Question

Resolve X^4/(x-1) (x-2)
into partial fractions


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Answer 1

Step-by-step explanation:

Given :-

x^4/(x-1)(x-2)

To find :-

Resolve it into partial fractions ?

Solution :-

Given that :

x^4/(x-1)(x-2)

Let x^4/(x-1)(x-2) = A/(x-1) + B(x-2)-------(1)

=> x^4/(x-1)(x-2) = [A(x-2)+B(x-1)]/(x-1)(x-2)

On cancelling (x-1)(x-2) both sides then

=> x^4 = A(x-2)+B(x-1) -------(2)

Put x = 1 then

=> 1^4 = A(1-2)+B(1-1)

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

=> 1 = -A +0

=> 1 = -A

=> - A = 1

=> A = -1 ----------(3)

Put x = 2 in (1) then

=> 2^4= A(2-2)+B(2-1)

=> 16 = A(0)+B(1)

=> 16=0+B

=> 16 = B

=>>B=16---------(4)

On Substituting the values of A and B in (1) then

=> x^4 /(x-1)(x-2)=[ (-1)/(x-1)]+[16/(x-2)]

Answer :-

x^4 /(x-1)(x-2)= [ (-1)/(x-1)]+[16/(x-2)]