Write the partial fraction decomposition of the following rational expression.
StartFraction 5 x squared minus 24 x plus 6 Over x left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis EndFraction
5x2−24x+6
x(x+2)(x+3)

We are asked to perform the partial fraction decomposition of the following rational expression:

\[
\frac{5x^2 - 24x + 6}{x(x+2)(x+3)}
\]

### Step-by-Step Process

1. **Set up the partial fraction decomposition:**

Since the denominator is a product of three linear factors \(x\), \(x+2\), and \(x+3\), the decomposition will take the form:

\[
\frac{5x^2 - 24x + 6}{x(x+2)(x+3)} = \frac{A}{x} + \frac{B}{x+2} + \frac{C}{x+3}
\]

where \(A\), \(B\), and \(C\) are constants that we need to determine.

2. **Multiply both sides by the denominator** \(x(x+2)(x+3)\) to eliminate the denominators:

\[
5x^2 - 24x + 6 = A(x+2)(x+3) + Bx(x+3) + Cx(x+2)
\]

3. **Expand the right-hand side:**

Expand each term on the right-hand side:

- \(A(x+2)(x+3) = A(x^2 + 5x + 6)\)
- \(B(x)(x+3) = B(x^2 + 3x)\)
- \(C(x)(x+2) = C(x^2 + 2x)\)

Thus, the equation becomes:

\[
5x^2 - 24x + 6 = A(x^2 + 5x + 6) + B(x^2 + 3x) + C(x^2 + 2x)
\]

4. **Combine like terms:**

Collect all terms involving \(x^2\), \(x\), and constants on the right-hand side:

\[
5x^2 - 24x + 6 = (A + B + C)x^2 + (5A + 3B + 2C)x + 6A
\]

5. **Set up a system of equations:**

Now, equate the coefficients of like powers of \(x\) on both sides:

- For \(x^2\): \(A + B + C = 5\)
- For \(x\): \(5A + 3B + 2C = -24\)
- For the constant: \(6A = 6\)

6. **Solve the system of equations:**

From the constant equation, \(6A = 6\), we find:

\[
A = 1
\]

Substitute \(A = 1\) into the other two equations:

- \(1 + B + C = 5\) → \(B + C = 4\)
- \(5(1) + 3B + 2C = -24\) → \(5 + 3B + 2C = -24\) → \(3B + 2C = -29\)

Now we have the system:

- \(B + C = 4\)
- \(3B + 2C = -29\)

Solve this system using substitution or elimination:

From \(B + C = 4\), we have \(C = 4 - B\). Substitute this into the second equation:

\[
3B + 2(4 - B) = -29
\]
\[
3B + 8 - 2B = -29
\]
\[
B + 8 = -29
\]
\[
B = -37
\]

Now, substitute \(B = -37\) into \(B + C = 4\):

\[
-37 + C = 4
\]
\[
C = 41
\]

7. **Write the partial fraction decomposition:**

Now that we have \(A = 1\), \(B = -37\), and \(C = 41\), the partial fraction decomposition is:

\[
\frac{5x^2 - 24x + 6}{x(x+2)(x+3)} = \frac{1}{x} - \frac{37}{x+2} + \frac{41}{x+3}
\]

---

Would you like more details on this decomposition or have any other questions?

Here are five related questions for practice:
1. How do you perform partial fraction decomposition when there are repeated factors in the denominator?
2. How would you decompose a rational expression with an irreducible quadratic factor in the denominator?
3. Can you find the partial fraction decomposition of \(\frac{x^2 - 5x + 4}{(x-1)(x+1)^2}\)?
4. How does partial fraction decomposition help in integration?
5. What is the difference between proper and improper fractions in partial fraction decomposition?

**Tip:** Always check if the numerator's degree is less than the denominator's when starting partial fraction decomposition. If not, perform polynomial long division first.

Write the partial fraction decomposition of the following rational expression: (5x^2 - 24x + 6) / (x(x + 2)(x + 3))

Explore how to find the partial fraction decomposition of (5x^2 - 24x + 6) / (x(x + 2)(x + 3)) with a step-by-step breakdown. Learn the process of setting up the decomposition, solving a system of equations for constants, and simplifying complex rational expressions. Ideal for Grades 10-12 students studying algebra.

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