Set up the partial fraction decomposition.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Set up the partial fraction decomposition.
The denominator has a linear factor $(x-3)$ and an irreducible quadratic factor $(x^2+9)$.
$$ \frac{x^2}{(x^2+9)(x-3)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+9} $$
Step 2: Multiply both sides by the common denominator $(x^2+9)(x-3)$.
$$ x^2 = A(x^2+9) + (Bx+C)(x-3) $$
Step 3: Solve for $A$, $B$, and $C$.
Substitute $x=3$:
$$ 3^2 = A(3^2+9) + (B(3)+C)(3-3) $$
$$ 9 = A(9+9) + 0 $$
$$ 9 = 18A $$
$$ A = \frac{9}{18} = \frac{1}{2} $$
Expand the equation:
$$ x^2 = Ax^2 + 9A + Bx^2 - 3Bx + Cx - 3C $$
$$ x^2 = (A+B)x^2 + (-3B+C)x + (9A-3C) $$
Equate coefficients:
Coefficient of $x^2$:
$$ 1 = A+B $$
Substitute $A=\frac{1}{2}$:
$$ 1 = \frac{1}{2} + B $$
$$ B = 1 - \frac{1}{2} = \frac{1}{2} $$
Coefficient of $x$:
$$ 0 = -3B+C $$
Substitute $B=\frac{1}{2}$:
$$ 0 = -3\left(\frac{1}{2}\right) + C $$
$$ C = \frac{3}{2} $$
Constant term (for verification):
$$ 0 = 9A-3C $$
Substitute $A=\frac{1}{2}$ and $C=\frac{3}{2}$:
$$ 0 = 9\left(\frac{1}{2}\right) - 3\left(\frac{3}{2}\right) $$
$$ 0 = \frac{9}{2} - \frac{9}{2} $$
$$ 0 = 0 $$
The values are consistent.
Step 4: Write the partial fraction decomposition.
$$ \frac{x^2}{(x^2+9)(x-3)} = \frac{\frac{1}{2}}{x-3} + \frac{\frac{1}{2}x+\frac{3}{2}}{x^2+9} $$
$$ \frac{x^2}{(x^2+9)(x-3)} = \frac{1}{2(x-3)} + \frac{x+3}{2(x^2+9)} $$
The final answer is $\boxed{\frac{1}{2(x-3)} + \frac{x+3}{2(x^2+9)}}$.
Step 1: Set up the partial fraction decomposition.
The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+4)$.
$$ \frac{1}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4} $$
Step 2: Multiply both sides by the common denominator $(x-1)(x^2+4)$.
$$ 1 = A(x^2+4) + (Bx+C)(x-1) $$
Step 3: Solve for $A$, $B$, and $C$.
Substitute $x=1$:
$$ 1 = A(1^2+4) + (B(1)+C)(1-1) $$
$$ 1 = A(1+4) + 0 $$
$$ 1 = 5A $$
$$ A = \frac{1}{5} $$
Expand the equation:
$$ 1 = Ax^2 + 4A + Bx^2 - Bx + Cx - C $$
$$ 1 = (A+B)x^2 + (-B+C)x + (4A-C) $$
Equate coefficients:
Coefficient of $x^2$:
$$ 0 = A+B $$
Substitute $A=\frac{1}{5}$:
$$ 0 = \frac{1}{5} + B $$
$$ B = -\frac{1}{5} $$
Coefficient of $x$:
$$ 0 = -B+C $$
Substitute $B=-\frac{1}{5}$:
$$ 0 = -\left(-\frac{1}{5}\right) + C $$
$$ 0 = \frac{1}{5} + C $$
$$ C = -\frac{1}{5} $$
Constant term (for verification):
$$ 1 = 4A-C $$
Substitute $A=\frac{1}{5}$ and $C=-\frac{1}{5}$:
$$ 1 = 4\left(\frac{1}{5}\right) - \left(-\frac{1}{5}\right) $$
$$ 1 = \frac{4}{5} + \frac{1}{5} $$
$$ 1 = \frac{5}{5} $$
$$ 1 = 1 $$
The values are consistent.
Step 4: Write the partial fraction decomposition.
$$ \frac{1}{(x-1)(x^2+4)} = \frac{\frac{1}{5}}{x-1} + \frac{-\frac{1}{5}x-\frac{1}{5}}{x^2+4} $$
$$ \frac{1}{(x-1)(x^2+4)} = \frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)} $$
The final answer is $\boxed{\frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)}}$.
Step 1: The given equation is:
$$ \frac{6x+7}{(x+3)(x+1)} = \frac{A}{x+3} + \frac{B}{x+1} $$
Step 2: Multiply both sides by the common denominator $(x+3)(x+1)$.
$$ 6x+7 = A(x+1) + B(x+3) $$
Step 3: Solve for $A$ and $B$.
To find $B$, substitute $x=-1$:
$$ 6(-1)+7 = A(-1+1) + B(-1+3) $$
$$ -6+7 = A(0) + B(2) $$
$$ 1 = 2B $$
$$ B = \frac{1}{2} $$
To find $A$, substitute $x=-3$:
$$ 6(-3)+7 = A(-3+1) + B(-3+3) $$
$$ -18+7 = A(-2) + B(0) $$
$$ -11 = -2A $$
$$ A = \frac{-11}{-2} = \frac{11}{2} $$
Step 4: Evaluate $(B-A)$.
$$ B-A = \frac{1}{2} - \frac{11}{2} $$
$$ B-A = \frac{1-11}{2} $$
$$ B-A = \frac{-10}{2} $$
$$ B-A = -5 $$
The final answer is $\boxed{-5}$.
