Questions: Find the partial fraction expansion. [ fracx^2+3(x+2)^3 ]

$Find the partial fraction expansion. [ fracx^2+3(x+2)^3 ]$

Transcript text: Find the partial fraction expansion. \[ \frac{x^{2}+3}{(x+2)^{3}} \]

Solution

Solution Steps

To find the partial fraction expansion of the given rational function, we need to express it as a sum of simpler fractions. Since the denominator $(x+2)^3$ is a repeated linear factor, we will decompose the fraction into terms of the form $\frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3}$. We then solve for the constants $A$, $B$, and $C$ by equating coefficients after multiplying through by the common denominator.

Step 1: Set Up the Partial Fraction Decomposition

We start with the expression:

\[ \frac{x^2 + 3}{(x + 2)^3} \]

We want to express this as a sum of simpler fractions:

\[ \frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{(x + 2)^3} \]

Step 2: Clear the Denominator

To eliminate the denominator, we multiply both sides by $(x + 2)^3$:

\[ x^2 + 3 = A(x + 2)^2 + B(x + 2) + C \]

Step 3: Expand and Collect Terms

Expanding the right-hand side gives:

\[ x^2 + 3 = A(x^2 + 4x + 4) + B(x + 2) + C \]

This simplifies to:

\[ x^2 + 3 = Ax^2 + (4A + B)x + (4A + 2B + C) \]

Step 4: Equate Coefficients

Now, we equate the coefficients from both sides:

For $x^2$: $A = 1$
For $x$: $4A + B = 0$
For the constant term: $4A + 2B + C = 3$

Step 5: Solve the System of Equations

Substituting $A = 1$ into the second equation:

\[ 4(1) + B = 0 \implies B = -4 \]

Now substituting $A = 1$ and $B = -4$ into the third equation:

\[ 4(1) + 2(-4) + C = 3 \implies 4 - 8 + C = 3 \implies C = 7 \]

Final Answer

Thus, the partial fraction decomposition is:

\[ \frac{x^2 + 3}{(x + 2)^3} = \frac{1}{x + 2} - \frac{4}{(x + 2)^2} + \frac{7}{(x + 2)^3} \]

The final answer is:

\[ \boxed{\frac{x^2 + 3}{(x + 2)^3} = \frac{1}{x + 2} - \frac{4}{(x + 2)^2} + \frac{7}{(x + 2)^3}} \]

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