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).

Suggested Topics to Practice in Advance

  1. Factoring using contracted multiplication
  2. Factorization
  3. Extracting the common factor in parentheses
  4. Factorization: Common factor extraction
  5. Factoring Trinomials

Practice Factorization and Algebraic Fractions

Examples with solutions for Factorization and Algebraic Fractions

Exercise #1

Select the field of application of the following fraction:

x16 \frac{x}{16}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

x16 \frac{x}{16} As we know, the only restriction that applies to division operation is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

However in the given expression:

x16 \frac{x}{16} the denominator is 16 and:

160 16\neq0 Therefore the fraction is well defined and thus the unknown, which is in the numerator, can take any value,

Meaning - the domain (definition range) of the given expression is:

all x

(This means that we can substitute any number for the unknown x and the expression will remain well defined),

Therefore the correct answer is answer B.

Answer

All X All~X

Exercise #2

Select the domain of the following fraction:

8+x5 \frac{8+x}{5}

Video Solution

Step-by-Step Solution

The domain depends on the denominator and we can see that there is no variable in the denominator.

Therefore, the domain is all numbers.

Answer

All numbers

Exercise #3

Select the the domain of the following fraction:

6x \frac{6}{x}

Video Solution

Step-by-Step Solution

The domain of a fraction depends on the denominator.

Since you cannot divide by zero, the denominator of a fraction cannot equal zero.

Therefore, for the fraction 6x \frac{6}{x} , the domain is "All numbers except 0," since the denominator cannot equal zero.

In other words, the domain is:

x0 x\ne0

Answer

All numbers except 0

Exercise #4

Select the field of application of the following fraction:

3x+2 \frac{3}{x+2}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

3x+2 \frac{3}{x+2}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

3x+2 \frac{3}{x+2} As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

x+20 x+2\neq0 We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

x+20x2 x+2\neq0 \\ \boxed{x\neq -2} Therefore, the domain (definition domain) of the given expression is:

x2 x\neq -2 (This means that if we substitute for the variable x any number different from(2) (-2) the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x2 x\neq-2

Exercise #5

Select the field of application of the following fraction:

82+x \frac{8}{-2+x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

82+x \frac{8}{-2+x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

82+x \frac{8}{-2+x} As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

2+x0 -2+x\neq0 We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

2+x0x2 -2+x\neq0 \\ \boxed{x\neq 2} Therefore, the domain (definition domain) of the given expression is:

x2 x\neq 2 (This means that if we substitute any number different from 2 2 for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x2 x\neq2

Exercise #6

Select the field of application of the following fraction:

713+x \frac{7}{13+x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

713+x \frac{7}{13+x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

713+x \frac{7}{13+x} As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

13+x0 13+x\neq0 We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

13+x0x13 13+x\neq0 \\ \boxed{x\neq -13} Therefore, the domain (definition domain) of the given expression is:

x13 x\neq -13 (This means that if we substitute any number different from (13) (-13) for the variable x, the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x13 x\neq-13

Exercise #7

Determine if the simplification below is correct:

5883=53 \frac{5\cdot8}{8\cdot3}=\frac{5}{3}

Video Solution

Step-by-Step Solution

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

88×53 \frac{8}{8}\times\frac{5}{3}

We simplify:

1×53=53 1\times\frac{5}{3}=\frac{5}{3}

Answer

Correct

Exercise #8

Determine if the simplification shown below is correct:

778=8 \frac{7}{7\cdot8}=8

Video Solution

Step-by-Step Solution

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

77×18 \frac{7}{7}\times\frac{1}{8}

We simplify:

1×18=18 1\times\frac{1}{8}=\frac{1}{8}

Therefore, the described simplification is false.

Answer

Incorrect

Exercise #9

Determine if the simplification below is correct:

484=18 \frac{4\cdot8}{4}=\frac{1}{8}

Video Solution

Step-by-Step Solution

We will divide the fraction exercise into two multiplication exercises:

44×81= \frac{4}{4}\times\frac{8}{1}=

We simplify:

1×81=8 1\times\frac{8}{1}=8

Therefore, the described simplification is false.

Answer

Incorrect

Exercise #10

Determine if the simplification below is correct:

3773=0 \frac{3\cdot7}{7\cdot3}=0

Video Solution

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:

77×33=1×1=1 \frac{7}{7}\times\frac{3}{3}=1\times1=1

Therefore, the simplification described is false.

Answer

Incorrect

Exercise #11

Determine if the simplification below is correct:

6363=1 \frac{6\cdot3}{6\cdot3}=1

Video Solution

Step-by-Step Solution

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

=?11=!1 \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.

Answer

Correct

Exercise #12

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Using the formula:

xy=zwxy=zy \frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y

We first convert the 8 into a fraction, and multiply

16ab?=81 \frac{16ab}{?}=\frac{8}{1}

16ab×1=8a 16ab\times1=8a

16ab=8a 16ab=8a

We then divide both sides by 8a:

16ab8a=8a8a \frac{16ab}{8a}=\frac{8a}{8a}

2b 2b

Answer

2b 2b

Exercise #13

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

16ab?=2b16ab?=2b1 \frac{16ab}{?}=2b \\ \downarrow\\ \frac{16ab}{?}=\frac{2b}{1} Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 16 to get the number 2, and reduce the term a a from the fraction's denominator since in the expression on the right side it does not appear, therefore we will choose the expression:

8a 8a because:

16=82 16=8\cdot 2 Let's verify that with this choice we indeed get the expression on the right side:

16ab?=2b11̸6b=?2b12b1=!2b1 \frac{16ab}{?}=\frac{2b}{1} \\ \downarrow\\ \frac{\not{16}\not{a}b}{\textcolor{red}{\not{8}\not{a}}}\stackrel{?}{= }\frac{2b}{1} \\ \downarrow\\ \boxed{\frac{2b}{1}\stackrel{!}{= }\frac{2b}{1} } therefore this choice is indeed correct.

In other words - the correct answer is answer B.

Answer

8a 8a

Exercise #14

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

27ab?=3ab27ab?=3ab1 \frac{27ab}{\text{?}}=3ab\\ \downarrow\\ \frac{27ab}{\text{?}}=\frac{3ab}{1} Now let's think logically, and remember the fraction reduction operation,

Note that both in the numerator of the expression on the right side and in the numerator of the expression on the left side exists the expression ab ab , therefore in the expression we are looking for there are no variables (since we are not interested in reducing them from the expression in the numerator on the left side),

Next, we ask which number was chosen to put in the denominator of the expression on the left side so that its reduction with the number 27 yields the number 3, the answer to this is of course - the number 9,

Because:

27=93 27=9\cdot 3 Let's verify that this choice indeed gives us the expression on the right side:

27ab?=3ab12̸7ab=?3ab13ab1=!3ab1 \frac{27ab}{\text{?}}=\frac{3ab}{1} \\ \downarrow\\ \frac{\not{27}ab}{\textcolor{red}{\not{9}}}\stackrel{?}{= }\frac{3ab}{1} \\ \downarrow\\ \boxed{\frac{3ab}{1}\stackrel{!}{= }\frac{3ab}{1} } Therefore this choice is indeed correct.

In other words - the correct answer is answer A.

Answer

9 9

Exercise #15

Complete the corresponding expression for the denominator

19ab?=a \frac{19ab}{?}=a

Video Solution

Step-by-Step Solution

Let's examine the problem, first we'll write down the expression on the right side as a fraction (using the fact that dividing a number by 1 doesn't change its value):

19ab?=a19ab?=a1 \frac{19ab}{?}=a \\ \downarrow\\ \frac{19ab}{?}=\frac{a}{1} Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 19 to get the number 1 and also reduce the term b b from the fraction's numerator since in the expression on the right side it doesn't appear, therefore we'll choose the expression:

19b 19b Let's verify that with this choice we indeed get the expression on the right side:

19ab?=a11̸9a1̸9=?a1a1=!a1 \frac{19ab}{?}=\frac{a}{1} \\ \downarrow\\ \frac{\not{19}a\not{b}}{\textcolor{red}{\not{19}\not{b}}}\stackrel{?}{= }\frac{a}{1} \\ \downarrow\\ \boxed{\frac{a}{1}\stackrel{!}{= }\frac{a}{1} } therefore this choice is indeed correct.

In other words - the correct answer is answer D.

Answer

19b 19b