Solutions - 0060-06

(1) Observe that $x^2+9=(x+3)(x-3)$:

$$\gg \gg \frac{x+2}{(x^2+2){(x-1)}^3(x-3)(x+3)}$$

On the other hand, $x^2+2$ cannot be factored further. (Its zeros are imaginary.)

Now all denominator factors are either linear or irreducible quadratic.

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(2) Write the partial fractions general form equation:

$$\begin{array}{c}\frac{x+2}{(x^2+2){(x-1)}^3(x^2-9)}\\ \\ \gg \gg \frac{Ax+B}{x^2+2}+\frac{C}{x-1}+\frac{D}{{(x-1)}^2}+\frac{E}{{(x-1)}^3}+\frac{F}{x-3}+\frac{G}{x+3}\end{array}$$

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(3) Notice a few things:

• Quadratic $x^2+2$ acquires linear term $Ax+B$ on top
• Linear $x-1$ is to 3rd power so it has repetition up to 3rd power
• Linear $x-3$ and $x+3$ are only to 1st power.