`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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