SOLUTION: Fill in the blanks with constants, to make a true equation: \frac{x^2 - 6x - 3}{x^3 - 4x} = ___/x + ___/(x - 2) + ___/(x + 2)

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Question 1209875: Fill in the blanks with constants, to make a true equation:
\frac{x^2 - 6x - 3}{x^3 - 4x} = ___/x + ___/(x - 2) + ___/(x + 2)
Found 2 solutions by CPhill, MathTherapy:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
To solve this problem, we'll use partial fraction decomposition.
**1. Factor the Denominator:**
The denominator of the left side is x³ - 4x. We can factor this as:
x³ - 4x = x(x² - 4) = x(x - 2)(x + 2)
**2. Set up the Partial Fraction Decomposition:**
Now we can write the given expression as a sum of simpler fractions:
(x² - 6x - 3) / \[x(x - 2)(x + 2)] = A/x + B/(x - 2) + C/(x + 2)
where A, B, and C are constants that we need to find.
**3. Clear the Denominators:**
Multiply both sides of the equation by x(x - 2)(x + 2) to get rid of the denominators:
x² - 6x - 3 = A(x - 2)(x + 2) + Bx(x + 2) + Cx(x - 2)
**4. Solve for the Constants:**
We can use various methods to solve for A, B, and C. Here are two common approaches:
* **Method 1: Substituting Convenient Values of x**
* **Let x = 0:**
* 0² - 6(0) - 3 = A(0 - 2)(0 + 2) + B(0)(0 + 2) + C(0)(0 - 2)
* -3 = A(-2)(2)
* -3 = -4A
* A = 3/4
* **Let x = 2:**
* 2² - 6(2) - 3 = A(2 - 2)(2 + 2) + B(2)(2 + 2) + C(2)(2 - 2)
* 4 - 12 - 3 = 0 + B(2)(4) + 0
* -11 = 8B
* B = -11/8
* **Let x = -2:**
* (-2)² - 6(-2) - 3 = A(-2 - 2)(-2 + 2) + B(-2)(-2 + 2) + C(-2)(-2 - 2)
* 4 + 12 - 3 = 0 + 0 + C(-2)(-4)
* 13 = 8C
* C = 13/8
* **Method 2: Expanding and Equating Coefficients**
* Expand the right side of the equation:
* x² - 6x - 3 = A(x² - 4) + B(x² + 2x) + C(x² - 2x)
* x² - 6x - 3 = Ax² - 4A + Bx² + 2Bx + Cx² - 2Cx
* x² - 6x - 3 = (A + B + C)x² + (2B - 2C)x - 4A
* Equate the coefficients of corresponding powers of x:
* Coefficient of x²: 1 = A + B + C
* Coefficient of x: -6 = 2B - 2C
* Constant term: -3 = -4A
* Solve the system of equations. From the third equation, A = 3/4. Substituting this into the first two equations and solving yields B = -11/8 and C = 13/8.
**5. Write the Result:**
Substitute the values of A, B, and C back into the partial fraction decomposition:
(x² - 6x - 3) / \[x(x - 2)(x + 2)] = (3/4)/x + (-11/8)/(x - 2) + (13/8)/(x + 2)
So the equation is:
(x² - 6x - 3) / (x³ - 4x) = 3/4 / x + (-11/8) / (x - 2) + 13/8 / (x + 2)
Therefore, the constants are:
* 3/4
* -11/8
* 13/8
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Fill in the blanks with constants, to make a true equation:

\frac{x^2 - 6x - 3}{x^3 - 4x} = ___/x + ___/(x - 2) + ___/(x + 2) 

 = 
 = 
 = 
 = 
 ---- Multiplying by LCD, x(x - 2)(x + 2)
 
 
 


(1)x^2 (- 6)x (- 3) = (A + B + C)x^2 (2B - 2C)x (- 4A)

 ----- EQUATING left-terms coefficients to right-terms coefficients

  - 3 = - 4A ---- eq (iii)
  

    1 = A + B + C -- eq (i)
     ------ Substituting  for A 
     ----- eq (iv)

  - 6 = 2B - 2C ---- eq (ii)
2(- 3) = 2(B - C)
   - 3 = B - C ----- eq (v)

  ----- Adding eqs (iv) & (iv)
 
 

      ----- eq (iv)
      -- Substituting  for B in eq (iv)
 

 =  now becomes:  =

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