Resolve Into Partical Fraction~

... Taking L.C.M. on R.H.S.

3x-1=A(2x+1)(x-1)+B (x-4)(x-1)+C(x-4)(2x+1). ....(2)

To Find of A, we have to make B & C zero so put x=4 in eqn (2)

we get,

3(4)-1= A(2(4)+1)(4-1)+B(4-4)(4-1)+C(4-4)(2(4)+1)

.·. 11 = A(9) (3) +B (0) + C (0)= A(27). .·.A=11/27

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Now, to find value of B, we have to make A nd C zero so the common bracket is

2x+1 .·.2x+1=0 .·.2x=-1. .·.x=-1/2

So that put x= -1/2 in eqn (2) we get,

Now, for find the value of C, we have to make A & B zero so common bracket between A & B is x-1

.·.put x= 1 in Eqn (2) we get,

3(1)-1=A(2(1)+1)(1-1)+B(1-4)(1-1)+C(1-4)(2(1)+1)

.·. 3-1 = A(0) + B=(0)+ C (-3)(3)

.·. Eqn (1) becomes

step-by-step explanation:

its 512 because parenthesis and to the 2nd power is 512 so it will be 512