Question

How do you write the partial fraction decomposition of the rational expression `(7 x^2 - 7 x+12)/(x^4+x^2-2)`?

Answer 1

`(7x^2-7x+12)/(x^4+x^2-2)=(7/3x+2/3)/(x^2+2)+(-13/3)/(x+1)+2/(x-1)`

Explanation:

From the given, we start from the factoring part of the denominator

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

so that

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

and

`(7x^2-7x+12)/((x^2+2)(x+1)(x-1))=(Ax+B)/(x^2+2)+C/(x+1)+D/(x-1)`

Let us now form the equations using the LCD we continue

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

`((Ax+B)(x^2-1)+C(x^2+2)(x-1)+D(x^2+2)(x+1))/((x^2+2)(x+1)(x-1))`

Expand and then simplify

`(Ax^3-Ax+Bx^2-B+Cx^3-Cx^2+2Cx-2C+Dx^3+Dx^2+2Dx+2D)/((x^2+2)(x+1)(x-1))`

`(A+C+D)x^3+(B-C+D)x^2+(-A+2C+2D)x^1+(-B-2C+2D)x^0`

Equate now the numerical coefficients of the numerators of left and right sides of the equation

`A+C+D=0" " "`first equation
`B-C+D=7" " "`second equation
`-A+2C+2D=-7" " "`third equation
`-B-2C+2D=12" " "`fourth equation

4 equations with 4 unknowns, solve the unknowns A,B,C,D by elimination of variables

by adding first and third equations the result will be
`3C+3D=-7" " "`fifth equation

by adding second and fourth equations the result will be
`-3C+3D=19" " "`sixth equation

by adding the fifth and sixth equations, the result will be
`6D=12`

`D=2`
it follows
`C=-13/3`
Solve A using first equation

`A=-C-D=-(-13/3)-2=(+13-6)/3=7/3`
`A=7/3`

Solve B using second equation

`B=C-D+7=-13/3-2+7=-13/3+5=2/3`

`B=2/3`

the final answer

`(7x^2-7x+12)/(x^4+x^2-2)=(7/3x+2/3)/(x^2+2)+(-13/3)/(x+1)+2/(x-1)`

God bless...I hope the explanation is useful..