`x/(x^3 - x^2 - 2x + 2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.
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Decompose the denominator:
`x^3-x^2-2x+2=x^2(x-1)-2(x-1)=(x^2-2)(x-1)=(x-1)(x-sqrt(2))(x+sqrt(2)).`
Therefore the fraction decomposition has the form
`x/(x^3-x^2-2x+2)=A/(x-1)+B/(x-sqrt(2))+C/(x+sqrt(2)).`
To find A,B and C multiply both sides by the original denominator:
`x=A(x^2-2)+B(x-1)(x+sqrt(2))+C(x-1)(x-sqrt(2)),` or
`x=x^2(A+B+C)+x(B(sqrt(2)-1)-C(sqrt(2)+1))+(-2A-Bsqrt(2)+Csqrt(2)).`
Thus `A+B+C=0,` `B(sqrt(2)-1)-C(sqrt(2)+1)=1` and `-2A-Bsqrt(2)+Csqrt(2)=0.`
`A=-(B+C),`
`B(sqrt(2)-1)-C(sqrt(2)+1)=1,`
`2(B+C)-Bsqrt(2)+Csqrt(2)=0,` or
`B(2-sqrt(2))+C(2+sqrt(2))=0,` or
`B(sqrt(2)-1)+C(sqrt(2)+1)=0.`
Add and subtract these two equations and obtain
`2B(sqrt(2)-1)=1,` or `B=1/(2(sqrt(2)-1))=(sqrt(2)+1)/2` and
`2C(sqrt(2)+1)=-1,` or `C=-1/(2(sqrt(2)+1))=-(sqrt(2)-1)/2.`
And `A= -(B+C)=-1.`
Now check this result:
`-1/(x-1)+(sqrt(2)+1)/2 1/(x-sqrt(2)) -(sqrt(2)-1)/2 1/(x+sqrt(2))=`
`=-1/(x-1)+1/2 ((sqrt(2)+1)x+sqrt(2)(sqrt(2)+1)-(sqrt(2)-1)x+sqrt(2)(sqrt(2)-1))/(x^2-2)=`
`=-1/(x-1) +1/2 (2x+4)/(x^2-2)=-1/(x-1)+(x+2)/(x^2-2)=`
`=(2-x^2+x^2+x-2)/((x+2)(x^2-2))=x/(x^3-x^2-2x+2),`
which is correct.
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