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))7x2−7x+12x4+x2−2=7x2−7x+12(x2+2)(x2−1)=7x2−7x+12(x2+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))7x2−7x+12x4+x2−2=7x2−7x+12(x2+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)7x2−7x+12(x2+2)(x+1)(x−1)=Ax+Bx2+2+Cx+1+Dx−1
Let us now form the equations using the LCD we continue
(7x^2-7x+12)/((x^2+2)(x+1)(x-1))=7x2−7x+12(x2+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))(Ax+B)(x2−1)+C(x2+2)(x−1)+D(x2+2)(x+1)(x2+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))Ax3−Ax+Bx2−B+Cx3−Cx2+2Cx−2C+Dx3+Dx2+2Dx+2D(x2+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(A+C+D)x3+(B−C+D)x2+(−A+2C+2D)x1+(−B−2C+2D)x0
Equate now the numerical coefficients of the numerators of left and right sides of the equation
A+C+D=0" " "A+C+D=0 first equation
B-C+D=7" " "B−C+D=7 second equation
-A+2C+2D=-7" " "−A+2C+2D=−7 third equation
-B-2C+2D=12" " "−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" " "3C+3D=−7 fifth equation
by adding second and fourth equations the result will be
-3C+3D=19" " "−3C+3D=19 sixth equation
by adding the fifth and sixth equations, the result will be
6D=126D=12
D=2D=2
it follows
C=-13/3C=−133
Solve A using first equation
A=-C-D=-(-13/3)-2=(+13-6)/3=7/3A=−C−D=−(−133)−2=+13−63=73
A=7/3A=73
Solve B using second equation
B=C-D+7=-13/3-2+7=-13/3+5=2/3B=C−D+7=−133−2+7=−133+5=23
B=2/3B=23
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)7x2−7x+12x4+x2−2=73x+23x2+2+−133x+1+2x−1
God bless...I hope the explanation is useful..
