Math Questions and Answers

Start Your Free Trial

`(2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x^3 + 4x^2 + 4x)` Write the partial fraction decomposition of the improper rational expression.

Expert Answers info

gsarora17 eNotes educator | Certified Educator

calendarEducator since 2015

write762 answers

starTop subjects are Math, Science, and Business

`(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)`

Since the above expression is an improper rational expression , the first step is to divide and express it as a sum of simpler fractions such that the degree of polynomial in the numerator is less than the degree of polynomial in the denominator.

Dividing the above expression using long division method yields,

`(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)=2x+(-x^2-7x-12)/(x^3+4x^2+4x)`

 Since the polynomials do not completely divide , we have to continue with the partial fractions of the remainder expression.

Now let's factorize the denominator of the remainder expression,

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

`=x(x+2)^2`

Let, `(-x^2-7x-12)/(x^3+4x^2+4x)=A/x+B/(x+2)+C/(x+2)^2`

`=(A(x+2)^2+B(x)(x+2)+Cx)/(x(x+2)^2)`

`=(A(x^2+4x+4)+B(x^2+2x)+Cx)/(x(x+2)^2)`

`=(x^2(A+B)+x(4A+2B+C)+4A)/(x(x+2)^2)`

`:.(-x^2-7x-12)=x^2(A+B)+x(4A+2B+C)+4A`

equating the coefficients of the like terms,

`A+B=-1`                ------- equation 1

`4A+2B+C=-7`   ------- equation 2

`4A=-12`

`A=-12/4`

`A=-3`

Plug the value of A in equation 1,

`-3+B=-1`

`B=-1+3`

`B=2`

Plug the value of A and B in equation 2 ,

`4(-3)+2(2)+C=-7`  

`-12+4+C=-7`

`C=-7+12-4`

`C=1`

`(-x^2-7x-12)/(x^3+4x^2+4x)=(-3)/x+2/(x+2)+1/(x+2)^2`

`:.(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)=2x-3/x+2/(x+2)+1/(x+2)^2`

 

check Approved by eNotes Editorial

Luca B. eNotes educator | Certified Educator

calendarEducator since 2011

write5,348 answers

starTop subjects are Math, Science, and Business

You need to decompose the fraction in simple irreducible fractions, such that:

`(2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x^3 + 4x^2 + 4x)  = (2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x(x^2 + 4x + 4))`

`(2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x^3 + 4x^2 + 4x) = (2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x(x+2)^2)`

`(2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x(x+2)^2)= A/x + B/(x+2) + C/((x+2)^2)`

You need to bring to the same denominator all fractions, such that:

`2x^4 + 8x^3 + 7x^2 - 7x - 12= A(x+2)^2 + Bx(x+2) + Cx`

`2x^4 + 8x^3 + 7x^2 - 7x - 12= Ax^2 + 4Ax + 4A + Bx^2 + 2Bx + Cx`

You need to group the terms having the same power of x:

`2x^4 + 8x^3 + 7x^2 - 7x - 12= x^2(A+B) + x(4A+2B+C) + 4A`

Comparing the expressions both sides yields:

`A + B =7 `

`4A+2B+C = -7`

`4A = -12 => A = -3 => B = 7+3 => B = 10 `

`C = -7 - 4A - 2B=> C = -7 +12 - 20 => C = -15` 

Hence, the partial fraction decomposition of the improper rational expression is `(2x^4 + 8x^3 + 7x^2 - 7x - 12)/(x(x+2)^2)= -3/x + 10/(x+2) - 15/((x+2)^2)`