Problem 31
Question
multiply or divide as indicated. $$ \frac{x^{2}+x-12}{x^{2}+x-30} \cdot \frac{x^{2}+5 x+6}{x^{2}-2 x-3} \div \frac{x+3}{x^{2}+7 x+6} $$
Step-by-Step Solution
Verified Answer
The simplified form of the expression is \( \frac{x+4}{x-5} \)
1Step 1: Simplify the Fractions
Let's start with simplifying the fractions by factoring the numerator and denominator of each fraction as follows: \n\( \frac{(x-3)(x+4)}{(x-5)(x+6)} \cdot \frac{(x+2)(x+3)}{(x-3)(x+1)} \div \frac{x+3}{(x+1)(x+6)} \)
2Step 2: Preparing for Multiplication and Division
We can rearrange the expression as the product of the first two fractions and the reciprocal of the last fraction. \n \( \frac{(x-3)(x+4)}{(x-5)(x+6)} \cdot \frac{(x+2)(x+3)}{(x-3)(x+1)} \cdot \frac{(x+1)(x+6)}{x+3} \)
3Step 3: Cancel out Common Factors
Now we can now cancel out the common factors from the numerators and denominators to simplify the expression: \n \( \frac{x+4}{x-5} \)
4Step 4: Simplify Further
The expression has already been simplified as much as possible.
Key Concepts
Factoring PolynomialsMultiplying and Dividing FractionsCanceling Common Factors
Factoring Polynomials
Understanding how to factor polynomials is like mastering a way to break down numbers into their multiplication components, but with variables added into the mix. For example, when we see a quadratic expression like ( x^2 + bx + c ), we want to find two binomials (expressions with two terms) that multiply to give us that original quadratic.
The key is to find two numbers that add up to b and multiply to c. It’s like a puzzle where the numbers have a secret relationship hidden in the sum and product. In our exercise, we had to factor polynomials in both the numerators and denominators. This step is crucial because it reveals common factors that we’ll eventually cancel out to simplify the algebraic fraction. Factoring makes everything that comes next so much easier!
The key is to find two numbers that add up to b and multiply to c. It’s like a puzzle where the numbers have a secret relationship hidden in the sum and product. In our exercise, we had to factor polynomials in both the numerators and denominators. This step is crucial because it reveals common factors that we’ll eventually cancel out to simplify the algebraic fraction. Factoring makes everything that comes next so much easier!
Multiplying and Dividing Fractions
When we are dealing with fractions, whether they’re numeric or algebraic, multiplication and division follow the same core principles. To multiply fractions, we take the numerators (the numbers on top) and multiply them together, and do the same with the denominators (the numbers on the bottom). It’s like stacking layers of a cake – the top with the top, and the bottom with the bottom.
For division, we flip the second fraction, making it the reciprocal, and then multiply as usual. Remember, dividing by a fraction is the same as multiplying by its opposite, or its 'upside-down' twin. So, to divide by a fraction, you just have to turn it over and multiply!
For division, we flip the second fraction, making it the reciprocal, and then multiply as usual. Remember, dividing by a fraction is the same as multiplying by its opposite, or its 'upside-down' twin. So, to divide by a fraction, you just have to turn it over and multiply!
Canceling Common Factors
Canceling common factors between the numerator and the denominator can feel like cleaning your room and finding out you’ve had a clear path to walk all along. It’s about simplifying; just like you can’t have more than one of the same number in a multiplication, you can't have the same factor in both the top and the bottom of a fraction.
When we cancel out these common factors, we're essentially dividing both the numerator and the denominator by the same number, which simplifies the expression without changing its value. It's like saying, 'I owe you 10 dollars, but you owe me 10 dollars,' so ultimately, we don’t owe each other anything.
When we cancel out these common factors, we're essentially dividing both the numerator and the denominator by the same number, which simplifies the expression without changing its value. It's like saying, 'I owe you 10 dollars, but you owe me 10 dollars,' so ultimately, we don’t owe each other anything.
Attention to Detail
Always keep an eye out for factors that look different but are actually the same, like (x+2) and 2+x. Sometimes factors are just playing hide and seek in different forms.Other exercises in this chapter
Problem 30
Simplify each exponential expression. $$ x^{-6} \cdot x^{12} $$
View solution Problem 30
Find the union of the sets. $$\\{1,3,7,8\\} \cup\\{2,3,8\\}$$
View solution Problem 31
Factor each trinomial, or state that the trinomial is prime. $$9 x^{2}-9 x+2$$
View solution Problem 31
In Exercises 15-58, find each product. $$ (x+3)(x-3) $$
View solution