`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x)` Write the partial fraction decomposition of the improper rational expression.
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calendarEducator since 2015
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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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calendarEducator since 2015
write1,115 answers
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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.`
calendarEducator since 2015
write178 answers
starTop subjects are Math, Science, and Business
`(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) `