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# Division of Rational Expressions

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Dividing Rational Expressions

The area of a rectangle is $\frac{12x^2yz^3}{5xy^2z}$ . The length of the rectangle is $\frac{2xy}{z^2}$ . What is the width of the rectangle?

### Guidance

Dividing rational expressions has one additional step than multiply them. Recall that when you divide fractions, you need to flip the second fraction and change the problem to multiplication. The same rule applies to dividing rational expressions.

#### Example A

Divide $\frac{5a^3b^4}{12ab^8} \div \frac{15b^6}{8a^6}$ .

Solution: Flip the second fraction, change the $\div$ sign to multiplication and solve.

$\frac{5a^3b^4}{12ab^8} \div \frac{15b^6}{8a^6} = \frac{5a^3b^4}{12ab^8} \cdot \frac{8a^6}{15b^6} = \frac{40a^9b^4}{180ab^{14}} = \frac{2a^8}{9b^{10}}$

#### Example B

Divide $\frac{x^4-3x^2-4}{2x^2+x-10} \div \frac{x^3-3x^2+x-3}{x-2}$

Solution: Flip the second fraction, change the $\div$ sign to multiplication and solve.

$\frac{x^4-3x^2-4}{2x^2+x-10} \div \frac{x^3-3x^2+x-3}{x-2} &= \frac{{\color{red}x^4-3x^2-4}}{2x^2+x-10} \cdot \frac{x-2}{{\color{blue}x^3-3x^2+x-3}} \\&= \frac{(x^2-4)(x^2+1)}{(2x-5)(x+2)} \cdot \frac{x-2}{(x^2+1)(x-3)} \\&= \frac{(x-2)\cancel{(x+2)} \cancel{(x^2+1)}}{(2x-5)\cancel{(x+2)}} \cdot \frac{x-2}{\cancel{(x^2+1)}(x-3)} \\&= \frac{(x-2)^2}{(2x-5)(x-3)}$

Review the Factoring by Grouping concept to factor the blue polynomial and the Factoring Polynomials in Quadratic Form concept to factor the red polynomial.

#### Example C

Perform the indicated operations: $\frac{{\color{blue}x^3-8}}{x^2-6x+9} \div (x^2+3x-10) \cdot \frac{x^2+x-12}{x^2+11x+30}$

Solution: Flip the second term, factor, and cancel. The blue polynomial is a difference of cubes. Review the Sum and Difference of Cubes concept for how to factor this polynomial.

$\frac{x^3-8}{x^2-6x+9} \div (x^2+3x-10) \cdot \frac{x^2+x-12}{x^2+11x+30} &= \frac{x^3-8}{x^2-6x+9} \cdot \frac{1}{x^2+3x-10} \cdot \frac{x^2+2x-15}{x^2+11x+30} \\&= \frac{\cancel{(x-2)}(x^2+2x+4)}{\cancel{(x-3)}(x-3)} \cdot \frac{1}{\cancel{(x-2)} \cancel{(x+5)}} \cdot \frac{\cancel{(x+5)} \cancel{(x-3)}}{(x+5)(x+6)} \\&= \frac{x^2+2x+4}{(x-3)(x+5)(x+6)}$

Intro Problem Revisit To find the width, divide the area by the length and simplify.

$\frac{12x^2yz^3}{5xy^2z} \div \frac{2xy}{z^2}\\\frac{12x^2yz^3}{5xy^2z} \cdot \frac{z^2}{2xy}\\\frac{12x^2yz^5}{10x^2y^3z}\\\frac{6z^4}{5y^2}$

Therefore, the width of the rectangle is $\frac{6z^4}{5y^2}$ .

### Guided Practice

Perform the indicated operations.

1. $\frac{a^5b^3c}{6a^2c^9} \div \frac{2a^7b^{11}}{24c^2}$

2. $\frac{x^2+12x-45}{x^2-5x+6} \div \frac{x^2+17x+30}{x^4-16}$

3. $(x^3+2x^2-9x-18) \div \frac{x^2+11x+24}{x^2-11x-24} \div \frac{x^2-6x-16}{x^2+5x-24}$

1. $\frac{a^5b^3c}{6a^2c^9} \div \frac{2a^7b^{11}}{24c^2} = \frac{a^5b^3c}{6a^2c^9} \cdot \frac{24c^2}{2a^7b^{11}} = \frac{24a^5b^3c^3}{12a^9b^{11}c^9} = \frac{2}{a^4b^8c^6}$

2. $\frac{x^2+12x-45}{x^2-5x+6} \div \frac{x^2+17x+30}{x^4-16} &= \frac{x^2+12x-45}{x^2-5x+6} \cdot \frac{x^4-16}{x^2+17x+30} \\&= \frac{\cancel{(x+15)}\cancel{(x-3)}}{\cancel{(x-3)} \cancel{(x-2)}} \cdot \frac{(x^2+4)\cancel{(x-2)} \cancel{(x+2)}}{\cancel{(x+15)} \cancel{(x+2)}} \\&= x^2+4$

3. $(x^3+2x^2-9x-18) \div \frac{x^2+11x+24}{x^2-11x+24} \div \frac{x^2-6x-16}{x^2+5x-24} &= \frac{x^3+2x^2-9x-18}{1} \cdot \frac{x^2-11x+24}{x^2+11x+24} \cdot \frac{x^2+5x-24}{x^2-6x-16} \\&= \frac{(x-3) \cancel{(x+3)} \cancel{(x+2)}}{1} \cdot \frac{\cancel{(x-8)}(x-3)}{\cancel{(x+8)}\cancel{(x+3)}} \cdot \frac{\cancel{(x+8)}(x-3)}{\cancel{(x-8)} \cancel{(x+2)}} \\&=(x-3)^2$

### Practice

Divide the following expressions. Simplify your answer.

1. $\frac{6a^4b^3}{8a^3b^6} \div \frac{3a^5}{4a^3b^4}$
2. $\frac{12x^5y}{xy^4} \div \frac{18x^3y^6}{3x^2y^3}$
3. $\frac{16x^3y^9z^3}{15x^5y^2z} \div \frac{42xy^7z^2}{45x^2yz^5}$
4. $\frac{x^2+2x-3}{x^2-3x+2} \div \frac{x^2+3x}{4x-8}$
5. $\frac{x^2-2x-3}{x^2+6x+5} \div \frac{4x-12}{x^2+8x+15}$
6. $\frac{x^2+6x+2}{12-3x} \div \frac{6x^2-13x-5}{x^2-4x}$
7. $\frac{x^2-5x}{x^2+x-6} \div \frac{x^2-2x-15}{x^3+3x^2-4x-12}$
8. $\frac{3x^3-3x^2-6x}{2x^2+15x-8} \div \frac{6x^2+18x-60}{2x^2+9x-5}$
9. $\frac{x^3+27}{x^2+5x-14} \div \frac{x^2-x-12}{2x^2+2x-40} \div \frac{1}{x-2}$
10. $\frac{x^2+2x-15}{2x^3+7x^2-4x} \div (5x+3) \div \frac{21-10x+x^2}{5x^3+23x^2+12x}$

We all know that when you divide fractions, you take the second fraction, flip it, and change it to a multiplication problem. But, do you know why? Let's investigate the why here.

1. What is $6 \div 2$ ?
2. What about $\frac{1 \div 1}{6 \div 2}$ ?
3. Is the problem above the same as $\frac{1}{6} \div \frac{1}{2}$ ? Why or why not?

Let's take a different approach. Let's write a division problem as a huge fraction: $\frac{\frac{30}{52}}{\frac{15}{13}}$

1. We know we cannot have fractions in the denominator of another fraction. What would we have to multiply the denominator by to cancel it out?
2. Multiply the top and bottom from your answer in #14. What did you multiply by?