A rational expression (or function), is any expression of the form \(f(x)=\large{\frac{N(x)}{D(x)}}\), where the numerator, \(N(x)\), and the denominator, \(D(x)\), are polynomials.
| Examples of REs | Non-examples of REs |
| \(f(x)=\large{\frac{4x^2-7x+13}{x-6}}\) | \(g(x)=\large{\frac{\sqrt{x-2}}{3x^4-8x^2+x-9}}\) |
| \(y=18-5x\iff y=\large{\frac{18-5x}{1}}\) | \(h(x)=\large{\frac{|6-11x|}{3x+1}}\) |
The domain consists of all numbers the are allowed into the function. Rational functions allow all numbers except those that make the denominator equal zero: \(D(x)=0\). As a quick illustration, in the first RE above: $$f(x)=\frac{4x^2-7x+13}{x-6}$$ the domain is \(\left(-\infty,6\right) \cup \left(6,\infty\right)\) since the denominator, \(D(x)=x-6\), equals zero when \(x-6=0\implies x=6\). For most problems, the REs will be trinomials that are factorable. All the factoring skills from the previous section will be needed.
Since REs act like fractions, a situation like reducing \(\large{\frac{25}{75}}\normalsize{=}\large{\frac{1}{3}}\) occurs often. The first thing is factor both \(N(x)\) and \(D(x)\) completely and cross out any common factors: $$\frac{x^2+4x+4}{x^2-6x-16}\implies \frac{(x+2)(x+2)}{(x-8)(x+2)}$$ Since the common factor of \(x+2\) appears in both \(N(x)\) and \(D(x)\), cancel it out: $$\frac{(x+2)\color{red}{(x+2)}}{(x-8)\color{red}{(x+2)}}\implies \frac{x+2}{x-8}$$ is now in reduced terms. When considering the domain, all values of the original function must be considered NOT the reduced. In this case, the domain would be \((-\infty,-2)\cup (-2,8)\cup (8,\infty)\)
This process is an extension of multiplication of fractions: $$\frac{20}{12}\cdot \frac{45}{36}$$ One method involves multiplying straight across and then reduce the fraction. The method discussed here will use canceling out factors and then multiplying the left over factors: $$\frac{4\cdot5}{4\cdot3}\cdot\frac{5\cdot9}{9\cdot4}\implies\frac{\color{red}{4}\cdot5}{\color{red}{4}\cdot3}\cdot\frac{5\cdot\color{red}{9}}{\color{red}{9}\cdot4}$$ The common factors of \(4\) and \(9\) cancel out since everything is written as multiplication, then multiply the left over factors. $$\frac{5\cdot5}{3\cdot4}=\frac{25}{12}$$ This idea is exactly how to handle multiplication and division of REs.
Multiply \(\large{\frac{x-4}{x^2-10x+24}}\normalsize{\cdot}\large{\frac{x^2-3x-18}{x^2+7x+12}}\).
| Factor completely numerator and denominator: | \(\large{\frac{x-4}{(x-6)(x-4)}}\normalsize{\cdot}\large{\frac{(x+3)(x-6)}{(x+4)(x+3)}}\) |
| Cancel out common factors: | \(\large{\frac{\color{red}{x-4}}{\color{red}{(x-4)}\color{blue}{(x-6)}}}\normalsize{\cdot}\large{\frac{\color{green}{(x+3)\color{blue}{(x-6)}}}{(x+4)\color{green}{(x+3)}}}\) |
| Leave in factored form, no need to multiply out. | \(\large{\frac{1}{x+4}}\;\color{green}{\checkmark}\) |
Multiply \(\large{\frac{6x^2+5xy-6y^2}{12x^2-11xy+2y^2}}\normalsize{\cdot}\large{\frac{8x^2-14xy+3y^2}{4x^2-12xy+9y^2}}\).
| Factor all parts completely: | \(\large{\frac{(2x+3y)(3x-2y)}{(3x-2y)(4x-1y)}}\normalsize{\cdot}\large{\frac{(2x-3y)(2x-3y)}{(4x-1y)(2x-3y)}}\) |
| Cancel out common factors: | \(\large{\frac{2x+3y}{4x-y}}\normalsize{\cdot}\large{\frac{2x-3y}{4x-y}}\normalsize{\implies}\large{\frac{(2x+3y)(2x-3y)}{\left(4x-y\right)^2}}\;\color{green}{\checkmark}\) |
In order to divide, we must understand that dividing fractions really turns into multiplying by the reciprocal of the second fraction: $$\frac{10}{4}\div \frac{5}{16} \implies \frac{10}{4}\cdot \color{green}{\frac{16}{5}}$$ Now simply apply the multiplication techniques above:
$$\frac{2\cdot \color{red}{5}}{\color{red}{4}}\cdot \frac{\color{red}{4}\cdot 4}{\color{red}{5}}\implies \frac{2\cdot 4}{1}\implies 8$$Divide \(\large{\frac{5m^2+17m-12}{3m^2+7m-20}}\normalsize{\div}\large{\frac{5m^2+2m-3}{15m^2-34m+15}}\).
| Factor all parts completely: | \(\large{\frac{(5m-3)(m+4)}{(m+4)(3m-5)}}\normalsize{\cdot}\large{\frac{(3m-5)(5m-3)}{(5m-3)(m+1)}}\) |
| Multiply by reciprocal of second fraction: | \(\large{\frac{(5m-3)(m+4)}{(m+4)(3m-5)}}\normalsize{\cdot}\large{\frac{(5m-3)(m+1)}{(3m-5)(5m-3)}}\) |
| Cancel out common factors: | \(\large{\frac{5m-3}{m+1}}\;\color{green}{\checkmark}\) |
I know this goes without saying, but this section really takes both good factoring skills, and then canceling out factored terms. One major mistake that I see a lot, is when students try to cancel BEFORE factoring. Canceling can only occur once there is multiplicaton throughout the numerator and denominator, so that is why factoring is important!!!