Problem 99 - Algebra for College Students

Question

Perform the indicated operations. $$ \frac{a^{2}-6 a+9}{a^{3}-8} \div \frac{a^{2}-a-6}{a^{2}-4} $$

Step-by-Step Solution

Answer

After simplifying, the final expression is \(\frac{a-3}{a^2+2a+4}\).

Step 1: Factor the Numerators and Denominators

Start by factoring the expressions in the numerators and denominators. For the numerator of the first fraction, \(\frac{a^{2}-6a+9}{a^{3}-8}\), factor \(a^{2}-6a+9\) which is \((a-3)^{2}\).For the denominator \(a^{3}-8\), recognize it as a difference of cubes: \(a^{3}-2^{3} = (a-2)(a^{2}+2a+4)\).For the second fraction, \(\frac{a^{2}-a-6}{a^{2}-4}\), factor the numerator \(a^{2}-a-6\) which is \((a-3)(a+2)\).For the denominator \(a^{2}-4\), recognize it as a difference of squares: \(a^{2}-4 = (a-2)(a+2)\).

Step 2: Rewrite the Factored Form

Rewrite the fractions with their factored forms:\[\frac{(a-3)^2}{(a-2)(a^2+2a+4)} \div \frac{(a-3)(a+2)}{(a-2)(a+2)}.\]

Step 3: Apply the Division Rule for Fractions

To divide by a fraction, multiply by its reciprocal. Rewriting the problem:\[\frac{(a-3)^2}{(a-2)(a^2+2a+4)} \times \frac{(a-2)(a+2)}{(a-3)(a+2)}.\]

Step 4: Simplify the Expression

Cancel common factors in the numerator and the denominator. Cancel \((a-3)\), \((a-2)\), and \((a+2)\):\[\frac{(a-3)^2}{(a-2)(a^2+2a+4)} \times \frac{(a-2)(a+2)}{(a-3)(a+2)} = \frac{a-3}{a^2+2a+4}.\]

Key Concepts

Factoring Polynomials

Factoring polynomials is a key skill in simplifying and solving algebraic expressions. It involves rewriting a polynomial as a product of its factors. For instance, consider the expression from our exercise: \(a^{2}-6a+9\). This can be factored into \((a-3)^{2}\), since \((-3) + (-3) = -6\) and \((-3) \times (-3) = 9\). Identifying patterns such as perfect square trinomials and differences of squares or cubes can make factoring quicker and easier.

• Recognize the format. For example, \(a^{3}-8\) is a difference of cubes because it can be rewritten as \(a^{3} - 2^{3}\).
• Use factorization formulas: \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\).
• Similarly, the quadratic expression \(a^{2} - 4\) can be factored using the difference of squares: \(a^{2} - 2^{2} = (a - 2)(a + 2)\).

Mastery of these techniques is crucial for simplifying complex expressions and solving polynomial equations.

Multiplication of Fractions

Once the polynomials are factored, the next step involves multiplying fractions. When dealing with fraction multiplication, you multiply the numerators together and the denominators together. For instance, in our exercise, after rewriting the division as multiplication, we have:
\ \ \(\frac{(a-3)^{2}}{(a-2)(a^{2}+2a+4)} \times \frac{(a-2)(a+2)}{(a-3)(a+2)}\)\

• This requires multiplying the numerators: \((a-3)^{2} \times (a-2)(a+2)\).
• And the denominators: \((a-2)(a^{2}+2a+4) \times (a-3)(a+2)\).

Multiplication of fractions can often reveal common factors that can be canceled out, simplifying the expression dramatically. This is essential for keeping solutions neat and manageable.

Simplification of Expressions

After multiplication of fractions, the next crucial step is simplification. The goal is to make the expression as simple as possible by canceling out common factors in the numerator and the denominator. In our exercise, we simplify: \(\frac{(a-3)^{2}}{(a-2)(a^{2}+2a+4)} \times \frac{(a-2)(a+2)}{(a-3)(a+2)}\).

• Cancel common terms like \((a-3)\), \((a-2)\), and \((a+2)\).
• We then simplify to get \(\frac{a-3}{a^{2}+2a+4}\).
• This simplified form is much easier to work with and understand.

Simplification steps are critical for solving algebra problems efficiently. They help in reducing complex expressions to their simplest form, which is often required in higher-level math problems.