Express 1/((x^2)(1-3x) in partial fractions.

This is a step above a partial fraction question seen in core 1 and core 2, made mkre difficult due to the double (repeated) root of x. To solve this we must consider that there is 3 separate roots, x, x^2 and (1-3x). We know that the partial fraction must be in the form of 1 = A/x +B/x^2 + C/(1-3x). To remove the denominators we times through every term on the right hand side of the equation by x^2 and (1-3x) to give us 1 = Ax(1-3x) + B(1-3x) + Cx^2 where A,B,C are constants we will now find. To find the constants we elimate terms by selecting values of x which will reduce the equation by making terms equal to zero.

Taking x=0 both the A and C terms are reduced to zero and we find B=1

Taking x=1/3 both the A and B terms are reduced to zero and we find C/9 = 1 which gives us C = 9

To find the A constant requires slightly more work as it contains both an x and (1-3x) this means we cannot remove the B and C term directly without also removing the A term. Therefore we choose any value of other than x=0 or x=1/3 and use the values for B and C found about to calculate A. The easiest value of x should always be taken to avoid extra cancellation, in this case it is x=1.

Taking x=1 and subbing in B = 1 and C = 9 we find that A =3

Now we have the 3 constants required we go back to what we said the form of the partial fractions is above:  Ax(1-3x) + B(1-3x) + Cx^2 and we can sub in these values. This gives us our final answer.

3/x + 1/(x^2) + 9/(1-3x)