Problem 11 - Precalculus

Question

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{3 x^{2}+17 x-38}{x^{3}-3 x^{2}-4 x+12}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{-2}{x-2} + \frac{8}{x-3} + \frac{-3}{x+2} \).

Step 1: Factor the Denominator

The given rational expression is \( \frac{3x^2 + 17x - 38}{x^3 - 3x^2 - 4x + 12} \). We need to factor the cubic polynomial in the denominator. To do this, we look for roots of the polynomial \( x^3 - 3x^2 - 4x + 12 \). Using the Rational Root Theorem and trying possible roots like \( x = 1, -1, 2, -2, 3, -3, 4 \), we find that \( x = 2 \) is a root. \( x = 2 \) gives us the factor \( (x - 2) \). Performing synthetic division, we divide \( x^3 - 3x^2 - 4x + 12 \) by \( x - 2 \) to find:\[x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6)\]Next, factor \( x^2 - x - 6 \) as well, which gives:\[x^2 - x - 6 = (x - 3)(x + 2)\]Thus, the full factorization is:\[x^3 - 3x^2 - 4x + 12 = (x - 2)(x - 3)(x + 2)\]

Step 2: Determine the Partial Fraction Form

With the denominator factored as \( (x - 2)(x - 3)(x + 2) \), we can express the rational expression in terms of partial fractions. The general form of the partial fraction decomposition is:\[\frac{3x^2 + 17x - 38}{(x - 2)(x - 3)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x - 3} + \frac{C}{x + 2}\]where \( A, B, \) and \( C \) are constants to be determined.

Step 3: Determine the Constants

To find the constants \( A, B, \) and \( C \), multiply through by the denominator \( (x - 2)(x - 3)(x + 2) \) to eliminate the fractions:\[3x^2 + 17x - 38 = A(x - 3)(x + 2) + B(x - 2)(x + 2) + C(x - 2)(x - 3)\]To find \( A \), substitute \( x = 2 \) into the equation:\[3(2)^2 + 17(2) - 38 = A((2 - 3)(2 + 2)) + 0 + 0\]\[12 + 34 - 38 = A(-1 \cdot 4)\]\[8 = -4A \Rightarrow A = -2\]To find \( B \), substitute \( x = 3 \):\[3(3)^2 + 17(3) - 38 = 0 + B((3 - 2)(3 + 2)) + 0\]\[27 + 51 - 38 = B(1 \cdot 5)\]\[40 = 5B \Rightarrow B = 8\]To find \( C \), substitute \( x = -2 \):\[3(-2)^2 + 17(-2) - 38 = 0 + 0 + C((-2 - 2)(-2 - 3))\]\[12 - 34 - 38 = C(-4 \cdot -5)\]\[-60 = 20C \Rightarrow C = -3\]

Step 4: Write the Final Partial Fraction Decomposition

Substitute the values of \( A, B, \) and \( C \) back into the partial fraction form:\[\frac{3x^2 + 17x - 38}{(x - 2)(x - 3)(x + 2)} = \frac{-2}{x - 2} + \frac{8}{x - 3} + \frac{-3}{x + 2}\]

Key Concepts

Rational Expression

A rational expression is similar to a fraction, but instead of plain numbers, it consists of polynomials in the numerator and denominator. Imagine it as a division problem where both the top and bottom are made of expressions like ax + b.
For instance, in the expression \(\frac{3x^2 + 17x - 38}{x^3 - 3x^2 - 4x + 12}\), the numerator is a quadratic polynomial, and the denominator is a cubic polynomial.

Rational expressions are convenient for solving equations involving polynomials. They can often be simplified by factoring out the polynomials. Remember, simplification can make it easier to perform additional operations like addition, subtraction, or finding partial fractions.

• Simplifying involves canceling out identical factors from the numerator and denominator.
• Always check for restrictions where the denominator equals zero because division by zero is undefined.
• In order to simplify or find decompositions, it's crucial to work with factored forms first.

Polynomial Factorization

Polynomial Factorization is all about expressing a polynomial as a product of its factors. It's like breaking down a complicated expression into pieces that multiply to give the original polynomial.
In our case, the factorization applies to \(x^3 - 3x^2 - 4x + 12\). First, we identify the roots of the polynomial using techniques like the Rational Root Theorem. We find out that \(x = 2\) is a root, which immediately gives us a factor, \((x - 2)\).

• After finding one root, use Synthetic Division to divide the polynomial by this factor.
• Continue to factor the resulting quotient until all parts are expressed in simple terms.
• Ultimately, \(x^3 - 3x^2 - 4x + 12\) becomes \((x - 2)(x - 3)(x + 2)\).

Polynomial Factorization is fundamental since it simplifies rational expressions and helps in finding solutions to polynomial equations. It can be viewed as reverse-engineering a multiplication problem. It is similar to converting a product back into its multipliers.

Synthetic Division

Synthetic Division is a simplified form of long division for polynomials, especially useful for dividing by linear expressions like \(x - c\). It saves time and reduces the complexity involved in regular polynomial division.
For example, when we found \(x = 2\) was a root of \(x^3 - 3x^2 - 4x + 12\), we needed to divide the polynomial by \(x - 2\) to fully factor it. Synthetic division became our tool of choice.

• Write the coefficients of the polynomial in a row.
• Use the root (here \(x = 2\), meaning shift by 2) to simplify step-by-step.
• Perform operations smoothly from left to right, essentially transforming the division into a compact series of arithmetic steps.

This method yields the remaining polynomial factors rapidly, as demonstrated, leaving us with \(x^2 - x - 6\), which we further factor. This makes synthetic division not just efficient but also easy to understand, enabling quick evaluations and verifications in polynomial contexts.