The partial fraction decomposition of $$ \frac{6 x^{2}+3 x+82}{x^{3}+2 x^{2}+16 x+32} $$ can be $$ \frac{5}{f(x)}+\frac{x+1}{g(x)} $$, where

Question

UpStudy AI · Accepted Answer

We start with the rational function

$$
\frac{6x^2+3x+82}{x^3+2x^2+16x+32}.
$$

**Step 1. Factor the Denominator**

Notice that the denominator can be factored by grouping. Write

$$
x^3+2x^2+16x+32 = x^2(x+2) + 16(x+2) = (x+2)(x^2+16).
$$

**Step 2. Set Up the Partial Fractions**

The standard partial fraction decomposition for

$$
\frac{6x^2+3x+82}{(x+2)(x^2+16)}
$$

is

$$
\frac{A}{x+2}+\frac{Bx+C}{x^2+16}.
$$

According to the problem statement, the decomposition is given by

$$
\frac{5}{f(x)}+\frac{x+1}{g(x)},
$$

which suggests that $$A=5$$ and that $$f(x)=x+2$$ and $$g(x)=x^2+16$$.

**Step 3. Verify the Decomposition**

We write

$$
\frac{6x^2+3x+82}{(x+2)(x^2+16)} = \frac{5}{x+2}+\frac{x+1}{x^2+16}.
$$

To verify, combine the right-hand side over a common denominator:

$$
\frac{5(x^2+16) + (x+1)(x+2)}{(x+2)(x^2+16)}.
$$

First, expand and simplify the numerator:

- Expand $$5(x^2+16)$$:

$$
5x^2 + 80.
$$

- Expand $$(x+1)(x+2)$$:

$$
x^2 + 3x + 2.
$$

- Sum the two results:

$$
5x^2 + 80 + x^2 + 3x + 2 = 6x^2 + 3x + 82.
$$

Thus, the numerator matches exactly with the numerator of the original fraction.

**Conclusion**

The functions $$f(x)$$ and $$g(x)$$ are:

$$
f(x)=x+2 \quad \text{and} \quad g(x)=x^2+16.
$$
$$ f(x) = x + 2 $$ and $$ g(x) = x^2 + 16 $$.

The partial fraction decomposition of \( \frac{6 x^{2}+3 x+82}{x^{3}+2 x^{2}+16 x+32} \) can be \( \frac{5}{f(x)}+\frac{x+1}{g(x)} \), where