Factoring Algebraic Fractions

Algebraic fractions are fractions with variables.

Ways to factor algebraic fractions:

We will find the most appropriate common factor to extract.
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.
If the formulas for abbreviated multiplication cannot be used, we will proceed to factorize with trinomials.
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).

Detailed explanation

Start practice

Test yourself on factorization and algebraic fractions!

Select the field of application of the following fraction:

$\frac{x}{16}$

Practice more now

Observe, you can factorize every expression included in your fraction separately in any way you desire and, in the end, you will arrive at the factorized expression.

Let's see an example of factoring algebraic fractions:
$\frac{x^2+7x+12}{x+3}=$

As you can see, in this fraction only the numerator can be factored.
We will factor it and obtain:
$\frac{(x+4)(x+3)}{(x+3)}=$
Now, we can reduce in the following way and we will obtain:

$x+4$

Examples and exercises with solutions on factoring algebraic fractions

Exercise #1

Select the field of application of the following fraction:

$\frac{x}{16}$

Video Solution

Step-by-Step Solution

Let's examine the given expression:

$\frac{x}{16}$

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:

$\frac{x}{16}$

the denominator is 16 and:

$16\neq0$

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

$All~X$

Exercise #2

Select the domain of the following fraction:

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

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:

$\frac{6}{x}$

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

In other words, the domain is:

$x\ne0$

Answer

All numbers except 0

Exercise #4

Select the field of application of the following fraction:

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

Video Solution

Step-by-Step Solution

Let's examine the given expression:

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

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.

In the given expression:

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

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:

$x+2\neq0$

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):

$x+2\neq0 \\ \boxed{x\neq -2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -2$

(This means that if we substitute for the variable x any number different from $(-2)$ 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

$x\neq-2$

Exercise #5

Select the field of application of the following fraction:

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

Video Solution

Step-by-Step Solution

Let's examine the given expression:

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

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.

In the given expression:

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

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:

$-2+x\neq0$

$-2+x\neq0 \\ \boxed{x\neq 2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 2$

(This means that if we substitute any number different from $2$ 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.