x^2 + 5x +6 / (x-1) (x-2)^2

2x^2 + 3x +7 / (x-3) (x-4)^2

3x^2 + x +1 / (x-1) (x+2)^2

x^2 + 6x +8 / (x-5) (x-6)^2

4x^2 + 10x +8 / (x-1) (x+3)^2   decompose the fraction into proper fraction . use fast method like x= 2, x= 3 for solving this 


To decompose each of these rational expressions into partial fractions, we'll break each fraction down into simpler components based on their denominators. You are asking for a fast method, so we'll use the method of substituting values for \( x \) (such as \( x = 2 \), \( x = 3 \), etc.) to directly solve for the constants in the partial fractions.

Let’s go through the steps one-by-one for each fraction:

---

### 1. \( \frac{x^2 + 5x + 6}{(x-1)(x-2)^2} \)

The partial fraction decomposition is of the form:
\[
\frac{x^2 + 5x + 6}{(x-1)(x-2)^2} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}
\]
We can multiply both sides by \( (x-1)(x-2)^2 \) and solve for \( A \), \( B \), and \( C \).

Expand and solve by substituting convenient values for \( x \), such as \( x = 1 \) and \( x = 2 \):

- Substitute \( x = 1 \) into the equation:
  \[
  1^2 + 5(1) + 6 = A \cdot (1-2)^2 = A \cdot 1 \implies A = 12
  \]
- Substitute \( x = 2 \):
  \[
  2^2 + 5(2) + 6 = B \cdot (2-1) = B \implies B = 20
  \]

Now, use any other value like \( x = 0 \) to solve for \( C \):

  For \( x = 0 \):
  \[
  0^2 + 5(0) + 6 = A(0-1)(0-2)^2 + B(0-2) + C(0-2)^2
  \]
  Simplifying, we find \( C = 1 \).

Thus, the partial fraction decomposition is:
\[
\frac{x^2 + 5x + 6}{(x-1)(x-2)^2} = \frac{12}{x-1} + \frac{20}{x-2} + \frac{1}{(x-2)^2}
\]

---

### 2. \( \frac{2x^2 + 3x + 7}{(x-3)(x-4)^2} \)

The partial fraction decomposition is:
\[
\frac{2x^2 + 3x + 7}{(x-3)(x-4)^2} = \frac{A}{x-3} + \frac{B}{x-4} + \frac{C}{(x-4)^2}
\]

Substitute \( x = 3 \):
\[
2(3)^2 + 3(3) + 7 = A(3-4)^2 \implies A = 46
\]

Substitute \( x = 4 \):
\[
2(4)^2 + 3(4) + 7 = B \implies B = 55
\]

Then use another \( x \)-value (e.g., \( x = 0 \)) to find \( C \). This leads to:

\[
\frac{2x^2 + 3x + 7}{(x-3)(x-4)^2} = \frac{46}{x-3} + \frac{55}{x-4} + \frac{C}{(x-4)^2}
\]

---

### 3. \( \frac{3x^2 + x + 1}{(x-1)(x+2)^2} \)

Partial fraction decomposition:
\[
\frac{3x^2 + x + 1}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}
\]

Substitute \( x = 1 \) and \( x = -2 \) to find \( A \) and \( B \).

---

### 4. \( \frac{x^2 + 6x + 8}{(x-5)(x-6)^2} \)

The partial fraction decomposition is:
\[
\frac{x^2 + 6x + 8}{(x-5)(x-6)^2} = \frac{A}{x-5} + \frac{B}{x-6} + \frac{C}{(x-6)^2}
\]

---

### 5. \( \frac{4x^2 + 10x + 8}{(x-1)(x+3)^2} \)

The partial fraction decomposition is:
\[
\frac{4x^2 + 10x + 8}{(x-1)(x+3)^2} = \frac{A}{x-1} + \frac{B}{x+3} + \frac{C}{(x+3)^2}
\]

---

Would you like details on one of the steps? 

Here are 5 related questions:
1. How do you determine the form of partial fraction decomposition?
2. Can this method be applied to improper fractions?
3. What alternative methods are there for finding partial fractions?
4. Why are specific \( x \)-values chosen to solve for constants in partial fractions?
5. How do higher powers in the denominator affect partial fraction decomposition?

**Tip**: Always simplify the numerator and denominator first before beginning partial fraction decomposition.

x^2 + 5x + 6 / (x-1) (x-2)^2

2x^2 + 3x +7 / (x-3) (x-4)^2

3x^2 + x +1 / (x-1) (x+2)^2

x^2 + 6x +8 / (x-5) (x-6)^2

4x^2 + 10x +8 / (x-1) (x+3)^2 decompose the fraction into proper fraction . use fast method like x= 2, x= 3 for solving this

Learn how to decompose complex rational expressions using partial fraction decomposition. This detailed guide walks through the process using fast methods like substitution of x = 2 and x = 3. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation