Problem 79 - Algebra for College Students

Question

Perform the indicated operations. $$\frac{2 h^{2}-5 h-3}{5 h^{2}-4 h-1} \div \frac{2 h^{2}+7 h+3}{h^{2}+2 h-3}$$

Step-by-Step Solution

Answer

\[\frac{h - 3}{5h + 1}\]

Step 1 - Rewrite the Division as Multiplication

Rewrite the division of fractions as multiplication by the reciprocal of the second fraction: \[\frac{2 h^{2}-5 h-3}{5 h^{2}-4 h-1} \times \frac{h^{2}+2 h-3}{2 h^{2}+7 h+3}\]

Step 2 - Factor all Polynomials

Factor each polynomial in the numerator and denominator: \[2h^2 - 5h - 3 = (2h + 1)(h - 3)\] \[5h^2 - 4h - 1 = (5h + 1)(h - 1)\] \[h^2 + 2h - 3 = (h + 3)(h - 1)\] \[2h^2 + 7h + 3 = (2h + 1)(h + 3)\]

Step 3 - Write the Factored Form

Rewrite the expression using the factored forms: \[\frac{(2h + 1)(h - 3)}{(5h + 1)(h - 1)} \times \frac{(h + 3)(h - 1)}{(2h + 1)(h + 3)}\]

Step 4 - Cancel Common Factors

Cancel out the common factors from the numerators and denominators: \[\frac{(2h + 1)(h - 3)}{(5h + 1)(h - 1)} \times \frac{(h + 3)(h - 1)}{(2h + 1)(h + 3)} = \frac{(h - 3)}{(5h + 1)}\]

Step 5 - Simplify the Expression

After canceling the common factors, the simplified expression is: \[\frac{h - 3}{5h + 1}\]

Key Concepts

Polynomial Factoring

To understand polynomial factoring, think of it as the process of breaking down a complex expression into simpler factors that, when multiplied, give you the original polynomial. This is similar to factoring numbers, but instead of numbers, we deal with algebraic expressions.

For example, consider the polynomial \(2h^2 - 5h - 3\). Using factoring techniques such as grouping or using the quadratic formula, we can break this down into \( (2h + 1)(h - 3)\). The same steps apply to other polynomials in our problem too.

Once you factor each polynomial, it becomes much easier to handle operations like multiplication and division.

Division of Fractions

When we divide fractions in algebra, the process is similar to dividing numerical fractions. The key rule is to multiply by the reciprocal of the second fraction. For example, \( \frac{a}{b} \div \frac{c}{d} \) becomes \( \frac{a}{b} \times \frac{d}{c} \).

In our exercise, we rewrite the division as: \( \frac{2 h^{2}-5 h-3}{5 h^{2}-4 h-1} \div \frac{2 h^{2}+7 h+3}{h^{2}+2 h-3} \)

This becomes: \( \frac{2 h^{2}-5 h-3}{5 h^{2}-4 h-1} \times \frac{h^{2}+2 h-3}{2 h^{2}+7 h+3} \). The reciprocal inversion makes it easier to proceed with factoring and canceling steps.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their simplest form. This means removing any redundant terms and making the expression as straightforward as possible.

In our specific problem, after factoring and rewriting, we get: \( \frac{(2h + 1)(h - 3)}{(5h + 1)(h - 1)} \times \frac{(h + 3)(h - 1)}{(2h + 1)(h + 3)} \).

We can then simplify by canceling out the common factors from the numerator and denominator across the multiplication sign. The goal is to have the least terms possible, which in this case simplifies to: \( \frac{h - 3}{5h + 1} \).

Common Factors

Finding and canceling common factors makes our calculations much easier. Common factors are terms that appear both in the numerator and the denominator.

Take, for instance, the factored form \( \frac{(2h + 1)(h - 3)}{(5h + 1)(h - 1)} \times \frac{(h + 3)(h - 1)}{(2h + 1)(h + 3)} \). Here, \( (2h + 1) \) and \( (h - 1) \) are common factors and can be canceled out.

This reduces the complexity and leads us to the simplified expression: \( \frac{h - 3}{5h + 1} \). Simplifying by canceling common factors is a crucial step in making algebraic fraction operations manageable.