Question

Find the sum or difference. \(\frac{x^2-5}{x^2+5 x-14}-\frac{x+3}{x+7}\)

Short answer

After performing the subtraction of the two fractions, the final answer is \(\frac{2x + 1}{(x-2)(x+7)}\).

Step-by-step solution

Factor the Denominators

The first step is to factor the denominators where possible. The denominator \(x^2+5x-14\) can be factored to \((x-2)(x+7)\) and \(x+7\) can stay as it is.

Find a Common Denominator

In order to subtract fractions, they must have the same denominator. In this case, the common denominator would be \((x-2)(x+7)\). So, multiply top and bottom of the second term by \((x-2)\) to get the common denominator.

Combine the Fractions

Under a common denominator, the two fractions can now be subtracted. The new equation becomes: \(\frac{(x^2-5) - (x+3)(x-2)}{(x-2)(x+7)}\). It's important to pay attention to the sign. Because we're subtracting the second fraction, all terms in the second fraction are subtracted.

Simplify the Numerator

Next, simplify the numerator by calculating \((x+3)(x-2)\) and combining like terms: \(\frac{x^2 - 5 - x^2 + 2x + 6}{(x-2)(x+7)}\), which simplifies further to \(\frac{2x + 1}{(x-2)(x+7)}\).

Final Simplification

Check for any common factors in the numerator and the denominator that could be cancelled out. As there is none, \(\frac{2x + 1}{(x-2)(x+7)}\) is the final answer.

Key concepts

Polynomials

Polynomials are mathematical expressions made up of variables and constants, combined using addition, subtraction, and multiplication. These expressions can have one or more terms, and each term consists of a variable raised to a power.
Examples include:

• Single-term polynomial: \(3x^2\)
• Two-term polynomial: \(x + 5\)
• Multi-term polynomial: \(2x^3 - 4x^2 + x - 5\)

Polynomials are foundational in algebra and are used in various mathematical operations, like factoring and solving equations. In algebraic fractions, polynomials often show up in the numerators and denominators, requiring us to manipulate them using algebraic methods to simplify or combine fractions.

Factoring

Factoring is the process of breaking down a polynomial into simpler components called factors, which, when multiplied together, give back the original polynomial.
To factor a quadratic polynomial such as \(x^2 + 5x - 14\), we look for two numbers that multiply to the constant term (-14) and add up to the coefficient of the linear term (5). These factors are \((x-2)\) and \((x+7)\).
This transformation is crucial because it allows us to manipulate and simplify expressions, especially when dealing with algebraic fractions. Factoring helps to identify common factors in denominators, which is necessary for adding or subtracting fractions, finding a common denominator, and simplifying expressions later on.

Common Denominator

To perform addition or subtraction of algebraic fractions, it is essential to have a common denominator. A common denominator is a shared multiple of the denominators of the fractions involved.
In the exercise, the common denominator is the product \((x-2)(x+7)\), which is found by factoring and using the highest power of each factor that appears in any denominator.

• The first fraction already has \((x-2)(x+7)\) as its denominator.
• The second fraction \(\frac{x+3}{x+7}\) needs to be adjusted by multiplying its numerator and denominator by \((x-2)\) to achieve this common denominator.

Having a common denominator is crucial for the next operation, which involves combining the fractions by performing addition or subtraction.

Simplification

Simplification is the process of making an algebraic expression as concise and straightforward as possible. In the case of our algebraic fractions, simplification involves:
1. Ensuring a common denominator is present, which has already been achieved.2. Simplifying the numerator by performing arithmetic operations like expansion and combining like terms.
For instance, when subtracting algebraic fractions:\(\frac{x^2-5 - (x+3)(x-2)}{(x-2)(x+7)}\), the numerator must be carefully calculated as \(x^2 - 5 - x^2 + 2x + 6\).
This further simplifies to \(\frac{2x + 1}{(x-2)(x+7)}\).
Finally, it is important to check if we can cancel any common factors between the numerator and the denominator. In this exercise, since no further reduction is possible, the simplified form of the expression remains \(\frac{2x + 1}{(x-2)(x+7)}\).