Problem 18 - Elementary Algebra

Question

For exercises 7-32, simplify. $$ \frac{n^{2}-9 n+18}{n^{2}-6 n+9} \cdot \frac{n^{2}-2 n-3}{n^{2}-9 n+18} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{(n + 1)}{(n - 3)} \).

Step 1 - Factor the expressions

Start by factoring all the quadratic expressions in the numerator and denominator. For the first fraction's numerator, factorize:\( n^2 - 9n + 18 = (n - 3)(n - 6) \).For the first fraction's denominator, factorize:\( n^2 - 6n + 9 = (n - 3)^2 \).For the second fraction's numerator, factorize:\( n^2 - 2n - 3 = (n - 3)(n + 1) \).For the second fraction's denominator, factorize:\( n^2 - 9n + 18 = (n - 3)(n - 6) \).

Step 2 - Rewrite the expression

Rewrite the rational expression using the factored forms:\[ \frac{(n - 3)(n - 6)}{(n - 3)^2} \times \frac{(n - 3)(n + 1)}{(n - 3)(n - 6)} \].

Step 3 - Cancel common factors

Cancel the common factors in the numerators and denominators. Notice that \((n - 3)(n - 6)\) appears in both the numerator and denominator of the fractions. After canceling, we have:\[ \frac{1}{(n - 3)} \times (n + 1) \].

Step 4 - Simplify the remaining expression

Multiply the remaining expressions to get the final simplified form:\[ \frac{(n + 1)}{(n - 3)} \].

Key Concepts

quadratic expressions

Quadratic expressions are polynomials of degree two. They are often written in the form: quadratic form: \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. To simplify rational expressions involving quadratic expressions, we need to factor them. Factoring is the process of rewriting the quadratic expression as a product of simpler expressions.

factoring

Factoring is breaking down a complex expression into simpler parts (factors) that, when multiplied together, give the original expression. For quadratic expressions, common methods of factoring include:

• Identifying common factors
• Using the quadratic formula
• Factoring by grouping

For the expressions in the original problem, the factoring steps were:
(\text{1}) Factor \(n^2 - 9n + 18\) to get \((n - 3)(n - 6)\).
(\text{2}) Factor \(n^2 - 6n + 9\) to get \((n - 3)^2\).
(\text{3}) Factor \(n^2 - 2n - 3\) to get \((n - 3)(n + 1)\). Factoring transforms the expression into a product of binomials that simplify the problem.

canceling common factors

After factoring, the next step in simplifying rational expressions is canceling common factors. This means removing identical factors from the numerator and denominator because \(xy/yz = x/z\).
In our problem, these common factors were (\text{1}) \((n-3)(n-6)/(n-3)^2\) simplifies to \(1/(n-3)\) (\text{2}) \((n-3)(n+1)/(n-3)(n-6)\) simplifies to \((n+1)/1\), which equals \((n+1)\). Removing common factors simplifies the expression greatly. This step leads to the final simplified rational expression.