Problem 75 - Elementary Algebra

Question

For exercises 39-82, simplify. $$ \frac{u^{2}+8 u+15}{u^{2}+2 u+1} \div \frac{u^{2}+7 u+10}{u^{2}+3 u+2} $$

Step-by-Step Solution

Answer

\( \frac{u + 3}{(u + 1)(u + 2)} \)

Step 1 - Factor the Numerators

First, factor the numerators of both fractions. The numerator of the first fraction is \(u^2 + 8u + 15\). This can be factored as \((u + 3)(u + 5)\). The numerator of the second fraction is \(u^2 + 7u + 10\), which can be factored as \((u + 2)(u + 5)\).

Step 2 - Factor the Denominators

Next, factor the denominators of both fractions. The denominator of the first fraction is \(u^2 + 2u + 1\). This can be factored as \((u + 1)^2\) or \((u + 1)(u + 1)\). The denominator of the second fraction is \(u^2 + 3u + 2\), which can be factored as \((u + 1)(u + 2)\).

Step 3 - Rewrite the Problem as Multiplication

Rewrite the division problem as a multiplication problem by taking the reciprocal of the second fraction: \[ \frac{(u + 3)(u + 5)}{(u + 1)(u + 1)} \times \frac{(u + 1)(u + 2)}{(u + 2)(u + 5)} \]

Step 4 - Simplify the Expression

Cancel out the common factors in the numerator and the denominator: \[ \frac{(u + 3) ot{(u + 5)}}{(u + 1) (u + 1)} \times \frac{(u + 1) (u + 2)}{(u + 2) ot{(u + 5)}} \] This simplifies to: \[ \frac{(u + 3)}{(u + 1)(u + 1)} \times \frac{(u + 1)}{(u + 2)} \]

Step 5 - Combine and Simplify Further

Combine the fractions: \[ \frac{(u + 3)(u + 1)}{(u + 1)^2 (u + 2)} \] Cancel out the common \((u + 1)\) term: \[ \frac{(u + 3)}{(u + 1)(u + 2)} \]

Step 6: Final Solution

The simplified form of the given expression is: \[ \frac{u + 3}{(u + 1)(u + 2)} \]

Key Concepts

Factoring Polynomials

When simplifying algebraic fractions, a key step is factoring polynomials. Factoring means writing a polynomial as a product of its simpler binomials or monomials. For example, the polynomial \( u^2 + 8u + 15 \) can be factored into \( (u + 3)(u + 5) \). This is because when you multiply \( (u + 3) \) and \( (u + 5) \), you get the original polynomial back. Factoring helps break down complex expressions, making them easier to work with.

Multiplying Fractions

Multiplying fractions in algebra is similar to multiplying numerical fractions. You multiply the numerators together and the denominators together. For instance, if you have two fractions \( \frac{a}{b} \) and \( \frac{c}{d} \), their product is \( \frac{a \cdot c}{b \cdot d} \). This step is crucial when simplifying algebraic fractions because it allows us to combine fractions after factoring, moving us closer to the simplest form.

Division of Fractions

When dividing fractions, you actually multiply by the reciprocal of the fraction you're dividing by. The reciprocal of a fraction \( \frac{a}{b} \) is \( \frac{b}{a} \). For our problem, we have to divide: \( \frac{u^{2}+8u+15}{u^{2}+2u+1} \div \frac{u^{2}+7u+10}{u^{2}+3u+2} \). We accomplish this by multiplying the first fraction by the reciprocal of the second: \( \frac{u^{2}+8u+15}{u^{2}+2u+1} \times \frac{u^{2}+3u+2}{u^{2}+7u+10} \). This step simplifies the division process by converting it into a multiplication problem.

Canceling Common Factors

To simplify an expression fully, you cancel out common factors in the numerator and denominator. For example, in the fraction \( \frac{(u + 3)}{(u + 1)(u + 1)} \times \frac{(u + 1)}{(u + 2)} \), the \( (u + 1) \) in both the numerator and the denominator cancels out. Removing these common factors reduces the fraction to its simplest form. Always check for and cancel these common terms to simplify the expression correctly.

Simplifying Expressions

The final goal in algebraic fractions is to simplify the expression. After factoring, multiplying, dividing, and canceling common factors, combine the fractions if possible. For example, by canceling common factors in our problem, we get the simplified form: \( \frac{u + 3}{(u + 1)(u + 2)} \). Simplifying means making the expression as straightforward as possible while maintaining its value. This process helps with easier interpretation and computation in further mathematical problems.