1/(X^3-6x^2+11x-6) = A/(x-a)+ B(x-b) + C/(x-c)?

Question

1/(X^3-6x^2+11x-6) = A/(x-a)+ B(x-b) + C/(x-c)

Answer 1

This is a problem in decomposition of partial fractions: If the denominator on the left side can be factored, then the fraction can be written as a sum of fractions containing the individual factors. The hardest part of this problem will be factoring that denominator.



x^3 - 6x^2 + 11x - 6. The first thing I try with a polynomial with four terms is factoring by grouping; but I don't think it's possible in this case. We can do simple trial and error, possibly using long polynomial division, or synthetic division. Testing x - 1, I found that



x^3 - 6x^2 + 11x - 6 = (x - 1) (x^2 - 5x + 6)



= (x - 1) (x - 2) (x - 3).



That means we can decompose your given equation into

1/(x^3 - 6x^2 + 11x - 6) = A/(x - 1) + B/(x - 2) + C/(x - 3).

We must find A, B, and C.



Multiply both sides of the equation by the common denominator [ (x - 1) (x - 2) (x - 3)]:



1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)



We're solving for A, B, and C, but the equation should hold true for ALL x. The trick at this point is to find the clever values of x that make it possible to identify A, B, and C. Note what happens if we let x = 1 in the equation above:



1 = A(-1)(-2) + B(0)(-2) + C(0)(-1)

1 = 2A

A = 1/2



Similarly, let x = 2 in order to isolate B:

1 = A(0)(-1) + B(1)(-1) + C(1)(0)

1 = -B

B = -1



And again, let x = 3 to isolate C:

1 = A(1)(0) + B(2)(0) + C(2)(1)

1 = 2C

C = 1/2



Thus, we have our final answer:



(1/2)/(x - 1) + -1/(x - 2) + (1/2)/(x - 3)



You can check this answer by finding the common denominator and adding the fractions back together.

Answer 2

First, factor the denominator.



By inspection, it is easy to see that x = 1 is a root of x^3 - 6x^2 + 11x - 6 = 0.

==> x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3).



Therefore,

1/(x^3 - 6x^2 + 11x - 6) = A/(x - 1) + B/(x - 2) + C/(x - 3).



Next, we clear denominators:

1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2).



x = 1 ==> A = 1/2

x = 2 ==> B = -1

x = 3 ==> C = 1/2.



In conclusion,

1/(x^3 - 6x^2 + 11x - 6) = (1/2)/(x - 1) + (-1)/(x - 2) + (1/2)/(x - 3).



I hope this helps!

Answer 3

you are able to look on your answer below the type (x^2 + 2px + p^2 + A)*(x^2+ 2x + a million + B)*(x^2 - 2*(p+a million)x + (p+a million)^2 + C) with the situations that p >= a million, A > 0, B > 0 C > 0. figuring out to 0 the coefficients of x^4, x^3, x^2 you get here equations A + B + C = 2*(a million+p+p^2) pA + B - (p+a million)C = -p(p+a million) AB + BC + CA + A(p^2 - 2p - 2) +B*(-2p^2+2p+a million) + C*(p^2+4p+a million) = 0 Edit: I forgot a time era +p^2 + (p^2+a million)*(p+a million)^2. for this reason what comes after must be marginally changed.... removing C one is decreased to (2p + a million)*A + (p + 2)*B = (p + a million)*(2p^2 + p + 2) pA^2 + 2(p+a million)AB + B^2 + A*(2p^3+4p^2-2p-2) + B*(-2p^3-2p^2+4p+2) + p*(p+a million)(p^2+4p+a million) = 0. be conscious that (p+a million)^2-p >0. So we are decreased to understanding while a hyperbola and a line, with coefficients finding on p, have rational intersections, considered one of which being with non damaging coordinates. The existence of an intersection factor with constructive coordinates looks ok for p sufficiently huge, say p>3. however the rationality area reduces to exhibiting that Q(p) is the sq. of a rational, the place Q is a polynomial of degree 10 with integer coefficients, bobbing up because of the fact the discriminant of a 2d order equation in A. So probable in basic terms Maple or Mathematica people have a huge gamble...