# Express the following as partial fractions: `(3)/((x+1)(x+2)(x+3))`

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

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

When x = -1

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

`3 = 2A+0+0`

`A = 3/2`

When x = -2

`3 = B(-2+1)(-2+3)`

`3 = -B`

`B = -3`

When x = -3

`3 = C(-3+1)(-3+2)`

`3 = 2C`

`C = 3/2`

*So by partial fractions;*

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

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