resolve into partial fraction X
(x²+1) (x-1)
Answers
Answer:
How do I solve this into a partial fraction x²+1/(x-1)²(x²+2x+2)?
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Reducing Fractions and/or Factorizations is always a Discretion, as truly, the Fraction and Numbers are oft supplied as they are to provide an Information.
Explanation
Although there is no proof for or against this due the actual vagueness of: No =□ was supplied to actually Infer, let place this into an Informative Perspective, and in a Real life circumstance, such of the like would not even place on your desk without scrutiny, except if there was an undertrained personnel gathering or journalling infos, and yourself were the Superior who reviewed Idiocies and Exhibits, from Disgruntled Staff so, Let's say this was supplied an =□ that was Beknown only at those High Level Supervisory Levels, and these were arrived to Staff to see if these Might have atall involved in a Breach of that ‘Really Big Secret' Info.
In event this =□ = Really Big Secret, was also suffice to suggest: 1=2x+2 by thus would enable: x2+1=x2+2x+2
We realize that also:
Per Condition of 1=2x+2, by x=(-1/2) or 2x/2 the given equation is thereby that condition met, reducible or factorizable to:
1/[(x−1)2]or−2x/[(x−1)2]
Otherwise you must realize it relies upon the Discretions as these pertain facets:
x2+1=[(x−(−1))(x+(−1))] has a ‘Construable as Mutual presency’ in the denominator in (x-1)^2 atwhich you may reason [x+(-1)]÷(x-1) has scaled this as a Factor to [x-(-1)]÷[(x-1)(x^2+2x+2)], provided you are reasonable that: may is a discretion not a must, nor a permit.
If there is ability for x^2+2x+2 to divide by the (x-(-1)), you may as well say that reliant upon the outcome, you may be fixating this to a precursory or preset more associable x=-1/2, by a reasonible deduction of the coefficiency involved as € in x=-€/2, atwhich, when that coefficiency and condition is met as €=1, we also must reckon that 1=2x+2 by thus would enable: x2+1=x2+2x+2
but as always:
Reducing Fractions and division of Superimposed as by factorizations, is always a Discretion, as truly, the Fraction and Numbers are oft supplied as they are to provide an Information which is Relevant.
Tense Matters
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x2+1(x−1)2(x2+2x+2)=Ax−1+B(x−1)2+Cx+Dx2+2x+2
x2+1(x−1)2(x2+2x+2)=A(x−1)(x2+2x+2)+B(x2+2x+2)+(Cx+D)(x−1)2(x−1)2(x2+2x+2)
x2+1=A(x−1)(x2+2x+2)+B(x2+2x+2)+(Cx+D)(x−1)2
x2+1=A(x3+x2−2)+B(x2+2x+2)+(Cx+D)(x2−2x+1)
x2+1=Ax3+Ax2−2A+Bx2+2Bx+2B+Cx3+(D−2C)x2+(C−2D)x+D
x2+1=(A+C)x3+(A+B−2C+D)x2+(2B+C−2D)x+(−2A+2B+D)
A+C=0⟹A=−C
A+B−2C+D=1⟹B−3C+D=1
2B+C−2D=0
−2A+2B+D=1⟹2B+2C+D=1
(2B+2C+D)−(2B+C−2D)=1−0⟹C+3D=1
2(B−3C+D)−(2B+C−2D)=2(1)−0⟹−7C+4D=2
7(C+3D)+(−7C+4D)=7(1)+2⟹25D=9⟹D=925
C+3D=1⟹C+2725=1⟹C=−225
A=−C⟹A=225
B−3C+D=1⟹B+625+925=1⟹B=25
x2+1(x−1)2(x2+2x+2)=225x−1+25(x−1)2+−225x+925x2+2x+2