Problem 14 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{9 x+21}{x^{2}+2 x-15}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression is \(\frac{3}{x+5} + \frac{2}{x-3}\).

Step 1: Factoring the Denominator

Firstly, the denominator (quadratic expression) needs to be factored. The equation to factor is \(x^{2} +2x -15\). To factor this, one needs to find two numbers that sum up to 2 (coefficient of \(x\)) and multiply to -15. These numbers are 5 and -3. So, the factorization of \(x^{2} +2x -15\) results in (x + 5)(x - 3)

Step 2: Forming the Basic Partial Fractions

To write the basic form of the partial fractions, the rational expression is split into separate fractions. In this case, it is split as: \(\frac{9x +21}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}\). A and B are constants that need to be found next.

Step 3: Solving for the Constants

To find the constants A and B, the right side of equation must be made identical to the left side. For this, the fractions on the right side have to be added together, i.e., they get a common denominator, giving the equation: \(9x + 21 = A(x - 3) + B(x + 5)\). For A, substitute \(x = 3\), the calculation results in A=3. For B, substitute \(x = -5\), the calculation results in B=2.

Key Concepts

Factoring Quadratic Expressions

In mathematical terms, factoring is breaking down a complex expression into simpler terms called factors. Let's consider how this applies specifically to quadratic expressions. A quadratic expression typically looks like this: \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
To factor this kind of expression, we're looking for two numbers that both:

• Multiply to give the product of \(a \times c\),
• And add up to \(b\).

Here's how it works using the given expression: \(x^2 + 2x - 15\). Here, \(a=1\), \(b=2\), and \(c=-15\). We need two numbers that multiply to \(-15\) and add to \(2\). The numbers \(5\) and \(-3\) satisfy these conditions.
Now, substitute these numbers back into the expression, giving us \((x+5)(x-3)\). By factoring the quadratic expression, we've broken it down into its individual binomial components, making it easier to work with further.

Equating Coefficients

Once the quadratic has been factored, the next step in partial fraction decomposition is to write the expression as a sum of simpler fractions. For our specific problem, this is:\[ \frac{9x + 21}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3} \]Now, we equate the coefficients of the terms on both sides. In essence, we're trying to match the whole expressions by comparing the coefficients of individual terms when both sides of the equation are expanded.
This leads to matching coefficients of terms with the same power of \(x\) on both sides. After expanding:\[ 9x + 21 = A(x - 3) + B(x + 5) \]You'll want to distribute \(A\) and \(B\) to get:\[ A \cdot x - 3A + B \cdot x + 5B \]Group common terms:\[ (A + B)x + (-3A + 5B) \]Here, equate the coefficients of \(x\) and the constant to those on the left side \(9x + 21\):

• For the coefficient of \(x\): \(A + B = 9\)
• For the constant: \(-3A + 5B = 21\)

This method structures the equation to solve for the constants \(A\) and \(B\).

Finding Constants

To determine values for \(A\) and \(B\), substitution can be applied for particular values of \(x\). This is often initiated by choosing strategic values that simplify the calculation.
For example, substitute \(x = 3\) into the equation \[ 9x + 21 = A(x - 3) + B(x + 5) \]This choice results in zeroing out \(B\) since \(x - 3 = 0\), giving: \[ 9(3) + 21 = A(3 - 3) + B(3 + 5) \]Simplifying gives \(A = 3\).
Similarly, substitute \(x = -5\), which eliminates \(A\) due to \(x + 5 = 0\), and you'll find:\[ 9(-5) + 21 = A(-5 - 3) + B(-5 + 5) \]This results in \(B = 2\).
By systematically choosing values that simplify the expressions, you can efficiently solve for the constants. This process is vital in partial fraction decomposition, making the original complex rational expression easier to handle.