Question

Resolve in partial fraction

3/ (x + 1)(x + 2) )(x+3)​

Answer 1

Given :-

3/[(x+1)(x+2)(x+3)]

To find :-

Partial factorisation

Solution :-

Given that 3/[(x+1)(x+2)(x+3)]

Let 3/[(x+1)(x+2)(x+3)]

= [A/(x(1)]+[B/(x+2)]+[c/(x+3)] -------(1)

=> 3/[(x+1)(x+2)(x+3)] = [A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)]/[(x+1)(x+2)(x+3)]

On cancelling [(x+1)(x+2)(x+3)] both sides then

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

i) Put x = -1 then (2) becomes

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

=> 3 = A(1)(2)+B(0)(2)+C(0)(1)

=> 3 = 2A+0+0

=> 3 = 2A

=> A = 3/2

ii) Put x = -2 then (2) becomes

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

=> 3 = A(0)(1)+B(-1)(1)+C(-1)(0)

=> 3 = A(0)+B(-1)+C(0)

=> 3 = 0-B+0

=> 3 = - B

=> B = -3

iii) Put x = -3 then (2) becomes

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

=> 3 = A(-1)(0)+B(-2)(0)+C(-2)(-1)

=> 3 = A(0)+B(0)+C(2)

=> 3 = 0+0+2C

=> 3 = 2C

=> C = 3/2

On substituting the value of A, B and C in (1) then

3/[(x+1)(x+2)(x+3)]

3/[(x+1)(x+2)(x+3)]=[(3/2)/(x(1)]+[(-3)/(x+2)]+[(3/2)/(x+3)]

(or)

3/[(x+1)(x+2)(x+3)]

= [3/2(x+1)]-[3/(x+2)]+[3/2(x+3)]

Answer 2

$\bold{ANSWER≈}$

Given :

3/[(x+1)(x+2)(x+3)]

To find :

Partial factorisation

Solution :

Given that 3/[(x+1)(x+2)(x+3)]

Let 3/[(x+1)(x+2)(x+3)]

= [A/(x(1)]+[B/(x+2)]+[c/(x+3)] ----(1)

=> 3/[(x+1)(x+2)(x+3)] = [A(x+2)(x+3)+B(x+1) (x+3)+C(x+1)(x+2)]/[(x+1)(x+2)(x+3)]

On cancelling [(x+1)(x+2)(x+3)] both sides

then

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

i) Put x = -1 then (2) becomes

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

=> 3 = A(1)(2)+B(0)(2)+C(O)(1)

=> 3 = 2A+0+0

=> 3 = 2A

=> A = 3/2

ii) Put x = -2 then (2) becomes 3 = A(-2+2)(-2+3)+B(-2+1)(-2+3)+C(-2+1) (-2+2)

=> 3 = A(0)(1)+B(-1)(1)+C(-1)(0)

=> 3 = A(0)+B(-1)+C(0)

=> 3 = 0-B+0

=> 3 = - B

=> B = -3

iii) Put x = -3 then (2) becomes

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

=> 3 = A(-1)(0)+B(-2)(0)+C(-2)(-1)

=> 3 = A(0)+B(0)+C(2)

=> 3 = 0+0+2C

=> 3 = 2C

=> C = 3/2

On substituting the value of A, B and C in (1) then

3/[(x+1)(x+2)(x+3)]

3/[(x+1)(x+2)(x+3)]=[(3/2)/(x(1)]+[(-3)/(x+2)] +[(3/2)/(x+3)]

(or)

3/[(x+1)(x+2)(x+3)]

= [3/2(x+1)]-[3/(x+2)]+[3/2(x+3)]