Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
