Chapter 6.2: Multiplication and Division of Rational Expressions

Multiplying and dividing rational expressions is very similar to the process used to multiply and divide fractions.

Example 1

Reduce and multiply \dfrac{15}{49} and \dfrac{14}{45}.

    \[\dfrac{15}{49}\cdot \dfrac{14}{45}\text{ reduces to }\dfrac{1}{7}\cdot \dfrac{2}{3}, \text { which equals }\dfrac{2}{21}\]

(15 and 45 reduce to 1 and 3, and 14 and 49 reduce to 2 and 7)

This process of multiplication is identical to division, except the first step is to reciprocate any fraction that is being divided.

Example 2

Reduce and divide \dfrac{25}{18} by \dfrac{15}{6}.

\dfrac{25}{18} \div \dfrac{15}{6} \text{ reciprocates to } \dfrac{25}{18}\cdot \dfrac{6}{15}, \text{ which reduces to }\dfrac{5}{3}\cdot \dfrac{1}{3}, \text{ which equals } \dfrac{5}{9}

(25 and 15 reduce to 5 and 3, and 6 and 18 reduce to 1 and 3)

When multiplying with rational expressions,  follow the same process: first, divide out common factors, then multiply straight across.

Example 3

Reduce and multiply \dfrac{25x^2}{9y^8} and \dfrac{24y^4}{55x^7}.

    \[\dfrac{25x^2}{9y^8}\cdot \dfrac{24y^4}{55x^7}\text{ reduces to }\dfrac{5}{3y^4}\cdot \dfrac{8}{11x^5}, \text{ which equals }\dfrac{40}{33x^5y^4}\]

(25 and 55 reduce to 5 and 11, 24 and 9 reduce to 8 and 3, x2 and x7 reduce to x5, y4 and y8 reduce to y4)

Remember: when dividing fractions, reciprocate the dividing fraction.

Example 4

Reduce and divide \dfrac{a^4b^2}{a} by \dfrac{b^4}{4}.

\dfrac{a^4b^2}{a} \div \dfrac{b^4}{4}\text{ reciprocates to } \dfrac{a^4b^2}{a}\cdot \dfrac{4}{b^4}, \text{ which reduces to }\dfrac{a^3}{1}\cdot \dfrac{4}{b^2}, \text{ which equals }\dfrac{4a^3}{b^2}

(After reciprocating, 4a4b2 and b4 reduce to 4a3 and b2)

In dividing or multiplying some fractions, the polynomials in the fractions must be factored first.

Example 5

Reduce, factor and multiply \dfrac{x^2-9}{x^2+x-20} and \dfrac{x^2-8x+16}{3x+9}.

    \[\dfrac{x^2-9}{x^2+x-20}\cdot \dfrac{x^2-8x+16}{3x+9}\text{ factors to }\dfrac{(x+3)(x-3)}{(x-4)(x+5)}\cdot \dfrac{(x-4)(x-4)}{3(x+3)}\]

Dividing or cancelling out the common factors (x + 3) and (x - 4) leaves us with \dfrac{x-3}{x+5}\cdot \dfrac{x-4}{3}, which results in \dfrac{(x-3)(x-4)}{3(x+5)}.

Example 6

Reduce, factor and multiply or divide the following fractions:

    \[\dfrac{a^2+7a+10}{a^2+6a+5}\cdot \dfrac{a+1}{a^2+4a+4}\div \dfrac{a-1}{a+2}\]

Factoring each fraction and reciprocating the last one yields:

    \[\dfrac{(a+5)(a+2)}{(a+5)(a+1)}\cdot \dfrac{(a+1)}{(a+2)(a+2)}\cdot \dfrac{(a+2)}{(a-1)}\]

Dividing or cancelling out the common polynomials leaves us with:

    \[\dfrac{1}{a-1}\]

Questions

Simplify each expression.

  1. \dfrac{8x^2}{9}\cdot \dfrac{9}{2}
  2. \dfrac{8x}{3}\div \dfrac{4x}{7}
  3. \dfrac{5x^2}{4}\cdot \dfrac{6}{5}
  4. \dfrac{10p}{5}\div \dfrac{8}{10}
  5. \dfrac{(m-6)}{7(7m-5)}\cdot \dfrac{5m(7m-5)}{m-6}
  6. \dfrac{7(n-2)}{10(n+3)}\div \dfrac{n-2}{(n+3)}
  7. \dfrac{7r}{7r(r+10)}\div \dfrac{r-6}{(r-6)^2}
  8. \dfrac{6x(x+4)}{(x-3)}\cdot \dfrac{(x-3)(x-6)}{6x(x-6)}
  9. \dfrac{x-10}{35x+21}\div \dfrac{7}{35x+21}
  10. \dfrac{v-1}{4}\cdot \dfrac{4}{v^2-11v+10}
  11. \dfrac{x^2-6x-7}{x+5}\cdot \dfrac{x+5}{x-7}
  12. \dfrac{1}{a-6}\cdot \dfrac{8a+80}{8}
  13. \dfrac{4m+36}{m+9}\cdot \dfrac{m-5}{5m^2}
  14. \dfrac{2r}{r+6}\div \dfrac{2r}{7r+42}
  15. \dfrac{n-7}{6n-12}\cdot \dfrac{12-6n}{n^2-13n+42}
  16. \dfrac{x^2+11x+24}{6x^3+18x^2}\cdot \dfrac{6x^3+6x^2}{x^2+5x-24}
  17. \dfrac{27a+36}{9a+63}\div \dfrac{6a+8}{2}
  18. \dfrac{k-7}{k^2-k-12}\cdot \dfrac{7k^2-28k}{8k^2-56k}
  19. \dfrac{x^2-12x+32}{x^2-6x-16}\cdot \dfrac{7x^2+14x}{7x^2+21x}
  20. \dfrac{9x^3+54x^2}{x^2+5x-14}\cdot \dfrac{x^2+5x-14}{10x^2}
  21. (10m^2+100m)\cdot \dfrac{18m^3-36m^2}{20m^2-40m}
  22. \dfrac{n-7}{n^2-2n-35}\div \dfrac{9n+54}{10n+50}
  23. \dfrac{x^2-1}{2x-4}\cdot \dfrac{x^2-4}{x^2-x-2}\div \dfrac{x^2+x-2}{3x-6}
  24. \dfrac{a^3+b^3}{a^2+3ab+2b^2}\cdot \dfrac{3a-6b}{3a^2-3ab+3b^2}\div \dfrac{a^2-4b^2}{a+2b}

Answers to odd questions

1. \dfrac{8x^2}{9}\cdot \dfrac{9}{2}\Rightarrow \dfrac{\cancel{2}\cdot 4\cdot x^2}{\cancel{9}}\cdot \dfrac{\cancel{9}}{\cancel{2}}\Rightarrow 4x^2

3. \dfrac{5x^2}{4}\cdot \dfrac{6}{5}\Rightarrow \dfrac{\cancel{5}\cdot x^2}{2\cdot \cancel{2}}\cdot \dfrac{\cancel{2}\cdot 3}{\cancel{5}}\Rightarrow \dfrac{3x^2}{2}

5. \dfrac{\cancel{(m-6)}}{7\cancel{(7m-5)}}\cdot \dfrac{5m\cancel{(7m-5)}}{\cancel{m-6}}\Rightarrow \dfrac{5m}{7}

7. \dfrac{7r}{7r(r+10)}\div \dfrac{r-6}{(r-6)^2}\Rightarrow \dfrac{7r}{7r(r+10)}\cdot \dfrac{(r-6)^2}{r-6}\Rightarrow \dfrac{\cancel{7r}}{\cancel{7r}(r+10)}\cdot \dfrac{(r-6)\cancel{(r-6)}}{\cancel{r-6}}\Rightarrow \\
\dfrac{r-6}{r+10}

9. \dfrac{x-10}{35x+21}\div \dfrac{7}{35x+21}\Rightarrow \dfrac{x-10}{7\cancel{(5x+3)}}\cdot \dfrac{\cancel{7}\cancel{(5x+3)}}{\cancel{7}}\Rightarrow \dfrac{x-10}{7}

11. \dfrac{x^2-6x-7}{x+5}\cdot \dfrac{x+5}{x-7}\Rightarrow \dfrac{\cancel{(x-7)}(x+1)}{\cancel{(x+5)}}\cdot \dfrac{\cancel{(x+5)}}{\cancel{(x-7)}}\Rightarrow x+1

13. \dfrac{4m+36}{m+9}\cdot \dfrac{m-5}{5m^2}\Rightarrow \dfrac{4\cancel{(m+9)}}{\cancel{m+9}}\cdot \dfrac{m-5}{5m^2}\Rightarrow \dfrac{4(m-5)}{5m^2}

15. \dfrac{n-7}{6n-12}\cdot \dfrac{12-6n}{n^2-13n+42}\Rightarrow \dfrac{\cancel{(n-7)}}{\cancel{6}(n-2)}\cdot \dfrac{\cancel{6}(2-n)}{(n-6)\cancel{(n-7)}}\Rightarrow \dfrac{-1\cancel{(n-2)}}{\cancel{(n-2)}(n-6)}\Rightarrow \\
\dfrac{-1}{n-6}

17. \dfrac{27a+36}{9a+63}\div \dfrac{6a+8}{2}\Rightarrow \dfrac{\cancel{9}\cancel{(3a+4)}}{\cancel{9}(a+7)}\cdot \dfrac{\cancel{2}}{\cancel{2}\cancel{(3a+4)}}\Rightarrow \dfrac{1}{a+7}

19. \dfrac{x^2-12x+32}{x^2-6x-16}\cdot \dfrac{7x^2+14x}{7x^2+21x}\Rightarrow \dfrac{\cancel{(x-8)}(x-4)}{\cancel{(x-8)}\cancel{(x+2)}}\cdot \dfrac{\cancel{7x}\cancel{(x+2)}}{\cancel{7x}(x+3)}\Rightarrow \dfrac{x-4}{x+3}

21. (10m^2+100m)\cdot \dfrac{18m^3-36m^2}{20m^2-40m}\Rightarrow \cancel{10m}(m+10)\cdot \dfrac{\cancel{2}\cdot 9m^2\cancel{(m-2)}}{\cancel{2}\cdot \cancel{10m}\cancel{(m-2)}}\Rightarrow
9m^2(m+10)

23. \\ \dfrac{x^2-1}{2x-4}\cdot \dfrac{x^2-4}{x^2-x-2}\div \dfrac{x^2+x-2}{3x-6}\Rightarrow \\
\dfrac{\cancel{(x-1)}\cancel{(x+1)}}{2\cancel{(x-2)}}\cdot \dfrac{\cancel{(x+2)}\cancel{(x-2)}}{\cancel{(x-2)}\cancel{(x+1)}}\cdot \dfrac{3\cancel{(x-2)}}{\cancel{(x+2)}\cancel{(x-1)}}\Rightarrow \dfrac{3}{2}

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