Problem 39 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. $$\frac{3 x^{2}-7 x+33}{x^{3}+27}$$

Step-by-Step Solution

Answer

The partial fractions are \(\frac{4.5}{x+3} + \frac{-1.5x - 2.5}{x^2 - 3x + 9}\).

Step 1: Factor the denominator

First, we need to factor the denominator. The denominator is the cubic expression \(x^3 + 27\). This is a sum of cubes, which can be factored using the formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Here, \(a = x\) and \(b = 3\), so the factorization is \((x + 3)(x^2 - 3x + 9)\).

Step 2: Set up the partial fraction decomposition

Since the denominator is now \((x+3)(x^2 - 3x + 9)\), we set up the partial fraction decomposition as \(\frac{3x^{2} - 7x + 33}{(x + 3)(x^2 - 3x + 9)} = \frac{A}{x + 3} + \frac{Bx + C}{x^2 - 3x + 9}\), where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Clear the denominators

Multiply both sides by the common denominator \((x+3)(x^2 - 3x + 9)\) to clear the fractions: \(3x^2 - 7x + 33 = A(x^2 - 3x + 9) + (Bx + C)(x + 3)\). Expand the right side to simplify further.

Step 4: Expand and collect like terms

Expand the right side: \(A(x^2 - 3x + 9) = Ax^2 - 3Ax + 9A\) and \((Bx + C)(x + 3) = Bx^2 + 3Bx + Cx + 3C\). Combine these to get \(Ax^2 - 3Ax + 9A + Bx^2 + 3Bx + Cx + 3C\). Collect like terms: \((A + B)x^2 + (-3A + 3B + C)x + (9A + 3C)\).

Step 5: Match coefficients to solve for constants

Match the coefficients of corresponding powers of \(x\) on both sides: 1. For \(x^2\): \(A + B = 3\).2. For \(x\): \(-3A + 3B + C = -7\).3. For constant term: \(9A + 3C = 33\).Solve these equations simultaneously to find \(A\), \(B\), and \(C\).

Step 6: Solve the system of equations

From the equations obtained: 1. \(A + B = 3\)2. \(-3A + 3B + C = -7\)3. \(9A + 3C = 33\)Solving these gives: - From equation 1: \(B = 3 - A\) - From equation 2: substitute \(B\), we have \(-3A + 3(3 - A) + C = -7\) simplifies to \(9 - 3A + C = -7\) or \(C = -7 - 9 + 3A = 3A - 16\) - Substitute \(C\) in equation 3: \(9A + 3(3A - 16) = 33\) gives \(9A + 9A - 48 = 33\) simplifies to \(18A = 81\), so \(A = 4.5\). - Substitute \(A\) into \(B = 3 - A\): \(B = 3 - 4.5 = -1.5\) - Substitute \(A\) into \(C = 3A - 16\): \(C = 3(4.5) - 16 = 13.5 - 16 = -2.5\).

Step 7: Write the partial fraction decomposition

The values of \(A\), \(B\), and \(C\) give us the decomposition: \[ \frac{3x^2 - 7x + 33}{x^3 + 27} = \frac{4.5}{x+3} + \frac{-1.5x - 2.5}{x^2 - 3x + 9} \].

Key Concepts

Irreducible Quadratic Factor

When dealing with partial fraction decomposition, it's essential to recognize irreducible quadratic factors within the denominator of a rational expression. In our problem, the denominator factored into two parts:

• A linear factor \(x + 3\)
• An irreducible quadratic factor \(x^2 - 3x + 9\)

An irreducible quadratic factor is a quadratic expression that cannot be factored further using real numbers. This means that there are no real roots or factors that simplify it further. In the context of partial fraction decomposition, when you encounter an irreducible quadratic factor in the denominator, it's set up with a numerator of one degree less, such as \(Bx + C\). Understanding this concept is vital when it comes to setting up the correct expression for partial fraction decomposition, enabling you to split a complex expression into simpler fractions.

Cubic Expression

A cubic expression is a polynomial where the highest power of the variable is three. For example, the expression \(x^3 + 27\) is a perfect cubic expression because it can be expressed as \(x^3 + 3^3\). Cubic expressions like these often appear in partial fraction decomposition, requiring factoring to simplify.To factor a cubic expression like \(x^3 + 27\), you use the formula for the sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Applying this formula, with \(a = x\) and \(b = 3\), lets you rewrite the denominator as \( (x + 3)(x^2 - 3x + 9) \). Factoring cubic expressions is crucial, as it breaks down the problem and makes handling each component with partial fraction decomposition easier.

Matching Coefficients

Matching coefficients is a method used to find unknown constants in polynomial expressions. This technique involves equating the coefficients of corresponding powers of variables from both sides of an equation. In the context of partial fraction decomposition, once you've expanded the equation after distributing terms, you'll use matching coefficients to solve for constants like \(A, B,\) and \(C\).Suppose you have:

• \(Ax^2 - 3Ax + 9A + Bx^2 + 3Bx + Cx + 3C\)

You combine like terms and then match these with the original polynomial's coefficients:

• For \(x^2\): \(A + B = 3\)
• For \(x\): \(-3A + 3B + C = -7\)
• For the constant term: \(9A + 3C = 33\)

By solving the resulting system, you can determine the values of \(A, B,\) and \(C\). This process is fundamental to breaking down complex expressions into simpler fractions.

Solving Equations System

Solving a system of equations is a mathematical process wherein you find the values of variables that satisfy multiple equations simultaneously. During partial fraction decomposition, this step comes after setting up expressions based on factored forms and applying the matching coefficients approach.In the exercise example, you ended up with a system of equations:

• \(A + B = 3\)
• \(-3A + 3B + C = -7\)
• \(9A + 3C = 33\)

The key is solving for \(A, B,\) and \(C\) by using substitution or elimination methods:

• Substitute \(B = 3 - A\) from the first equation into the others.
• Use simplification and substitute values back to solve for one variable at a time.
• Continue substituting into the remaining equations until all values of \(A, B,\) and \(C\) are known.

Solving the system of equations accurately is critical for determining the correct constants, leading to the valid final form of the partial fraction decomposition.