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Sure, let's break down the process of simplifying and decomposing the expression [tex]\(\frac{x+12}{(x+1)^2(x-2)}\)[/tex] into partial fractions.
### Step 1: Express the given function
Given the expression:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} \][/tex]
### Step 2: Simplify if possible
First, we check if the expression can be simplified further. Here, since no common factors exist between the numerator and the denominator, the expression is already simplified.
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} \][/tex]
### Step 3: Partial Fraction Decomposition
We aim to break down the given expression into simpler fractions. For this, we assume:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2} \][/tex]
Our goal is to find constants [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex].
### Step 4: Combine the fractions on the right-hand side
To combine the partial fractions into a single fraction, the common denominator is [tex]\((x+1)^2(x-2)\)[/tex]:
[tex]\[ \frac{A(x+1)(x-2) + B(x-2) + C(x+1)^2}{(x+1)^2(x-2)} \][/tex]
Equating the numerators from both sides, we get:
[tex]\[ x + 12 = A(x+1)(x-2) + B(x-2) + C(x+1)^2 \][/tex]
### Step 5: Expand and solve for [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]
Expand the right-hand side:
[tex]\[ x + 12 = A(x^2 - x - 2) + B(x - 2) + C(x^2 + 2x + 1) \][/tex]
[tex]\[ x + 12 = A x^2 - A x - 2 A + B x - 2 B + C x^2 + 2 C x + C \][/tex]
Combine like terms:
[tex]\[ x + 12 = (A + C) x^2 + (-A + B + 2 C) x + (-2A - 2B + C) \][/tex]
### Step 6: Set up equations for the coefficients
By comparing coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term on both sides, we get:
1. [tex]\(A + C = 0\)[/tex]
2. [tex]\(-A + B + 2C = 1\)[/tex]
3. [tex]\(-2A - 2B + C = 12\)[/tex]
### Step 7: Solve the system of equations
From equation (1):
[tex]\[ C = -A \][/tex]
Substitute [tex]\( C = -A \)[/tex] into equations (2) and (3):
[tex]\[ -A + B + 2(-A) = 1 \][/tex]
[tex]\[ -A + B - 2A = 1 \][/tex]
[tex]\[ -3A + B = 1 \quad \text{(4)} \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -2A - 2B + (-A) = 12 \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -3A - 2B = 12 \quad \text{(5)} \][/tex]
Solve the system of linear equations (4) and (5):
From equation (4),
[tex]\[ B = 3A + 1 \quad \text{(6)} \][/tex]
Substitute (6) into (5):
[tex]\[ -3A - 2(3A + 1) = 12 \][/tex]
[tex]\[ -3A - 6A - 2 = 12 \][/tex]
[tex]\[ -9A - 2 = 12 \][/tex]
[tex]\[ -9A = 14 \][/tex]
[tex]\[ A = -\frac{14}{9} \][/tex]
Using [tex]\( A = -\frac{14}{9} \)[/tex],
[tex]\[ C = -A = \frac{14}{9} \][/tex]
From equation (6),
[tex]\[ B = 3A + 1 \][/tex]
[tex]\[ B = 3(-\frac{14}{9}) + 1 \][/tex]
[tex]\[ B = -\frac{42}{9} + 1 \][/tex]
[tex]\[ B = -\frac{42}{9} + \frac{9}{9} \][/tex]
[tex]\[ B = -\frac{33}{9} \][/tex]
[tex]\[ B = -\frac{11}{3} \][/tex]
### Step 8: Construct the partial fractions
Substituting [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] back into our partial fractions:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} = \frac{-\frac{14}{9}}{x+1} + \frac{-\frac{11}{3}}{(x+1)^2} + \frac{\frac{14}{9}}{x-2} \][/tex]
Simplifying the fractions:
[tex]\[ = -\frac{14}{9(x+1)} - \frac{11}{3(x+1)^2} + \frac{14}{9(x-2)} \][/tex]
Thus, the partial fraction decomposition of [tex]\( \frac{x+12}{(x+1)^2(x-2)} \)[/tex] is:
[tex]\[ \boxed{-\frac{14}{9(x+1)} - \frac{11}{3(x+1)^2} + \frac{14}{9(x-2)}} \][/tex]
### Step 1: Express the given function
Given the expression:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} \][/tex]
### Step 2: Simplify if possible
First, we check if the expression can be simplified further. Here, since no common factors exist between the numerator and the denominator, the expression is already simplified.
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} \][/tex]
### Step 3: Partial Fraction Decomposition
We aim to break down the given expression into simpler fractions. For this, we assume:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2} \][/tex]
Our goal is to find constants [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex].
### Step 4: Combine the fractions on the right-hand side
To combine the partial fractions into a single fraction, the common denominator is [tex]\((x+1)^2(x-2)\)[/tex]:
[tex]\[ \frac{A(x+1)(x-2) + B(x-2) + C(x+1)^2}{(x+1)^2(x-2)} \][/tex]
Equating the numerators from both sides, we get:
[tex]\[ x + 12 = A(x+1)(x-2) + B(x-2) + C(x+1)^2 \][/tex]
### Step 5: Expand and solve for [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]
Expand the right-hand side:
[tex]\[ x + 12 = A(x^2 - x - 2) + B(x - 2) + C(x^2 + 2x + 1) \][/tex]
[tex]\[ x + 12 = A x^2 - A x - 2 A + B x - 2 B + C x^2 + 2 C x + C \][/tex]
Combine like terms:
[tex]\[ x + 12 = (A + C) x^2 + (-A + B + 2 C) x + (-2A - 2B + C) \][/tex]
### Step 6: Set up equations for the coefficients
By comparing coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term on both sides, we get:
1. [tex]\(A + C = 0\)[/tex]
2. [tex]\(-A + B + 2C = 1\)[/tex]
3. [tex]\(-2A - 2B + C = 12\)[/tex]
### Step 7: Solve the system of equations
From equation (1):
[tex]\[ C = -A \][/tex]
Substitute [tex]\( C = -A \)[/tex] into equations (2) and (3):
[tex]\[ -A + B + 2(-A) = 1 \][/tex]
[tex]\[ -A + B - 2A = 1 \][/tex]
[tex]\[ -3A + B = 1 \quad \text{(4)} \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -2A - 2B + (-A) = 12 \][/tex]
[tex]\[ -2A - 2B - A = 12 \][/tex]
[tex]\[ -3A - 2B = 12 \quad \text{(5)} \][/tex]
Solve the system of linear equations (4) and (5):
From equation (4),
[tex]\[ B = 3A + 1 \quad \text{(6)} \][/tex]
Substitute (6) into (5):
[tex]\[ -3A - 2(3A + 1) = 12 \][/tex]
[tex]\[ -3A - 6A - 2 = 12 \][/tex]
[tex]\[ -9A - 2 = 12 \][/tex]
[tex]\[ -9A = 14 \][/tex]
[tex]\[ A = -\frac{14}{9} \][/tex]
Using [tex]\( A = -\frac{14}{9} \)[/tex],
[tex]\[ C = -A = \frac{14}{9} \][/tex]
From equation (6),
[tex]\[ B = 3A + 1 \][/tex]
[tex]\[ B = 3(-\frac{14}{9}) + 1 \][/tex]
[tex]\[ B = -\frac{42}{9} + 1 \][/tex]
[tex]\[ B = -\frac{42}{9} + \frac{9}{9} \][/tex]
[tex]\[ B = -\frac{33}{9} \][/tex]
[tex]\[ B = -\frac{11}{3} \][/tex]
### Step 8: Construct the partial fractions
Substituting [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] back into our partial fractions:
[tex]\[ \frac{x+12}{(x+1)^2(x-2)} = \frac{-\frac{14}{9}}{x+1} + \frac{-\frac{11}{3}}{(x+1)^2} + \frac{\frac{14}{9}}{x-2} \][/tex]
Simplifying the fractions:
[tex]\[ = -\frac{14}{9(x+1)} - \frac{11}{3(x+1)^2} + \frac{14}{9(x-2)} \][/tex]
Thus, the partial fraction decomposition of [tex]\( \frac{x+12}{(x+1)^2(x-2)} \)[/tex] is:
[tex]\[ \boxed{-\frac{14}{9(x+1)} - \frac{11}{3(x+1)^2} + \frac{14}{9(x-2)}} \][/tex]
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