Find the partial fraction decomposition of each. 6) $$ \frac{12}{x^{3}+8 x} $$

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UpStudy AI · Accepted Answer

The given expression is

$$
\frac{12}{x^3+8x}.
$$

**Step 1. Factor the denominator.**

Factor out $$x$$:

$$
x^3+8x = x(x^2+8).
$$

So, the expression becomes

$$
\frac{12}{x(x^2+8)}.
$$

**Step 2. Set up the partial fractions.**

Since the denominator is $$x(x^2+8)$$, we write

$$
\frac{12}{x(x^2+8)} = \frac{A}{x} + \frac{Bx+C}{x^2+8},
$$

where $$A$$, $$B$$, and $$C$$ are constants to be determined.

**Step 3. Clear the denominators by multiplying both sides by $$x(x^2+8)$$:**

$$
12 = A(x^2+8) + (Bx+C)x.
$$

**Step 4. Expand and collect like terms.**

Expanding the right-hand side:

$$
12 = Ax^2 + 8A + Bx^2 + Cx.
$$

Combine like terms:

$$
12 = (A+B)x^2 + Cx + 8A.
$$

**Step 5. Set up the system of equations.**

Since the left-hand side is a constant (with no $$x$$ and $$x^2$$ terms), the coefficients of $$x^2$$ and $$x$$ must be zero. Thus, we have:

1. Coefficient of $$x^2$$: $$A+B = 0$$
2. Coefficient of $$x$$: $$C = 0$$
3. Constant term: $$8A = 12$$

**Step 6. Solve for $$A$$, $$B$$, and $$C$$.**

From the constant term:

$$
8A = 12 \implies A = \frac{12}{8} = \frac{3}{2}.
$$

From $$A+B=0$$:

$$
B = -A = -\frac{3}{2}.
$$

And from the coefficient of $$x$$:

$$
C = 0.
$$

**Step 7. Write the final partial fraction decomposition.**

Substitute the values of $$A$$, $$B$$, and $$C$$ into the partial fractions:

$$
\frac{12}{x(x^2+8)} = \frac{\frac{3}{2}}{x} + \frac{-\frac{3}{2}x}{x^2+8}.
$$

This can be simplified to:

$$
\frac{12}{x(x^2+8)} = \frac{3}{2x} - \frac{3x}{2(x^2+8)}.
$$
The partial fraction decomposition of $$ \frac{12}{x^{3}+8x} $$ is: $$ \frac{12}{x^{3}+8x} = \frac{3}{2x} - \frac{3x}{2(x^{2}+8)} $$

Find the partial fraction decomposition of each.
6)

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