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Sagot :
Sure, let's go through the partial fraction decomposition step by step for the rational expression
[tex]\[ \frac{x^2+3x+5}{x^3+x}. \][/tex]
### Step 1: Factor the Denominator
First, we need to factor the denominator [tex]\(x^3 + x\)[/tex]. Notice that we can factor out an [tex]\(x\)[/tex], giving us:
[tex]\[ x^3 + x = x(x^2 + 1). \][/tex]
So, the factored form of the denominator is [tex]\(x(x^2 + 1)\)[/tex].
### Step 2: Write the Partial Fraction Decomposition
Next, we express the rational function as the sum of partial fractions. Since the denominator factors as [tex]\(x(x^2 + 1)\)[/tex], we write:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}. \][/tex]
Here, [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are constants that need to be determined.
### Step 3: Combine the Partial Fractions
To find [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we start by combining the partial fractions over a common denominator [tex]\(x(x^2 + 1)\)[/tex]:
[tex]\[ \frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)}. \][/tex]
Simplify the numerator:
[tex]\[ A(x^2 + 1) + (Bx + C)x = A x^2 + A + B x^2 x + C x = (A + B) x^2 + C x + A. \][/tex]
### Step 4: Equate Numerators
Now, we equate the numerator of this expression to the numerator of the original rational function:
[tex]\[ x^2 + 3x + 5 = (A + B)x^2 + Cx + A. \][/tex]
By comparing coefficients, we get the following system of equations:
1. [tex]\(A + B = 1\)[/tex]
2. [tex]\(C = 3\)[/tex]
3. [tex]\(A = 5\)[/tex]
### Step 5: Solve for A, B, and C
From the third equation, we have [tex]\(A = 5\)[/tex].
Substituting [tex]\(A = 5\)[/tex] into the first equation, we get:
[tex]\[ 5 + B = 1 \implies B = -4. \][/tex]
Thus, the coefficients are:
[tex]\[ A = 5, \quad B = -4, \quad C = 3. \][/tex]
### Step 6: Write the Decomposed Form
So, the partial fraction decomposition of the rational expression is:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{5}{x} + \frac{-4x + 3}{x^2 + 1}. \][/tex]
### Step 7: Check the Result Algebraically
To verify, we need to recombine these fractions and check if we get back the original rational function:
[tex]\[ \frac{5}{x} + \frac{-4x + 3}{x^2 + 1} = \frac{5(x^2 + 1) + (-4x + 3)x}{x(x^2 + 1)}. \][/tex]
Simplify the numerator:
[tex]\[ 5(x^2 + 1) + (-4x + 3)x = 5x^2 + 5 - 4x^2 + 3x = x^2 + 3x + 5. \][/tex]
So we have:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)}, \][/tex]
which confirms our original rational function.
Thus, the correct partial fraction decomposition is
[tex]\[ \boxed{\frac{5}{x} + \frac{-4x + 3}{x^2 + 1}}. \][/tex]
[tex]\[ \frac{x^2+3x+5}{x^3+x}. \][/tex]
### Step 1: Factor the Denominator
First, we need to factor the denominator [tex]\(x^3 + x\)[/tex]. Notice that we can factor out an [tex]\(x\)[/tex], giving us:
[tex]\[ x^3 + x = x(x^2 + 1). \][/tex]
So, the factored form of the denominator is [tex]\(x(x^2 + 1)\)[/tex].
### Step 2: Write the Partial Fraction Decomposition
Next, we express the rational function as the sum of partial fractions. Since the denominator factors as [tex]\(x(x^2 + 1)\)[/tex], we write:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}. \][/tex]
Here, [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are constants that need to be determined.
### Step 3: Combine the Partial Fractions
To find [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we start by combining the partial fractions over a common denominator [tex]\(x(x^2 + 1)\)[/tex]:
[tex]\[ \frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)}. \][/tex]
Simplify the numerator:
[tex]\[ A(x^2 + 1) + (Bx + C)x = A x^2 + A + B x^2 x + C x = (A + B) x^2 + C x + A. \][/tex]
### Step 4: Equate Numerators
Now, we equate the numerator of this expression to the numerator of the original rational function:
[tex]\[ x^2 + 3x + 5 = (A + B)x^2 + Cx + A. \][/tex]
By comparing coefficients, we get the following system of equations:
1. [tex]\(A + B = 1\)[/tex]
2. [tex]\(C = 3\)[/tex]
3. [tex]\(A = 5\)[/tex]
### Step 5: Solve for A, B, and C
From the third equation, we have [tex]\(A = 5\)[/tex].
Substituting [tex]\(A = 5\)[/tex] into the first equation, we get:
[tex]\[ 5 + B = 1 \implies B = -4. \][/tex]
Thus, the coefficients are:
[tex]\[ A = 5, \quad B = -4, \quad C = 3. \][/tex]
### Step 6: Write the Decomposed Form
So, the partial fraction decomposition of the rational expression is:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{5}{x} + \frac{-4x + 3}{x^2 + 1}. \][/tex]
### Step 7: Check the Result Algebraically
To verify, we need to recombine these fractions and check if we get back the original rational function:
[tex]\[ \frac{5}{x} + \frac{-4x + 3}{x^2 + 1} = \frac{5(x^2 + 1) + (-4x + 3)x}{x(x^2 + 1)}. \][/tex]
Simplify the numerator:
[tex]\[ 5(x^2 + 1) + (-4x + 3)x = 5x^2 + 5 - 4x^2 + 3x = x^2 + 3x + 5. \][/tex]
So we have:
[tex]\[ \frac{x^2 + 3x + 5}{x(x^2 + 1)}, \][/tex]
which confirms our original rational function.
Thus, the correct partial fraction decomposition is
[tex]\[ \boxed{\frac{5}{x} + \frac{-4x + 3}{x^2 + 1}}. \][/tex]
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