Factoring Algebraic Fractions

πŸ†Practice factorization and algebraic fractions

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).
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Test yourself on factorization and algebraic fractions!

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Complete the corresponding expression for the denominator

\( \frac{12ab}{?}=1 \)

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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:
x2+7x+12x+3=\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:
(x+4)(x+3)(x+3)=\frac{(x+4)(x+3)}{(x+3)}=
Now, we can reduce in the following way and we will obtain:

Factorization of algebraic fractions


x+4x+4


Examples and exercises with solutions on factoring algebraic fractions

Exercise #1

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Let's examine the problem:

12ab?=1 \frac{12ab}{?}=1 Now let's think logically, and remember the known fact that dividing any number by itself always yields the result 1,

Therefore, in order to get the result 1 from dividing two numbers, the only way is to divide the number by itself, meaning-

The missing expression in the denominator of the fraction on the left side is the complete expression that appears in the numerator of the same fraction:

12ab 12ab .

Therefore- the correct answer is answer D.

Answer

12ab 12ab

Exercise #2

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

After examining the problem, proceed to 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?=2b↓16ab?=2b1 \frac{16ab}{?}=2b \\ \downarrow\\ \frac{16ab}{?}=\frac{2b}{1}

Remember the fraction reduction operation,

In order for the fraction on the left side to be deemed reducible, we want all the terms in its denominator to have a common factor. Additionally, we want to reduce the number 16 in order to obtain the number 2. Furthermore we want to reduce the term a a from the fraction's denominator given that in the expression on the right side it does not appear. Therefore we will choose the expression:

8a 8a

Due to the fact that:

16=8β‹…2 16=8\cdot 2

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

16ab?=2b1↓1ΜΈ6aΜΈb8ΜΈaΜΈ=?2b1↓2b1=!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 #3

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Using the formula:

xy=zw→x⋅y=z⋅y \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}

16abΓ—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 #4

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Upon examining the problem, proceed to 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?=a↓19ab?=a1 \frac{19ab}{?}=a \\ \downarrow\\ \frac{19ab}{?}=\frac{a}{1}
Remember the fraction reduction operation,

In order for the fraction on the left side to be deemed reducible, we want all the terms in its denominator to have a common factor. Additionally, we want to reduce the number 19 in order to obtain the number 1 as well as reducing the term b b from the fraction's numerator given that in the expression on the right side it doesn't appear. Therefore we'll choose the expression:

19b 19b

Let's verify that this choice results in the expression on the right side:

19ab?=a1↓1ΜΈ9abΜΈ1ΜΈ9bΜΈ=?a1↓a1=!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

Exercise #5

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Upon examining the problem, proceed to 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?=3ab↓27ab?=3ab1 \frac{27ab}{\text{?}}=3ab\\ \downarrow\\ \frac{27ab}{\text{?}}=\frac{3ab}{1}

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 the expression ab ab is present. 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, determine which number was chosen to be in the denominator of the expression on the left side in order that its reduction with the number 27 yields the number 3. The answer to this - the number 9,

Due to the fact that:

27=9β‹…3 27=9\cdot 3

Let's verify that this choice indeed gives us the expression on the right side:

27ab?=3ab1↓2ΜΈ7ab9ΜΈ=?3ab1↓3ab1=!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

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