Problem 31
Question
For the following problems, perform the multiplications and divisions. $$ \frac{x+8}{x+1} \cdot \frac{x+2}{x+8} $$
Step-by-Step Solution
Verified Answer
Question: Multiply the given fractions and simplify the expression:
$$
\frac{x+8}{x+1} \cdot \frac{x+2}{x+8}
$$
Answer: After multiplying the fractions and simplifying, the expression is:
$$
\frac{x^2+10x+16}{x^2+9x+8}
$$
1Step 1: Multiply the numerators
To multiply the fractions, first multiply the numerators:
$$
(x+8)(x+2)
$$
To find the result, use the distributive property:
$$
x^2+2x+8x+16=x^2+10x+16
$$
2Step 2: Multiply the denominators
Now, multiply the denominators:
$$
(x+1)(x+8)
$$
To find the result, use the distributive property:
$$
x^2+8x+x+8=x^2+9x+8
$$
3Step 3: Combine the numerators and denominators
Now, we've got the expression:
$$
\frac{x^2+10x+16}{x^2+9x+8}
$$
Since there are no common factors between the numerator and the denominator that can be cancelled, this is the final simplified expression.
Key Concepts
Distributive PropertySimplifying ExpressionsPolynomial Multiplication
Distributive Property
When multiplying algebraic expressions, the distributive property offers a systematic way to ensure all parts of one expression are multiplied by all parts of another. For example, when we multiply \((x + 8)(x + 2)\), we apply the distributive property. Imagine that the first term, \(x+8\), needs to distribute itself to each term in the second expression, \(x+2\). Similarly, the \(x+2\) expression must distribute or 'give' each of its terms to \(x+8\).
This leads to the process of multiplying each term of the first binomial with each term of the second binomial:
This leads to the process of multiplying each term of the first binomial with each term of the second binomial:
- \(x\) multiplies with \(x\) to give \(x^2\)
- \(x\) also multiplies with 2 to give \(2x\)
- 8 multiplies with \(x\) to produce \(8x\)
- Finally, 8 multiplies with 2 to give 16.
Simplifying Expressions
Simplifying expressions is like cleaning out a closet; the goal is to make it as organized and as simple as possible. After using the distributive property, we combine like terms, which are terms that have the same variable raised to the same power. In the exercise, \(2x\) and \(8x\) are like terms and when combined, they tidy up to \(10x\), trimming our expression to \(x^2 + 10x + 16\).
Sometimes, simplifying can involve cancelling out common factors between the numerator and the denominator but only if they are identical in both—not just similar. This isn't the case in our problem as \(x+8\) doesn't appear in both the numerator and denominator after multiplication. Hence, the final answer \(\frac{x^2+10x+16}{x^2+9x+8}\) is the simplest form we can achieve without any more terms cancelling out. Understanding how to simplify expressions helps ensure that solutions to algebraic problems are presented in their most basic, understandable form.
Sometimes, simplifying can involve cancelling out common factors between the numerator and the denominator but only if they are identical in both—not just similar. This isn't the case in our problem as \(x+8\) doesn't appear in both the numerator and denominator after multiplication. Hence, the final answer \(\frac{x^2+10x+16}{x^2+9x+8}\) is the simplest form we can achieve without any more terms cancelling out. Understanding how to simplify expressions helps ensure that solutions to algebraic problems are presented in their most basic, understandable form.
Polynomial Multiplication
Layered Multiplication
Multiplying polynomials is like dealing with layers. Each term from one polynomial interacts with each term from the other, just like layers of fabric stitched together. In our problem, two binomials are multiplied to create a new polynomial. The result has more terms because each term from the first binomial, \(x+8\), multiplies with each term from the second, \(x+2\), to create four products.Symmetry in Multiplication
It's important to realize that the order in which you multiply doesn't affect the product. Whether you start by multiplying the terms of \((x+2)\) with \((x+8)\) or vice versa, the outcome is the same.To ensure no terms are missed and mistakes are avoided, a structured approach, like the FOIL (First, Outside, Inside, Last) method, is often useful. That helps in keeping track of all the terms that need to be multiplied together, especially as the polynomials get longer and more complex in advanced mathematics.
Other exercises in this chapter
Problem 31
For the following problems, solve the rational equations. $$ \frac{3 y}{y-1}+\frac{2 y}{y-6}=\frac{5 y^{2}-15 y+20}{y^{2}-7 y+6} $$
View solution Problem 31
For the following problems, show that the fractions are equivalent. $$ \frac{-3}{5} \text { and }-\frac{3}{5} $$
View solution Problem 31
For the following problems, add or subtract the rational expressions. $$ \frac{y+4}{y-6}+\frac{y+8}{y-6} $$
View solution Problem 31
For the following problems, reduce each rational expression to lowest terms. $$ \frac{(a+1)^{5}(a-1)^{7}}{(a+1)^{3}(a-1)^{4}} $$
View solution