Problem 52 - Prealgebra and Introductory Algebra

Question

Multiply. $$\frac{25-n^{2}}{n^{2}-2 n-35} \cdot \frac{n^{2}-8 n-20}{n^{2}-3 n-10}$$

Step-by-Step Solution

Answer

- (n - 10) / ((n - 7) * (n + 2))

Step 1: Factor each Quadratic Expression

Factor the quadratic expressions \(25-n^{2}\), \(n^{2}-2 n-35\), \(n^{2}- 8n -20\), and \(n^{2}- 3n -10\). These become \(-(n-5)(n+5)\), \((n - 7)(n +5)\), \((n -10)(n +2)\), and \((n - 5)(n +2)\) respectively.

Step 2: Simplify and Cancel Common Factors

Observe and cancel out the common factors in the numerator and the denominator, which are \(n-5\) and \(n + 5\). After that, we obtain \( \frac{-(n -10)}{(n - 7) \cdot (n +2)}\). Notice the negative sign in the numerator, it could be written as \( - \frac{n - 10}{(n - 7) \cdot (n +2)} \)

Step 3: Write the Final Answer

The final result is \( - \frac{n - 10}{(n - 7) \cdot (n +2)}\)

Key Concepts

Simplifying Rational Expressions

Simplifying rational expressions involves reducing complex fractions to simpler forms. The process is akin to simplifying regular numerical fractions.

For example, in algebraic expressions, variables take the place of numbers. If we have an expression such as \(\frac{25-n^{2}}{n^{2}-2n-35}\), our goal is to rewrite it in a more simplified manner.

First, we factor all quadratic expressions. This can be done by finding two numbers that multiply to the constant term and add up to the coefficient of the linear term. Once factored, we might get \(25-n^{2} = -(n-5)(n+5)\) and \(n^{2}-2n-35 = (n-7)(n+5)\), just as we do with \(n^{2}-8n-20\) and \(n^{2}-3n-10\).

After factoring, we identify and cancel out common factors between the numerator and the denominator. In our example, \(n-5\) and \(n+5\) are common factors that can be eliminated, which considerably simplifies the expression.

Canceling Common Factors

Canceling common factors is fundamental in simplifying rational expressions. It's based on the property that anytime you have a factor that appears both in the numerator and in the denominator, you can divide both by that factor, thus 'canceling' it out.

Looking closely at our exercise, after factoring, we see similar factors: \(n-5\) and \(n+5\) appear in both the numerator and the denominator. So, we strike them out because \(\frac{n-5}{n-5} = 1\) and \(\frac{n+5}{n+5} = 1\).

It's important to remember that we can only cancel factors—not terms. Terms are separated by addition or subtraction. Only factors, which are separated by multiplication, can be canceled. If a term contains a common factor, factor it out first, then cancel if possible.

By eliminating these common factors, we're left with a much more manageable expression: \(\frac{-(n-10)}{(n-7)(n+2)}\).

Multiplying Fractions in Algebra

Multiplying fractions in algebra follows the same rules as multiplying numerical fractions: multiply the numerators together and the denominators together. The complication with algebraic fractions comes with the presence of variables and the need to factorize.

In the given exercise, after simplifying the individual fractions by canceling common factors, we're prepared to multiply the remaining factors.

Step-by-Step Multiplication

• First, ensure each fraction is fully simplified, as we've done previously.
• Then, multiply the numerators together. In our exercise, we combined \( (n-10) \) from the first fraction with the numeral \( -1 \) from the negative sign.
• Similarly, multiply the denominators together. Here, we'd combine \( (n-7) \) and \( (n+2) \) from the second fraction.
• Finally, write down the simplified product as the result, which is \( - \frac{n-10}{(n-7)(n+2)}\).

In the multiplication of algebraic fractions, it's also essential to watch out for the possibility to further simplify the product after multiplication if any common factors emerge.