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