How do you express $\frac{3 x}{{(x - 3)}^{2}}$ $(3x) / (x-3)^2$ in partial fractions?

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How do you express $\frac{3 x}{{(x - 3)}^{2}}$ $(3x) / (x-3)^2$ in partial fractions?

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Answer 1 · 2016-10-24T16:29:10

$\frac{3 x}{{(x - 3)}^{2}} \Leftrightarrow \frac{3}{x - 3} + \frac{9}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2$

Explanation:

Partial fractions of $\frac{3 x}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2$ are in the form $\frac{A}{x - 3} + \frac{B}{{(x - 3)}^{2}}$ $A/(x-3)+B/(x-3)^2$

i.e. $\frac{3 x}{{(x - 3)}^{2}} \Leftrightarrow \frac{A}{x - 3} + \frac{B}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2hArrA/(x-3)+B/(x-3)^2$

or $\frac{3 x}{{(x - 3)}^{2}} \Leftrightarrow \frac{A (x - 3) + B}{{(x - 3)}^{2}} = \frac{A x - 3 A + B}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2hArr(A(x-3)+B)/(x-3)^2=(Ax-3A+B)/(x-3)^2$

Comparing like terms $A = 3$ $A=3$ and $- 3 A + B = 0$ $-3A+B=0$ i.e. $B = 3 A = 9$ $B=3A=9$

Hence $\frac{3 x}{{(x - 3)}^{2}} \Leftrightarrow \frac{3}{x - 3} + \frac{9}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2$

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How do you express $\frac{3 x}{{(x - 3)}^{2}}$ $(3x) / (x-3)^2$ in partial fractions?

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