1)/(x^3 + 2x^2 + x)`
It is as follows

`(2x-1)/(x^3 + 2x^2 + x) = (2x-1)/(x(x+1)^2)`

`(2x-1)/(x(x+1)^2)= (a/x) + (b/(x+1))+(c/(x+1)^2))`

on simplification we get
`(2x-1)= (a(x+1)^2)+(bx(x+1))+cx`

As the roots of the denominator `(x(x+1)^2)` are` 0 , -1` . We can solve the unknown parameters by
plugging the values of `x` .

when `x=0` , we get
`a=-1 `
when `x=(-1)` we get
`c=3`

As we know the `a,c` values , we can find the value of `b` as

`2x-1 = (-1)(x+1)^2 + bx(x+1)+3x`
`2x-1 = bx^2+x+bx-x^2-1`
`2x-1 = x^2(b-1)+x(b+1)-1`
on comparing we get
`b+1 =2`
=> `b=1 `
so, `(2x-1)/(x^3 + 2x^2 + x) = ((-1)/x)+(1/(x+1))+(3/(x+1^2))`
so, the partial fraction for

`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x) = x+ (3/x)+ (2x-1)/(x^3 + 2x^2 + x)`

`=x+ (3/x)+ ((-1)/x)+(1/(x+1))+(3/(x+1^2))`
`= x+(2/x)+(1/(x+1))+(3/(x+1)^2).`

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First we have to make the fraction a proper one, for this it is necessary to divide the numerator by the denominator with the remainder.

`x^4+2x^3+4x^2+8x+2 =x(x^3+2x^2+x) + 3x^2 +8x+2.`

So `(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x) = x+(3x^2+8x+2)/(x^3 + 2x^2 + x)=x+(3x^2+8x+2)/(x(x+1)^2).`

The proper part has the decomposition form of

`(3x^2+8x+2)/(x(x+1)^2)=A/x+B/(x+1)+C/(x+1)^2.`

Multiply both sides by `x(x+1)^2` and obtain

`3x^2+8x+2=A(x^2+2x+1)+Bx(x+1)+Cx=`

`=x^2*(A+B)+x*(2A+B+C)+A.`

So `A+B=3,` `2A+B+C=8` and `A=2,` from this we get `B=1` and `C=3.`

The answer: `(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x) = x+2/x+1/(x+1)+3/(x+1)^2.`

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`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x)`

on long division we get

= >

`(x^3 + 2x^2 + x)`   divides `(x^4 + 2x^3 + 4x^2 + 8x + 2)` we get the

Quotient` 'x'` with remainder `3x^2 +8x+2`

Now,

when we divide `3x^2 +8x+2 ` with `(x^3 + 2x^2 + x)`  we get the quotient `'3/x' ` leaving the remainder `2x-1`

so, the partial fraction of

`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x) ` =

`x+ (3/x)+ (2x-1)/(x^3 + 2x^2 + x) `

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