It let's you achieve the different powers on the numerator. Eg To get the same denominator, for the first part we must multiply by (x-3)^2, giving us an x^2, then for the 2nd part we must multiply by (x-3), giving us the x, and 3rd part gives us our constant. Subtracting and multiplying these in the end gives up our original expression.
Lets see how this works out in this example.
\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{k_1}{(x-3)} + \frac{k_2}{(x-3)^2} + \frac{k_3}{(x-3)^3}
Multiply to get a common denominator.
\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot(x-3)^2 + k_2\cdot(x-3) + k_3}{(x-3)^3}
Expand.
\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot x^2 - k_1 \cdot 6x +9 \cdot k_1 +k_2 \cdot x -k_2 \cdot 3 + k_3}{(x-3)^3}
Simplify Like terms etc.
\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1\cdot x^2 + (k_2 -6k_1) \cdot x + (k_3 -3k_2 + 9)}{(x-3)^3}
Phew, that was a bit of confusing tex[/tex]. <br />
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Anyway, The simplest way to solve is to equate co-efficients on both sides :D.<br />
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So we get:<br />
k_1 = 1<br />
(k_2 -6k_1)=4<br />
(k_3 -3k_2 + 9) =7<br />
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YAY! Simultaneous Equations! <br />
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For the 2nd equation, since k_1=1[/tex], the equation simplifies to<br />
k_2 -6=4<br />
k_2 =10<br />
Put that into equation 3. <br />
k_3 -30 + 9 = 7<br />
k_3=28<br />
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YAY we have our question solved!<br />
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\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{1}{(x-3)} + \frac{10}{(x-3)^2} + \frac{28}{(x-3)^3}<br />
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HOORAH!
