Problem 71 - Elementary Algebra

Question

For exercises 39-82, simplify. $$ \frac{2 k^{2}+3 k}{2 k^{2}-13 k-24} \div \frac{k^{4}-k^{3}}{k^{2}-9 k+8} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{k^{2}} \).

Step 1 - Rewrite Division as Multiplication

Rewrite the division operation as multiplication by the reciprocal. This changes the expression to:\[ \frac{2 k^{2} + 3 k}{2 k^{2} - 13 k - 24} \times \frac{k^{2} - 9 k + 8}{k^{4} - k^{3}}. \]

Step 2 - Factor All Polynomials

Factor each polynomial in the expression:\[ 2 k^{2} + 3 k = k(2 k + 3), \]\[ 2 k^{2} - 13 k - 24 = (2 k + 3)(k - 8), \]\[ k^{4} - k^{3} = k^{3}(k - 1), \]\[ k^{2} - 9 k + 8 = (k - 1)(k - 8). \]

Step 3 - Substitute Factored Forms

Substitute the factors into the expression:\[ \frac{k(2 k + 3)}{(2 k + 3)(k - 8)} \times \frac{(k - 1)(k - 8)}{k^{3}(k - 1)}. \]

Step 4 - Cancel Common Factors

Cancel out the common factors in the numerator and the denominator:\[ \frac{k \, \cancel{(2 k + 3)}}{\cancel{(2 k + 3)} \,(k - 8)} \times \frac{\cancel{(k - 1)} \,(k - 8)}{k^{3} \times \cancel{(k - 1)}}. \]This further simplifies to:\[ \frac{k}{k^{3}} = \frac{1}{k^{2}}. \]

Key Concepts

Polynomial Factorization

Polynomial factorization is a crucial step in simplifying algebraic expressions. It involves breaking down a polynomial into a product of simpler polynomials that, when multiplied together, give the original polynomial. For example, in the expression given, we have several polynomials to factor. Starting with:

• \(2k^{2} + 3k\), which factors to \(k(2k + 3)\)
• \(2k^{2} - 13k - 24\), which factors to \((2k + 3)(k - 8)\)
• \(k^{4} - k^{3}\), which factors to \(k^{3}(k - 1)\)
• \(k^{2} - 9k + 8\), which factors to \((k - 1)(k - 8)\)

Factoring these polynomials helps in simplifying the problem by allowing us to cancel identical terms later.

Rational Expressions

A rational expression is a fraction in which the numerator and the denominator are polynomials. For instance, the given expression starts with:\[ \frac{2k^{2} + 3k}{2k^{2} - 13k - 24} \times \frac{k^{2} - 9k + 8}{k^{4} - k^{3}}. \]Rational expressions often need to be simplified by factoring the polynomials, as described earlier. After factoring, the expression becomes:\[ \frac{k(2k + 3)}{(2k + 3)(k - 8)} \times \frac{(k - 1)(k - 8)}{k^{3}(k - 1)}. \]Understanding how to handle these expressions is essential for simplifying complex algebraic fractions.

Reciprocal Multiplication

When dividing rational expressions, it's convenient to multiply by the reciprocal of the divisor. This transforms the division problem into a multiplication problem, which is often easier to manage. For example, in the given problem, we rewrite the division:\[ \frac{2k^{2} + 3k}{2k^{2} - 13k - 24} \times \frac{k^{2} - 9k + 8}{k^{4} - k^{3}}. \]This step is important because it allows us to treat the division of fractions as multiplication, simplifying the overall process of solving the problem.

Canceling Common Factors

One of the key steps in simplifying rational expressions is canceling common factors from the numerator and the denominator. In the problem, after factoring, we have:\[ \frac{k(2k + 3)}{(2k + 3)(k - 8)} \times \frac{(k - 1)(k - 8)}{k^{3}(k - 1)}. \]After canceling the common factors \((2k + 3)\) and \((k - 1)\), the expression simplifies to:\[ \frac{k \times \text{cancel}{(2k + 3)}}{\text{cancel}{(2k + 3)} \times (k - 8)} \times \frac{\text{cancel}{(k - 1)} \times (k - 8)}{k^{3} \times \text{cancel}{(k - 1)}}. \]Simplifying further, we get:\[ \frac{k}{k^{3}} = \frac{1}{k^{2}}. \]Canceling common factors reduces the expression to its simplest form, making it easier to work with.