# Factorization and Algebraic Fractions - Examples, Exercises and Solutions

Algebraic fractions are fractions with variables.

## Ways to factor algebraic fractions:

1. We will find the most appropriate common factor to extract.
2. If we do not see a common factor that we can extract, we will move on to factorization with formulas for abbreviated multiplication as we have studied.
3. If the formulas for abbreviated multiplication cannot be used, we will proceed to factorize with trinomials.
4. We will reduce according to the rules of reduction (we can only reduce when there is multiplication between the terms unless they are within parentheses, in which case, we will consider them independent terms).

## Examples with solutions for Factorization and Algebraic Fractions

### Exercise #1

Determine if the simplification shown below is correct:

$\frac{7}{7\cdot8}=8$

### Step-by-Step Solution

Let's consider the fraction and break it down into two multiplication exercises:

$\frac{7}{7}\times\frac{1}{8}$

We simplify:

$1\times\frac{1}{8}=\frac{1}{8}$

Therefore, the described simplification is false.

Incorrect

### Exercise #2

Determine if the simplification below is correct:

$\frac{4\cdot8}{4}=\frac{1}{8}$

### Step-by-Step Solution

We will divide the fraction exercise into two multiplication exercises:

$\frac{4}{4}\times\frac{8}{1}=$

We simplify:

$1\times\frac{8}{1}=8$

Therefore, the described simplification is false.

Incorrect

### Exercise #3

Determine if the simplification below is correct:

$\frac{3\cdot7}{7\cdot3}=0$

### Step-by-Step Solution

We will divide the fraction exercise into two different multiplication exercises.

As this is a multiplication exercise, you can use the substitution property:

$\frac{7}{7}\times\frac{3}{3}=1\times1=1$

Therefore, the simplification described is false.

Incorrect

### Exercise #4

Determine if the simplification below is correct:

$\frac{5\cdot8}{8\cdot3}=\frac{5}{3}$

### Step-by-Step Solution

Let's consider the fraction and break it down into two multiplication exercises:

$\frac{8}{8}\times\frac{5}{3}$

We simplify:

$1\times\frac{5}{3}=\frac{5}{3}$

Correct

### Exercise #5

Determine if the simplification below is correct:

$\frac{6\cdot3}{6\cdot3}=1$

### Step-by-Step Solution

We simplify the expression on the left side of the approximate equality:

$\frac{\textcolor{red}{\not{6}}\cdot\textcolor{blue}{\not{3}}}{\textcolor{red}{\not{6}}\cdot\textcolor{blue}{\not{3}}}\stackrel{?}{= }1\\ \downarrow\\ 1\stackrel{!}{= }1$therefore, the described simplification is correct.

Therefore, the correct answer is A.

Correct

### Exercise #6

Select the field of application of the following fraction:

$\frac{8+x}{5}$

### Step-by-Step Solution

Since the domain depends on the denominator, we note that there is no variable in the denominator.

Therefore, the domain is all numbers.

All numbers

### Exercise #7

Select the field of application of the following fraction:

$\frac{6}{x}$

### Step-by-Step Solution

Since the domain of definition depends on the denominator, and X appears in the denominator

All numbers will be suitable except for 0.

In other words, the domain of definition:

$x\ne0$

All numbers except 0

### Exercise #8

Complete the corresponding expression for the denominator

$\frac{16ab}{?}=8a$

### Step-by-Step Solution

Using the formula:

$\frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y$

We first convert the 8 into a fraction, and multiply

$\frac{16ab}{?}=\frac{8}{1}$

$16ab\times1=8a$

$16ab=8a$

We then divide both sides by 8a:

$\frac{16ab}{8a}=\frac{8a}{8a}$

$2b$

$2b$

### Exercise #9

Determine if the simplification described below is correct:

$\frac{x+6}{y+6}=\frac{x}{y}$

### Step-by-Step Solution

We use the formula:

$\frac{x+z}{y+z}=\frac{x+z}{y+z}$

$\frac{x+6}{y+6}=\frac{x+6}{y+6}$

Therefore, the simplification described is incorrect.

Incorrect

### Exercise #10

Determine if the simplification below is correct:

$\frac{3-x}{-x+3}=0$

### Step-by-Step Solution

$\frac{z-x}{-x+z}=1$

Incorrect

### Exercise #11

Indicate whether true or false

$\frac{c\cdot a}{a\cdot c}=0$

### Step-by-Step Solution

Let's simplify the expression on the left side of the proposed equation:

$\frac{\not{c}\cdot \not{a}}{\not{a}\cdot \not{c}}\stackrel{?}{= }0 \\ 1 \stackrel{?}{= }0$Clearly, we get a false statement because: 1 is different from: 0

$\boxed{ 1 \stackrel{!}{\neq }0}$Therefore, the proposed equation is not correct,

Not true

### Exercise #12

Determine if the simplification below is correct:

$\frac{3\cdot4}{8\cdot3}=\frac{1}{2}$

### Step-by-Step Solution

We simplify the expression on the left side of the approximate equality.

First let's consider the fact that the number 8 is a multiple of the number 4:

$8=2\cdot4$
Therefore, we will return to the problem in question and present the number 8 as a multiple of the number 4, then we will simplify the fraction:

$\frac{3\cdot4}{\underline{8}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{3\cdot4}{\underline{2\cdot4}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{\textcolor{blue}{\not{3}}\cdot\textcolor{red}{\not{4}}}{2\cdot\textcolor{red}{\not{4}}\cdot\textcolor{blue}{\not{3}}}\stackrel{?}{= }\frac{1}{2} \\ \downarrow\\ \frac{1}{2}\stackrel{!}{= }\frac{1}{2}$
Therefore, the described simplification is correct.

That is, the correct answer is A.

True

### Exercise #13

Complete the corresponding expression for the denominator

$\frac{12ab}{?}=1$

### Video Solution

$12ab$

### Exercise #14

Complete the corresponding expression for the denominator

$\frac{16ab}{?}=2b$

### Video Solution

$8a$

### Exercise #15

Complete the corresponding expression for the denominator

$\frac{27ab}{\text{?}}=3ab$

### Video Solution

$9$