## Math 8 Chapter 2 Lesson 4: Reducing the denominator of many fractions

## 1. Theoretical Summary

### 1.1. Find the common denominator

– Factorize the denominators of the given fractions.

– The common denominator to be found is a product whose factors are chosen as follows:

- The numerical factor of the common denominator is the product of the numerical factors in the denominators of the learned fractions. (If the numerical factors of the denominators are positive integers, the numerical factor of the common denominator is their BCNN.)
- For each base of the power present in the denominators we choose the power with the highest exponent

### 1.2. Formal uniformity

To reduce the denominator of many fractions, we can do the following:

– Factorize the denominators and find the common denominator.

– Find the sub-factor of each denominator.

– Factor and denominator of each fraction with corresponding sub-factor.

## 2. Illustrated exercise

### 2.1. Exercise 1

The common denominator of the following fractions:

a.\(\frac{7}{{x + 4}},\frac{1}{{3{x^2} – 48}}\)

b.\(\frac{4}{{{x^3} – 7{x^2}}},\frac{1}{{2{x^2} – 14}}\)

**Solution guide**

a.

\(\begin{array}{l} \frac{7}{{x + 4}}\\ = \frac{{7.3.\left( {x – 4} \right)}}{{3\left( {x + 4} \right)\left( {x – 4} \right)}}\\ = \frac{{21\left( {x – 4} \right)}}{{3\left( {x + 4} \right)\left( {x – 4} \right)}} \end{array}\)

\(\begin{array}{l} \frac{1}{{3{x^2} – 48}}\\ = \frac{1}{{3\left( {x + 4} \right)\ left( {x – 4} \right)}} \end{array}\)

b.

\(\begin{array}{l} \frac{4}{{{x^3} – 7{x^2}}}\\ = \frac{4}{{x\left( {{x^2 } – 7} \right)}}\\ = \frac{{4.2}}{{2x\left( {{x^2} – 7} \right)}}\\ = \frac{8}{{2x \left( {{x^2} – 7} \right)}} \end{array}\)

\(\begin{array}{l} \frac{1}{{2{x^2} – 14}}\\ = \frac{1}{{2\left( {{x^2} – 7} \right)}}\\ = \frac{x}{{2x\left( {{x^2} – 7} \right)}} \end{array}\)

### 2.2. Exercise 2

The common denominator of the following fractions:

a.\(\frac{1}{{{x^6}{y^4}}},\frac{{29}}{{{x^{17}}{y^3}}}\)

b.\(\frac{3}{{25{x^3}{y^2}}},\frac{2}{{15{x^5}{y^2}}}\)

**Solution guide**

a.

\(\begin{array}{l} \frac{1}{{{x^6}{y^4}}}\\ = \frac{{{x^{11}}}}{{{x^) {17}}{y^4}}} \end{array}\)

\(\begin{array}{l} \frac{{29}}{{{x^{17}}{y^3}}}\\ = \frac{{29y}}{{{x^{17) }}{y^4}}} \end{array}\)

b.

\(\begin{array}{l} \frac{3}{{25{x^3}{y^2}}}\\ = \frac{{3.3{x^2}}}{{3{x ^2}.25{x^3}{y^2}}}\\ = \frac{{9{x^2}}}{{75{x^5}{y^2}}}\\ \ frac{2}{{15{x^5}{y^2}}}\\ = \frac{{2.5}}{{5.15{x^5}{y^2}}}\\ = \frac{ {10}}{{75{x^5}{y^2}}} \end{array}\)

### 2.3. Exercise 3

Given two fractions: \(\frac{1}{{{x^2} + 3x – 10}},\frac{x}{{{x^2} + 7x + 10}}\)

Prove \({x^3} + 5{x^2} – 4x – 20\) is the common denominator of the above two fractions and the common denominator of the two fractions above.

**Solution guide**

Prove:

\(\begin{array}{l} {x^3} + 5{x^2} – 4x – 20\\ = \left( {{x^2} + 3x – 10} \right)\left( { x + 2} \right)\\ {x^3} + 5{x^2} – 4x – 20\\ = \left( {{x^2} + 7x + 10} \right)\left( {x – 2} \right) \end{array}\)

Formal uniformity:

\(\begin{array}{l} \frac{1}{{{x^2} + 3x – 10}}\\ = \frac{{x + 2}}{{\left( {{x^2 } + 3x – 10} \right)\left( {x + 2} \right)}}\\ \frac{x}{{{x^2} + 7x + 10}}\\ = \frac{{x) \left( {x – 2} \right)}}{{\left( {{x^2} + 7x + 10} \right)\left( {x – 2} \right)}} \end{array} \)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **The common denominator of the fractions:

a) \(\displaystyle {{25} \over {14{x^2}y}},{{14} \over {21x{y^5}}}\)

b) \(\displaystyle {{11} \over {102{x^4}y}},{3 \over {34x{y^3}}}\)

c) \(\displaystyle {{3x + 1} \over {12x{y^4}}},{{y – 2} \over {9{x^2}{y^3}}}\)

d) \(\displaystyle {1 \over {6{x^3}{y^2}}},{{x + 1} \over {9{x^2}{y^4}}},{{ x – 1} \over {4x{y^3}}}\)

**Verse 2: **The common denominator of the fractions:

a) \(\displaystyle {{7x – 1} \over {2{x^2} + 6x}},{{5 – 3x} \over {{x^2} – 9}}\)

b) \(\displaystyle {{x + 1} \over {x – {x^2}}},{{x + 2} \over {2 – 4x + 2{x^2}}}\)

c) \(\displaystyle {{4{x^2} – 3x + 5} \over {{x^3} – 1}},{{2x} \over {{x^2} + x + 1}} ,{6 \over {x – 1}}\)

d) \(\displaystyle {7 \over {5x}},{4 \over {x – 2y}},{{x – y} \over {8{y^2} – 2{x^2}}} \)

**Question 3: **Given the polynomial \(B = 2{x^3} + 3{x^2} – 29x + 30\) and two fractions \(\displaystyle {x \over {2{x^2} + 7x – 15} }\), \(\displaystyle {{x + 2} \over {{x^2} + 3x – 10}}\)

a) Divide polynomial B by the denominators of the two given fractions in turn.

b) Reducing the denominator of the two given fractions.

**Question 4: **Given two fractions \(\displaystyle {1 \over {{x^2} – 4x – 5}}\) and \(\displaystyle {2 \over {{x^2} – 2x – 3}}\)

Show that it is possible to choose the polynomial \({x^3} – 7{x^2} + 7x + 15\) as a common denominator in order to congruent denominators of the two given fractions. Let’s reduce the denominator.

### 3.2. Multiple choice exercises

**Question 1:** When denominating the two fractions \(\frac{2}{{4{x^2}{y^2}}}\) and \(\,\frac{1}{{8{x^3} {y^3}}}\) the simplest minor factor of the first fraction is

A. 2x

B. 4xy

C. 8xy

D. 32xy

**Verse 2:** Fill in the blanks \(\frac{x}{{x + 2}} = \frac{{x\left( {x + 3} \right)}}{{…….}}\)

A. \({{{x^2} + 5x + 6}}\)

B. \(x+3\)

C. \(x+2\)

D. \(x(x+2)\)

**Question 3:** The common denominator of the two fractions \(\frac{{3x}}{{{x^2} – 4}}\) and \(\frac{x}{{x + 2}}\) is

A. \({{x^2} – 4}\)

B. \(x+2\)

C. \(x-2\)

D. \(\left( {{x^2} – 4} \right)\left( {x + 2} \right)\)

**Question 4:** when denominating the two fractions \(\frac{x}{{2x – 6}}\,\) and \(\frac{{x – 2}}{{{x^2} – 9}}\ ) get which of the following results?

A. \(\frac{x}{{2\left( {x – 3} \right)}};\,\,\frac{{x – 2}}{{{x^2} – 9}} \)

B. \(\frac{{x\left( {x + 3} \right)}}{{2\left( {x – 3} \right)\left( {x + 3} \right)}}\ ,;\,\frac{{2\left( {x – 2} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\)

C. \(\frac{{x\left( {x + 3} \right)}}{{2\left( {x – 3} \right)\left( {x + 3} \right)}}\ ,;\,\frac{{x – 2}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\)

D. \(\frac{{x\left( {x + 3} \right)}}{{2\left( {x – 3} \right)}}\,;\,\frac{{x – 2) }}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\)

**Question 5: **The value of \(A = \frac{{{x^2} + 3x + 1}}{{x + 2}}\) at x = 1 is

A. \(\frac{5}{3}\)

B. \(\frac{{11}}{4}\,\,\,\)

C. \(\frac{5}{4}\)

D. \(\frac{4}{{11}}\)

## 4. Conclusion

Through this lesson, you will learn some of the main topics as follows:

- Know how to find the common denominator after factoring the denominators.
- Grasp the process of homogenization
- Know how to find the minor factor and how to do the exercises to return to the common denominator.

.

=============

## Leave a Reply