In the last section, we learned some rational function applications. In this section, we will learn how to simplify rational expressions, and how to multiply and divide them.
Let's review how to reduce fractions:
\begin{align*} \frac{2}{4}\amp=\frac{1\cdot2}{2\cdot2}\\ \amp=\frac{1\cdot\cancel{2}}{2\cdot\cancel{2}}\\ \amp=\frac{1}{2} \end{align*}Because those two \(2\)'s are connected to the numerator and denominator by multiplication, we can cross them out. Can we cross out numbers/variables if they are connected to the numerator and denominator by addition/subtraction? Let's see whether the following reduction can be done:
\begin{align*} \frac{3}{4}\amp=\frac{2+1}{2+2}\\ \amp\stackrel{?}{=}\frac{\cancel{2}+1}{\cancel{2}+2}\\ \amp\stackrel{?}{=}\frac{1}{2} \end{align*}Since \(\frac{3}{4}\ne\frac{1}{2}\text{,}\) we cannot cross out the same numbers/variables if they are connected to the numerator and denominator by addition/subtraction. In summary, we have:
\begin{align*} \frac{ax}{ay}\amp=\frac{x}{y}\\ \frac{a+x}{a+y}\amp\ne\frac{x}{y}\\ \frac{a-x}{a-y}\amp\ne\frac{-x}{-y} \end{align*}This is not the first time a property works for multiplication/division, but not for addition/subtraction. The same is true for square root:
\begin{align*} \sqrt{xy}\amp=\sqrt{x}\sqrt{y}\\ \sqrt{\frac{x}{y}}\amp=\frac{\sqrt{x}}{\sqrt{y}}\\ \sqrt{x+y}\amp\ne\sqrt{x}+\sqrt{y}\\ \sqrt{x-y}\amp\ne\sqrt{x}-\sqrt{y} \end{align*}Let's look at a few examples.
Simplify \(\frac{4x^3y}{2x}\)
When all numbers and variables are cancelled out in the numerator or denominator, we fill it with \(1\text{.}\)
We can do \(\frac{x}{x^2}=\frac{1}{x}\text{,}\) but can we cross out \((x+1)\) in \(\frac{2(x+1)}{3(x+1)}\text{?}\) Let's use numbers to test out:
\begin{align*} \frac{4}{6}\amp=\frac{2\cdot2}{3\cdot2}\\ \amp=\frac{2(1+1)}{3(1+1)}\\ \amp\stackrel{?}{=}\frac{2\cancel{(1+1)}}{3\cancel{(1+1)}}\\ \amp\stackrel{?}{=}\frac{2}{3} \end{align*}Although \((1+1)\) is an addition, the number \((1+1)\) is actually connected to both the numerator and denominator by multiplication. We can cross out \((1+1)\) in this case and reduce \(\frac{4}{6}\) to \(\frac{2}{3}\text{.}\) This implies we can reduce \(\frac{2(x+1)}{3(x+1)}\) to \(\frac{2}{3}\text{.}\) Compare the following:
Incorrect:
\begin{align*} \frac{x+1}{x+2}\amp=\frac{\cancel{x}+1}{\cancel{x}+2}\\ \amp=\frac{1}{2} \end{align*}Correct:
\begin{align*} \frac{x+1}{2(x+1)}\amp=\frac{1\cdot\cancel{(x+1)}}{2\cdot\cancel{(x+1)}}\\ \amp=\frac{1}{2} \end{align*}Simplify \(\frac{x+1}{2x+2}\)
We cannot simply cross out \(x\) because, in the numerator, \(x\) is connected to \(1\) by addition. However, we learned how to factor, which can turn addition/subtraction to multiplication:
\begin{equation*} \frac{x+1}{2x+2}=\frac{x+1}{2(x+1)} \end{equation*}Next, we can cross out \((x+1)\text{,}\) because it's connected to the numerator and denominator by multiplication:
\begin{align*} \frac{x+1}{2x+2}\amp=\frac{x+1}{2(x+1)}\\ \amp=\frac{1\cdot\cancel{(x+1)}}{2\cdot\cancel{(x+1)}}\\ \amp=\frac{1}{2} \end{align*}The result implies that, no matter what value \(x\) has, the expression \(\frac{x+1}{2x+2}\) can always be reduced to \(\frac{1}{2}\text{.}\) Let's verify this by substituting \(x\) with some random numbers, say, \(1\text{,}\) \(2\) and \(3\text{:}\)
If \(x=1\text{,}\)
\begin{align*} \frac{x+1}{2x+2}\amp=\frac{1+1}{2(1)+2}\\ \amp=\frac{2}{4}\\ \amp=\frac{1}{2} \end{align*}If \(x=2\text{,}\)
\begin{align*} \frac{x+1}{2x+2}\amp=\frac{2+1}{2(2)+2}\\ \amp=\frac{3}{6}\\ \amp=\frac{1}{2} \end{align*}If \(x=3\text{,}\)
\begin{align*} \frac{x+1}{2x+2}\amp=\frac{3+1}{2(3)+2}\\ \amp=\frac{4}{8}\\ \amp=\frac{1}{2} \end{align*}When we factor a polynomial, we should first check whether we can factor out any common factor. In the next example, when we factor \(x^2-4x\text{,}\) we should factor out the common factor \(x\) and have \(x^2-4x=x(x-4)\text{.}\) Avoid mixing up the following factoring:
\begin{align*} x^2-4x\amp=x(x-4)\\ x^2-4\amp=(x+2)(x-2) \end{align*}Simplify \(\frac{x^2-4x}{x^2-16}\)
We must avoid making this common mistake:
\begin{equation*} \frac{x}{x+4}=\frac{1}{4}\quad\text{incorrect!} \end{equation*}When a polynomial is not in order of degrees, we should re-order the polynomial before factoring. In the next example, we should change \(9y+2y^2-5\) to \(2y^2+9y-5\) first.
Simplify \(\frac{9y+2y^2-5}{y^2-25}\)
The formula \(b-a=-(a-b)\) is often used. Let's verify this formula:
\begin{align*} 3-2\amp=-(2-3)\amp\text{Both sides equal }1\\ 4-2\amp=-(2-4)\amp\text{Both sides equal }2\\ 2-4\amp=-(4-2)\amp\text{Both sides equal }-2 \end{align*}Here is a quick proof:
\begin{align*} b-a\amp=-a+b\\ \amp=-1(a-b)\\ \amp=-(a-b) \end{align*}Let's look at an example using this formula.
Simplify \(\frac{-48z+24z^2-3z^3}{4-z}\)
Don't forget the following variations of factoring trinomials:
\begin{align*} x^2+6x+8\amp=(x+2)(x+4)\\ x^2+6xy+8y^2\amp=(x+2y)(x+4y)\\ x^2y^2+6xy+8\amp=(xy+2)(xy+4) \end{align*}The following examples use those two variations.
Simplify \(\frac{3y-x}{x^2-xy-6y^2}\)
In the example above, notice we moved the negative sign from the numerator to the front of the fraction:
\begin{equation*} \frac{-1}{x+2y}=-\frac{1}{x+2y} \end{equation*}To understand why we can do this, let's find a pattern:
\begin{align*} \frac{-4}{2}\amp=-2\\ \frac{4}{-2}\amp=-2\\ -\frac{4}{2}\amp=-2\\ \frac{-4}{-2}\amp=2 \end{align*}We can see:
\begin{equation*} \frac{-4}{2}=\frac{4}{-2}=-\frac{4}{2} \end{equation*}This is why we can do \(\frac{-1}{x+2y}=-\frac{1}{x+2y}\text{.}\)
Next, we will also review the sum and difference of cubes formulas:
\begin{equation*} a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2) \end{equation*}They are often used when we see perfect cube numbers like \(1, 8, 27, 64\) and \(125\text{.}\)
Simplify \(\frac{x^3y^3-27}{12-7xy+x^2y^2}\)
Simplifying Rational Expressions
The domain of rational function \(f(x)=\frac{1}{x}\) is {\(x \mid x\ne0\)}, because we may not divide by \(0\text{.}\) Let's find the domain of \(g(x)=\frac{x-1}{x(x-1)}\text{.}\)
To find the domain of \(g(x)\text{,}\) we set its denominator to be \(0\) and solve for \(x\text{:}\)
\begin{align*} x(x-1)\amp=0\\ x=0\amp\text{ or }x-1=0\\ x=0\amp\text{ or }x=1 \end{align*}The domain of \(g(x)\) is {\(x \mid x\ne0, x\ne1\)}.
We just learned how to reduce rational expressions. If we reduce \(g(x)\text{,}\) we have:
\begin{equation*} g(x)=\frac{x-1}{x(x-1)}=\frac{1}{x} \end{equation*}After the reduction, the \(g(x)\)'s domain has become {\(x \mid x\ne0\)}. However, a function's domain shouldn't change due to rational expression reduction. The correct way to reduce \(g(x)\) is:
\begin{align*} g(x)\amp=\frac{x-1}{x(x-1)}\\ \amp=\frac{1}{x}\amp x\ne1 \end{align*}We must add the domain condition \(x\ne1\) because \(g(1)=\frac{1-1}{1(1-1)}\) is undefined. Let's look at another example.
Reduce the rational function \(R(y)=\frac{-y-2y^2}{2y^3-y^2-y}\text{.}\) Don't forget to list lost domain conditions.
We removed \(y(2y+1)\) from the fraction's denominator. To find lost domain conditions, we can solve \(y(2y+1)=0\) to find \(y\) values which will cause \(R(y)\) to be undefined:
\begin{align*} y(2y+1)\amp=0\\ y=0\amp\text{ or }2y+1=0\\ y=0\amp\text{ or }2y=-1\\ y=0\amp\text{ or }y=-\frac{1}{2} \end{align*}The result shows \(y\ne0\) and \(y\ne-\frac{1}{2}\) were removed from the function's domain restrictions. The reduced function is:
\begin{equation*} R(y)=-\frac{1}{y-1}\text{ where }y\ne0, y\ne-\frac{1}{2} \end{equation*}The values \(y=0\) and \(y=-\frac{1}{2}\) would cause \(R(y)\) to be undefined.
Simplifying Rational Expressions
Let's review how to multiply fractions:
\begin{align*} \frac{2}{9}\cdot\frac{3}{10}\amp=\frac{2}{3^2}\cdot\frac{3}{2\cdot5}\\ \amp=\frac{\cancel{2}}{3\cdot\cancel{3}}\cdot\frac{\cancel{3}}{\cancel{2}\cdot5}\\ \amp=\frac{1}{15} \end{align*}We can multiply rational expressions in a similar way. Let's look at a few examples.
Multiply rational expressions \(\frac{p^2q^4}{3r}\cdot\frac{9r^2}{pq^2}\)
Multiply rational expressions \(\frac{x^2-4x}{x^2-4}\cdot\frac{4-4x+x^2}{20-x-x^2}\)
Let's review how to divide fractions:
\begin{align*} \frac{2}{9}\div\frac{10}{3}\amp=\frac{2}{9}\cdot\frac{3}{10}\\ \amp=\frac{2}{3^2}\cdot\frac{3}{2\cdot5}\\ \amp=\frac{\cancel{2}}{3\cdot\cancel{3}}\cdot\frac{\cancel{3}}{\cancel{2}\cdot5}\\ \amp=\frac{1}{15} \end{align*}We can divide rational expressions in a similar way. Let's look at a few examples.
Divide rational expressions \(\frac{3x^2}{x^2-9y^2}\div\frac{6x^3}{x^2-2xy-15y^2}\)
Divide rational expressions \((x^2-5x-14)\div(x-7)\)
Divide rational expressions \(\frac{m^2n^2-3mn-4}{2mn}\div(m^2n^2-16)\)
Divide rational expressions \((p^4-16)\div\frac{p^4-2p^3}{2p}\)
Multiplying/Dividing Rational Expressions
