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To perform a partial fraction decomposition of the rational expression [tex]\(\frac{x+4}{x^2-x-30}\)[/tex], we first need to factor the denominator. Let's follow the steps.
### 1. Factor the Denominator
The first step is to find the roots of the quadratic polynomial in the denominator [tex]\(x^2 - x - 30\)[/tex].
We solve the equation:
[tex]\[x^2 - x - 30 = 0\][/tex]
This equation can be factored as:
[tex]\[ x^2 - x - 30 = (x - 6)(x + 5) \][/tex]
### 2. Decompose the Rational Expression
Now we can express the given rational expression [tex]\(\frac{x+4}{(x-6)(x+5)}\)[/tex] as a sum of two simpler fractions:
[tex]\[ \frac{x+4}{(x-6)(x+5)} = \frac{A}{x-6} + \frac{B}{x+5} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants that we need to determine.
### 3. Set Up the Equation
Next, we write:
[tex]\[ \frac{A}{x-6} + \frac{B}{x+5} = \frac{x+4}{(x-6)(x+5)} \][/tex]
Multiply through by the denominator [tex]\((x-6)(x+5)\)[/tex] to clear the fractions:
[tex]\[ A(x+5) + B(x-6) = x + 4 \][/tex]
### 4. Combine Like Terms
Distribute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] on the left side:
[tex]\[ A(x + 5) + B(x - 6) = Ax + 5A + Bx - 6B = x + 4 \][/tex]
Combine the [tex]\(x\)[/tex] terms and the constant terms:
[tex]\[ (A + B)x + (5A - 6B) = x + 4 \][/tex]
### 5. Form a System of Equations
By comparing coefficients, we obtain the following system of equations:
1. [tex]\(A + B = 1\)[/tex]
2. [tex]\(5A - 6B = 4\)[/tex]
### 6. Solve the System of Equations
To solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can use substitution or elimination.
#### Using Substitution:
From the first equation [tex]\(A + B = 1\)[/tex], solve for [tex]\(A\)[/tex]:
[tex]\[ A = 1 - B \][/tex]
Substitute [tex]\(A = 1 - B\)[/tex] into the second equation:
[tex]\[ 5(1 - B) - 6B = 4 \][/tex]
[tex]\[ 5 - 5B - 6B = 4 \][/tex]
[tex]\[ 5 - 11B = 4 \][/tex]
[tex]\[ -11B = -1 \][/tex]
[tex]\[ B = \frac{1}{11} \][/tex]
Substitute [tex]\(B = \frac{1}{11}\)[/tex] back into [tex]\(A = 1 - B\)[/tex]:
[tex]\[ A = 1 - \frac{1}{11} = \frac{11}{11} - \frac{1}{11} = \frac{10}{11} \][/tex]
### 7. Write the Partial Fractions
Having found [tex]\(A\)[/tex] and [tex]\(B\)[/tex], the partial fraction decomposition is:
[tex]\[ \frac{x + 4}{(x-6)(x+5)} = \frac{\frac{10}{11}}{x-6} + \frac{\frac{1}{11}}{x+5} \][/tex]
Simplify the expression:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Or,
[tex]\[ \frac{10}{11} \cdot \frac{1}{x-6} + \frac{1}{11} \cdot \frac{1}{x+5} = \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Which can be further simplified as:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Thus, the partial fraction decomposition of [tex]\(\frac{x+4}{(x-6)(x+5)}\)[/tex] is:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Check:
Multiplying back the partial fractions should give us the original expression:
[tex]\[ \frac{10(x+5) + 1(x-6)}{11(x-6)(x+5)} = \frac{10x + 50 + x - 6}{11(x-6)(x+5)} = \frac{11x + 44}{11(x-6)(x+5)} = \frac{x + 4}{(x-6)(x+5)} \][/tex]
So the decomposition is verified to be correct:
[tex]\[ \boxed{\frac{1}{11(x+5)} + \frac{10}{11(x-6)}} \][/tex]
### 1. Factor the Denominator
The first step is to find the roots of the quadratic polynomial in the denominator [tex]\(x^2 - x - 30\)[/tex].
We solve the equation:
[tex]\[x^2 - x - 30 = 0\][/tex]
This equation can be factored as:
[tex]\[ x^2 - x - 30 = (x - 6)(x + 5) \][/tex]
### 2. Decompose the Rational Expression
Now we can express the given rational expression [tex]\(\frac{x+4}{(x-6)(x+5)}\)[/tex] as a sum of two simpler fractions:
[tex]\[ \frac{x+4}{(x-6)(x+5)} = \frac{A}{x-6} + \frac{B}{x+5} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants that we need to determine.
### 3. Set Up the Equation
Next, we write:
[tex]\[ \frac{A}{x-6} + \frac{B}{x+5} = \frac{x+4}{(x-6)(x+5)} \][/tex]
Multiply through by the denominator [tex]\((x-6)(x+5)\)[/tex] to clear the fractions:
[tex]\[ A(x+5) + B(x-6) = x + 4 \][/tex]
### 4. Combine Like Terms
Distribute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] on the left side:
[tex]\[ A(x + 5) + B(x - 6) = Ax + 5A + Bx - 6B = x + 4 \][/tex]
Combine the [tex]\(x\)[/tex] terms and the constant terms:
[tex]\[ (A + B)x + (5A - 6B) = x + 4 \][/tex]
### 5. Form a System of Equations
By comparing coefficients, we obtain the following system of equations:
1. [tex]\(A + B = 1\)[/tex]
2. [tex]\(5A - 6B = 4\)[/tex]
### 6. Solve the System of Equations
To solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can use substitution or elimination.
#### Using Substitution:
From the first equation [tex]\(A + B = 1\)[/tex], solve for [tex]\(A\)[/tex]:
[tex]\[ A = 1 - B \][/tex]
Substitute [tex]\(A = 1 - B\)[/tex] into the second equation:
[tex]\[ 5(1 - B) - 6B = 4 \][/tex]
[tex]\[ 5 - 5B - 6B = 4 \][/tex]
[tex]\[ 5 - 11B = 4 \][/tex]
[tex]\[ -11B = -1 \][/tex]
[tex]\[ B = \frac{1}{11} \][/tex]
Substitute [tex]\(B = \frac{1}{11}\)[/tex] back into [tex]\(A = 1 - B\)[/tex]:
[tex]\[ A = 1 - \frac{1}{11} = \frac{11}{11} - \frac{1}{11} = \frac{10}{11} \][/tex]
### 7. Write the Partial Fractions
Having found [tex]\(A\)[/tex] and [tex]\(B\)[/tex], the partial fraction decomposition is:
[tex]\[ \frac{x + 4}{(x-6)(x+5)} = \frac{\frac{10}{11}}{x-6} + \frac{\frac{1}{11}}{x+5} \][/tex]
Simplify the expression:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Or,
[tex]\[ \frac{10}{11} \cdot \frac{1}{x-6} + \frac{1}{11} \cdot \frac{1}{x+5} = \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Which can be further simplified as:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Thus, the partial fraction decomposition of [tex]\(\frac{x+4}{(x-6)(x+5)}\)[/tex] is:
[tex]\[ \frac{10}{11(x-6)} + \frac{1}{11(x+5)} \][/tex]
Check:
Multiplying back the partial fractions should give us the original expression:
[tex]\[ \frac{10(x+5) + 1(x-6)}{11(x-6)(x+5)} = \frac{10x + 50 + x - 6}{11(x-6)(x+5)} = \frac{11x + 44}{11(x-6)(x+5)} = \frac{x + 4}{(x-6)(x+5)} \][/tex]
So the decomposition is verified to be correct:
[tex]\[ \boxed{\frac{1}{11(x+5)} + \frac{10}{11(x-6)}} \][/tex]
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