(x+5)/(x-1)(x-2)(x-3). Do the partial fraction

To decompose the rational function \(\frac{x+5}{(x-1)(x-2)(x-3)}\) into partial fractions, we need to express it as a sum of simpler fractions whose denominators are the factors of the original denominator.

The steps for partial fraction decomposition are as follows:

1. **Set Up the Partial Fraction Decomposition:**
   Since the denominator has distinct linear factors, we can write:
   \[
   \frac{x+5}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}
   \]
   where \(A\), \(B\), and \(C\) are constants to be determined.

2. **Combine the Fractions:**
   Combine the right-hand side over a common denominator:
   \[
   \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} = \frac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}
   \]

3. **Equate the Numerators:**
   Since the denominators are the same, equate the numerators:
   \[
   x+5 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
   \]

4. **Expand and Collect Like Terms:**
   Expand the right-hand side:
   \[
   x+5 = A(x^2 - 5x + 6) + B(x^2 - 4x + 3) + C(x^2 - 3x + 2)
   \]
   \[
   = (A+B+C)x^2 + (-5A - 4B - 3C)x + (6A + 3B + 2C)
   \]

5. **Form a System of Equations:**
   Equate the coefficients of corresponding powers of \(x\):
   \[
   \begin{cases}
   A + B + C = 0 \\
   -5A - 4B - 3C = 1 \\
   6A + 3B + 2C = 5
   \end{cases}
   \]

6. **Solve the System of Equations:**
   Solve for \(A\), \(B\), and \(C\):

   From the first equation:
   \[
   A + B + C = 0 \implies C = -A - B
   \]

   Substitute \(C = -A - B\) into the other two equations:

   Second equation:
   \[
   -5A - 4B - 3(-A - B) = 1 \implies -5A - 4B + 3A + 3B = 1 \implies -2A - B = 1
   \]
   \[
   B = -2A - 1
   \]

   Third equation:
   \[
   6A + 3B + 2(-A - B) = 5 \implies 6A + 3B - 2A - 2B = 5 \implies 4A + B = 5
   \]
   Substitute \(B = -2A - 1\):
   \[
   4A + (-2A - 1) = 5 \implies 2A - 1 = 5 \implies 2A = 6 \implies A = 3
   \]

   Substitute \(A = 3\) into \(B = -2A - 1\):
   \[
   B = -2(3) - 1 = -6 - 1 = -7
   \]

   Substitute \(A = 3\) and \(B = -7\) into \(C = -A - B\):
   \[
   C = -3 - (-7) = -3 + 7 = 4
   \]

7. **Write the Partial Fraction Decomposition:**
   Now we have:
   \[
   A = 3, \quad B = -7, \quad C = 4
   \]
   \[
   \frac{x+5}{(x-1)(x-2)(x-3)} = \frac{3}{x-1} + \frac{-7}{x-2} + \frac{4}{x-3}
   \]
   \[
   = \frac{3}{x-1} - \frac{7}{x-2} + \frac{4}{x-3}
   \]

This is the partial fraction decomposition of the given rational function.

Would you like more details on any part of this process, or do you have any specific questions? 

Here are some related questions you might find interesting:
1. How do you solve a system of linear equations using matrix methods?
2. What are the applications of partial fraction decomposition in calculus?
3. How can you verify the partial fraction decomposition is correct?
4. What happens if the denominator has repeated linear factors?
5. How do you handle partial fractions with irreducible quadratic factors in the denominator?
6. Can partial fraction decomposition be applied to complex functions?
7. What is the role of partial fraction decomposition in Laplace transforms?
8. How do partial fractions relate to polynomial division?

**Tip:** When solving systems of equations, consistently check your work by substituting your solutions back into the original equations to ensure they satisfy all conditions.

Learn how to perform partial fraction decomposition for the rational function (x+5)/(x-1)(x-2)(x-3) with step-by-step instructions. This problem covers key concepts in partial fraction decomposition and systems of equations. Suitable for advanced high school students and calculus enthusiasts.

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