Explore detailed step-by-step solutions and explanations on how to multiply, divide, factor, and simplify rational expressions. Each question comes with a complete worked solution, including factoring techniques, cancellation of common factors, and domain restrictions. These examples are designed to help students, parents, and teachers better understand rational expressions and prepare for algebra practice.
We multiply two rational expressions by multiplying their numerators and denominators as follows: \[ \frac{A}{B} \cdot \frac{C}{D} = \frac{A \cdot C}{B \cdot D} \] We divide two rational expressions by multiplying the first by the reciprocal of the second as follows: \[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{A \cdot D}{B \cdot C} \]
The graph shows a simple rational curve. Divide and simplify:
\[ \frac{-3}{2} \div \frac{6x - 9}{2x - 3} \]The division of two rational expressions is done by multiplying the first by the reciprocal of the second as follows (see division rule above). Hence
\[ -\frac{3}{2} \div \frac{6x - 9}{2x - 3} = \frac{-3}{2} \cdot \frac{2x - 3}{6x - 9} \]Multiply numerators and denominators (multiplication rule).
\[ = \frac{-3(2x - 3)}{2(6x - 9)} \]Factor the terms 6x-9 included in the denominator as follows:
\[ 6x - 9 = 3(2x - 3) \]and use the factored form in the rational expression
\[ = \frac{-3(2x - 3)}{2 \cdot 3(2x - 3)} \]simplify
\[ = \frac{\cancel{-3}\cancel{(2x - 3)}}{2 \cdot \cancel{3}\cancel{(2x - 3)}} = -\frac{1}{2} \text{ for } x \neq \frac{3}{2} \]The graph shows a rational curve with vertical asymptotes. Factor and simplify:
\[ \frac{2x - 5}{2x + 2} \cdot \frac{10x + 10}{4x - 10} \]Apply the multiplication rule (see above)
\[ \frac{2x - 5}{2x + 2} \cdot \frac{10x + 10}{4x - 10} = \frac{(2x - 5)(10x + 10)}{(2x + 2)(4x - 10)} \]Factor terms in the numerator and the denominator:
\[ 10 x + 10 = 10(x + 1) \qquad 2 x + 2 = 2(x + 1) \qquad 4 x - 10 = 2(2x - 5) \]and use in factored form
\[ = \frac{(2x-5) \cdot 10(x+1)}{2(x+1) \cdot 2(2x-5)} \]Simplify if possible
\[ = \frac{\cancel{(2x-5)} \cdot 10 \cancel{(x+1)}}{2 \cancel{(x+1)} \cdot 2 \cancel{(2x-5)}} = \frac{10}{4} = \frac{2 \cdot 5}{2 \cdot 2} = \frac{5}{2} \text{ for } x \neq -1 \text{ and } x \neq \frac{5}{2} \]The graph shows intersecting rational functions.
\[ \frac{2x^2 - 7x - 15}{x^2 + 3x - 4} \div \frac{x^2 + x - 30}{x^2 - 1} \]The division of two rational expressions is done by multiplying the first rational expression by the reciprocal of the second rational expression as follows (see division rule above). Hence
\[ \frac{2x^2 - 7x - 15}{x^2 + 3x - 4} \div \frac{x^2 + x - 30}{x^2 - 1} = \frac{2x^2 - 7x - 15}{x^2 + 3x - 4} \cdot \frac{x^2 - 1}{x^2 + x - 30} \]Multiply numerators and denominators (multiplication rule) but do not expand as we might be able to simplifty.
\[ = \frac{(2x^2 - 7x - 15)(x^2 - 1)}{(x^2 + 3x - 4)(x^2 + x - 30)} \]factor terms in the numerator and denominator if possible.
\[ 2 x^2 - 7 x - 15 = (2x + 3)(x - 5) ; x^2 - 1 = (x - 1)(x + 1) \] \[ x^2 + 3 x - 4 = (x + 4)(x - 1) ; x^2 + x - 30 = (x + 6)(x - 5) \]and use in rational expression
\[ = \frac{(2x+3)(x-5)(x-1)(x+1)}{(x+4)(x-1)(x+6)(x-5)} \]and simplify.
\[ = \frac{\cancel{(2x+3)}\cancel{(x-5)}\cancel{(x-1)}(x+1)}{(x+4)\cancel{(x-1)}(x+6)\cancel{(x-5)}} = \frac{(2x+3)(x+1)}{(x+4)(x+6)} \text{ for } x\neq -6,-4,-1,1,5 \]The graph shows a rational curve with multiple asymptotes. Divide, factor and simplify:
\[ \left( \frac{x-1}{x+2} + \frac{x^2-4}{x^2-1} \right) + \frac{x-2}{x+5} \]We have the multiplication of two rational expressions inside the parentheses and we then apply multiplication rule. We also have a division by a rational expression which is done by multiplying by the reciprocal. Hence
\[ \left( \frac{x-1}{x+2} \cdot \frac{x^2-4}{x^2-1} \right) \div \frac{x-2}{x+5} = \frac{(x-1)(x^2-4)}{(x+2)(x^2-1)} \cdot \frac{x+5}{x-2} \]Multiply numerators and denominators (multiplication rule) but do not expand as we might be able to simplify.
\[ = \frac{(x-1)(x^2-4)(x+5)}{(x+2)(x^2-1)(x-2)} \]Factor terms in numerator and denominator if possible and use in rational expression
\[ x^2 - 4 = (x - 2)(x + 2) \text{ and } x^2 - 1 = (x - 1)(x + 1)\]Simplify
\[ = \frac{\cancel{(x-1)}\cancel{(x-2)}\cancel{(x+2)}(x+5)}{\cancel{(x+2)}\cancel{(x-1)}(x+1)\cancel{(x-2)}} = \frac{x+5}{x+1} \text{ for } x\neq -2,-1,1,-5,2 \]The graph shows a rational function with a cubic numerator. Divide, factor and simplify:
\[ \frac{x^3 - 27}{x+3} \div \frac{x-3}{(x+3)^2} \]The division of two rational expressions is done by multiplying the first rational expression by the reciprocal of the second rational expression as follows (see division rule above). Hence
\[ \frac{x^3 - 27}{x + 3} \div \frac{x - 3}{(x + 3)^2} = \frac{x^3 - 27}{x + 3} \cdot \frac{(x + 3)^2}{x - 3} \]Multiply numerators and denominators (multiplication rule) but do not expand.
\[ = \frac{(x^3 - 27)(x + 3)^2}{(x + 3)(x - 3)} \]factor term \(x^3 - 27\) in the numerator and use it.
\[ x^3 - 27 = (x - 3)(x^2 + 3 x + 9) \]and use in rational expression
\[ = \frac{(x - 3)(x^2 + 3x + 9)(x + 3)^2}{(x + 3)(x - 3)} \]and simplify.
\[ = \frac{\cancel{(x - 3)}(x^2 + 3x + 9)(x + 3)^{\cancel{2}}}{\cancel{(x + 3)}\cancel{(x - 3)}} = (x + 3)(x^2 + 3x + 9) \] \[ \text{for } x \neq -3 \text{ and } x \neq 3 \]The graph shows a rational function with multiple asymptotes. Multiply, factor and simplify:
\[ \frac{2y - x}{4x^2 - 9y^2} \cdot \frac{4x + 6y}{y - \frac{1}{2}x} \]Apply the multiplication rule (see above)
\[ \frac{2y - x}{4x^2 - 9y^2} \cdot \frac{4x + 6y}{y - \frac{1}{2}x} = \frac{(2y - x)(4x + 6y)}{(4x^2 - 9y^2)(y - \frac{1}{2}x)} \]Factor terms in the numerator and the denominator:
\[ 2 y - x = 2 (y - (1/2) x) ; 4 x + 6 y = 2(2 x + 3) ; 4 x^2 - 9 y^2 = (2x - 3y)(2x + 3y) \]and use in actored form
\[ = \frac{2\left(y - \frac{1}{2}x\right) \cdot 2(2x + 3y)}{(2x - 3y)(2x + 3y)\left(y - \frac{1}{2}x\right)} \]Simplify if possible
\[ = \frac{\cancel{2}\cancel{\left(y - \frac{1}{2}x\right)} \cdot 2\cancel{(2x + 3y)}}{\cancel{(2x - 3y)}\cancel{(2x + 3y)}\cancel{\left(y - \frac{1}{2}x\right)}} = \frac{4}{2x - 3y} \text{ for } x \neq 2y , -\frac{3y}{2} , \frac{3y}{2} \]