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To express \(\frac{3x - 1}{3x^2 + 4x + 1}\) in partial fractions, we can follow these steps:
1. Factor the Denominator:
First, we factor the denominator \(3x^2 + 4x + 1\). We look for two numbers whose product is \(3 \cdot 1 = 3\) and whose sum is \(4\). These numbers are \(3\) and \(1\). Therefore, the quadratic can be factored as:
[tex]\[ 3x^2 + 4x + 1 = (3x + 1)(x + 1) \][/tex]
2. Set Up the Partial Fractions:
Now we express \(\frac{3x - 1}{3x^2 + 4x + 1}\) as a sum of partial fractions. The denominator factors into \((3x + 1)(x + 1)\), so we can write:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{A}{3x + 1} + \frac{B}{x + 1} \][/tex]
where \(A\) and \(B\) are constants to be determined.
3. Combine the Terms:
To find \(A\) and \(B\), we combine the fractions on the right-hand side over the common denominator:
[tex]\[ \frac{A}{3x + 1} + \frac{B}{x + 1} = \frac{A(x + 1) + B(3x + 1)}{(3x + 1)(x + 1)} \][/tex]
4. Equate the Numerators:
The denominators are now the same, so we equate the numerators:
[tex]\[ 3x - 1 = A(x + 1) + B(3x + 1) \][/tex]
Expanding the right-hand side gives:
[tex]\[ 3x - 1 = Ax + A + 3Bx + B \][/tex]
[tex]\[ 3x - 1 = (A + 3B)x + (A + B) \][/tex]
5. Solve for \(A\) and \(B\):
We now equate the coefficients from both sides of the equation:
[tex]\[ 3 = A + 3B \quad \text{(coefficient of } x\text{)} \][/tex]
[tex]\[ -1 = A + B \quad \text{(constant term)} \][/tex]
We solve this system of equations:
- From the second equation: \(A + B = -1\)
- Substitute \( A = -1 - B \) into the first equation:
[tex]\[ 3 = (-1 - B) + 3B \][/tex]
[tex]\[ 3 = -1 + 2B \][/tex]
[tex]\[ 4 = 2B \][/tex]
[tex]\[ B = 2 \][/tex]
- Substitute \( B = 2 \) back into \( A + B = -1 \):
[tex]\[ A + 2 = -1 \][/tex]
[tex]\[ A = -3 \][/tex]
6. Write the Partial Fraction Decomposition:
With \(A = -3\) and \(B = 2\), we can write the partial fractions as:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{A}{3x + 1} + \frac{B}{x + 1} = \frac{-3}{3x + 1} + \frac{2}{x + 1} \][/tex]
Thus, the partial fraction decomposition of \(\frac{3x - 1}{3x^2 + 4x + 1}\) is:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{-3}{3x + 1} + \frac{2}{x + 1} \][/tex]
1. Factor the Denominator:
First, we factor the denominator \(3x^2 + 4x + 1\). We look for two numbers whose product is \(3 \cdot 1 = 3\) and whose sum is \(4\). These numbers are \(3\) and \(1\). Therefore, the quadratic can be factored as:
[tex]\[ 3x^2 + 4x + 1 = (3x + 1)(x + 1) \][/tex]
2. Set Up the Partial Fractions:
Now we express \(\frac{3x - 1}{3x^2 + 4x + 1}\) as a sum of partial fractions. The denominator factors into \((3x + 1)(x + 1)\), so we can write:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{A}{3x + 1} + \frac{B}{x + 1} \][/tex]
where \(A\) and \(B\) are constants to be determined.
3. Combine the Terms:
To find \(A\) and \(B\), we combine the fractions on the right-hand side over the common denominator:
[tex]\[ \frac{A}{3x + 1} + \frac{B}{x + 1} = \frac{A(x + 1) + B(3x + 1)}{(3x + 1)(x + 1)} \][/tex]
4. Equate the Numerators:
The denominators are now the same, so we equate the numerators:
[tex]\[ 3x - 1 = A(x + 1) + B(3x + 1) \][/tex]
Expanding the right-hand side gives:
[tex]\[ 3x - 1 = Ax + A + 3Bx + B \][/tex]
[tex]\[ 3x - 1 = (A + 3B)x + (A + B) \][/tex]
5. Solve for \(A\) and \(B\):
We now equate the coefficients from both sides of the equation:
[tex]\[ 3 = A + 3B \quad \text{(coefficient of } x\text{)} \][/tex]
[tex]\[ -1 = A + B \quad \text{(constant term)} \][/tex]
We solve this system of equations:
- From the second equation: \(A + B = -1\)
- Substitute \( A = -1 - B \) into the first equation:
[tex]\[ 3 = (-1 - B) + 3B \][/tex]
[tex]\[ 3 = -1 + 2B \][/tex]
[tex]\[ 4 = 2B \][/tex]
[tex]\[ B = 2 \][/tex]
- Substitute \( B = 2 \) back into \( A + B = -1 \):
[tex]\[ A + 2 = -1 \][/tex]
[tex]\[ A = -3 \][/tex]
6. Write the Partial Fraction Decomposition:
With \(A = -3\) and \(B = 2\), we can write the partial fractions as:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{A}{3x + 1} + \frac{B}{x + 1} = \frac{-3}{3x + 1} + \frac{2}{x + 1} \][/tex]
Thus, the partial fraction decomposition of \(\frac{3x - 1}{3x^2 + 4x + 1}\) is:
[tex]\[ \frac{3x - 1}{3x^2 + 4x + 1} = \frac{-3}{3x + 1} + \frac{2}{x + 1} \][/tex]
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