# `3/(x^2 - 3x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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`3/(x^2-3x)`

To decompose a fraction, factor the denominator.

`3/(x(x-3))`

Then,  write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction.

`A/x`  and   `B/(x-3)`

Add these two fractions and set it equal to the given rational expression.

`3/(x(x-3)) = A/x + B/(x-3)`

To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.

`x(x-3)*3/(x(x-3))=(A/x+B/(x-3))*x(x-3)`

`3=A(x-3) + Bx`

Then, plug-in the roots of each factor.

For the factor (x-3), its root is x=3.

`3=A(3-3) + B(3)`

`3=3B`

`3/3=(3B)/3`

`1=B`

For the factor x, its root is x=0.

`3=A(0-3)+B(0)`

`3=-3A`

`3/(-3)=(-3A)/(-3)`

`-1=A`

So the given rational expression decomposes to:

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

This can be re-written as:

`1/(x-3) - 1/x`

To check, express the two fractions with same denominators.

`1/(x-3)-1/x=1/(x-3)*x/x - 1/x*(x-3)/(x-3)=x/(x(x-3)) - (x-3)/(x(x-3))`

Now that they have same denominators, proceed to subtract them.

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

Therefore,  `3/(x^2-3x) = 1/(x-3)-1/x` .

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