1.24: Simplifying, Multiplying and Dividing Rational Expressions

We recall that a rational number is one which can be written as a ratio \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0 .\) A rational expression (also called an algebraic fraction) is one which can be written as a ratio \(\frac{P}{Q}\) where \(P\) and \(Q\) are polynomials and \(Q \neq 0 .\) Just as we can write a number in simplest (or reduced) form, we can do the same for rational expressions.

The Fundamental Principle of Rational Expressions

For polynomials \(P, Q\) and \(R\) with \(Q \neq 0\) and \(R \neq 0\)

\[\frac{P}{Q}=\frac{P \cdot R}{Q \cdot R}\nonumber\]

To Simplify a Rational Expression

• Step 1. Find the Greatest Common Factor (GCF) of both numerator and denominator, if you can.
• Step 2. Factor completely both numerator and denominator.
• Step 3. Use the Fundamental Principle of Rational Expressions to divide out the common factor from the numerator and denominator.

Example 22.1

Simplify the following rational expression \(\dfrac{25 a^{6} b^{3}}{5 a^{3} b^{5}}\).

We begin by noting that the GCF of the numerator and denominator is \(5 a^{3} b^{3} .\) Using this, we can factor the numerator and denominator to get

\[\frac{25 a^{6} b^{3}}{5 a^{3} b^{5}}=\frac{\left(5 a^{3} b^{3}\right)\left(5 a^{3}\right)}{\left(5 a^{3} b^{3}\right)\left(b^{2}\right)}\nonumber\]

Lastly we can use the Fundamental Principle of Rational Expressions to simplify

\[\frac{\left(\not{5 a^{3} b^{3}}\right)\left(5 a^{3}\right)}{\left(\not{5 a^{3} b^{3}}\right)\left(b^{2}\right)}=\frac{5 a^{3}}{b^{2}}\nonumber\]

Another way to approach this same problem is to factor both numerator and denominator completely and then cancel where appropriate

\[\begin{align*} \frac{25 a^{6} b^{3}}{5 a^{3} b^{5}} &=\frac{\not 5 \cdot 5 \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot a \cdot a \cdot a \cdot \cancel{b} \cdot \cancel{b} \cdot \cancel{b}}{\not 5 \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{b} \cdot \cancel{b} \cdot \cancel{b} \cdot b \cdot b} \\[4pt] &=\frac{5 a^{3}}{b^{2}}\end{align*}\]

Example 22.2

Simplify:

\[\frac{2 x+2}{x^{2}-1}\nonumber\]

First we factor both numerator and denominator, then, once in factored form, we can use The Fundamental Principle of Rational Expressions to simplify.

\[\frac{2(x+1)}{(x-1)(x+1)}\nonumber\]

Now notice that \((x+1)\) is in common, so, we can cancel it out. Finally, our simplified rational expression is:

\[\frac{2}{(x-1)}\nonumber\]

Example 22.3

Simplify:

\[\frac{4 x-16}{x^{2}-x-12}=\frac{4(x-4)}{(x-4)(x+3)}=\frac{4}{(x+3)}\nonumber\]

Multiplying Rational Expressions

Multiplying Rational Expressions

Let \(P, Q, R, S\) be polynomials with \(Q \neq 0\) and \(S \neq 0\) then

\[\frac{P}{Q} \cdot \frac{R}{S}=\frac{P \cdot R}{Q \cdot S}\nonumber\]

Example 22.4

Multiply and write your answer in simplest form:

\[\frac{10 x^{2}}{2 y^{2}} \cdot \frac{14 y^{5}}{5 x^{3}}\nonumber\]

\[\frac{10 x^{2}}{2 y^{2}} \cdot \frac{14 y^{5}}{5 x^{3}}=\frac{\left(10 x^{2}\right)\left(14 y^{5}\right)}{\left(2 y^{2}\right)\left(5 x^{3}\right)}=\frac{140 x^{2} y^{5}}{10 x^{3} y^{2}}=\frac{14 y^{3}}{x}\nonumber\]

Alternatively, we could do some canceling before multiplying and achieve the same result

\[\frac{10 x^{2}}{2 y^{2}} \cdot \frac{14 y^{5}}{5 x^{3}}=\frac{\not 2 \cdot \not 5 \cdot \not {x^{2}}}{\not 2 \cdot \not{y^{2}}} \cdot \frac{14 \cdot \not{y^{2}} \cdot y^{3}}{5 \cdot \not{x^{2}} \cdot x}=\frac{14 y^{3}}{x}\nonumber\]

Example 22.5

Multiply and write the answer in simplest form

\[\left(\frac{11 x^{5}}{-7 y^{7} z^{2}}\right)\left(\frac{-10 y^{5}}{33 x^{3} z}\right)\left(\frac{21 z^{5}}{-6 x^{2}}\right)\nonumber\]

An organized way of doing this problem is to rearrange so that we can handle the “numerical portion” and “variable portion” separately. We are allowed to do this because multiplication of real numbers is commutative, i.e. we can multiply in any order we wish. The rearrangement results in:

\[\left(\frac{11}{-7} \cdot \frac{-10}{33} \cdot \frac{21}{-6}\right)\left(\frac{x^{5}}{y^{7} z^{2}} \cdot \frac{y^{5}}{x^{3} z} \cdot \frac{z^{5}}{x^{2}}\right)\nonumber\]

Now we can go ahead and do some canceling, carefully, one portion at a time:

\[\frac{\not 11}{-\not 7} \cdot \frac{\not{-2} \cdot 5}{\not 3 \cdot \not{11}} \cdot \frac{\not 3 \cdot \not 7}{\not {-2} \cdot 3}=-\frac{5}{3}\nonumber\]

and

\[\frac{\not{x^{3}} \cdot \not{x^{2}}}{y^{2} \cdot \not{y^{5}} \not{z^{2}}} \cdot \frac{\not{y^{5}}}{\not{x^{3}} \cdot \not{z}} \cdot \frac{\not{z^{2}} \cdot \not{z} \cdot z^{2}}{\not{x^{2}}}=\frac{z^{2}}{y^{2}}\nonumber\]

Lastly, we put our pieces back together to get the final solution of our problem which is

\[-\frac{5 z^{2}}{3 y^{2}}\nonumber\]

Example 22.6

Multiply and simplify:

\[\frac{4}{27 x+18 y} \cdot \frac{9 x+6 y}{6}=\frac{4}{9(3 x+2 y)} \cdot \frac{3(3 x+2 y)}{6}\nonumber\]

Notice that \((3 x+2 y)\) is in common in both numerator and denominator and thus can be simplified. So:

\[\frac{4}{27 x+18 y} \cdot \frac{9 x+6 y}{6}=\frac{4}{9} \cdot \frac{3}{6}=\frac{4 \cdot 3}{9 \cdot 6}=\frac{2}{9}\nonumber\]

We divide rational expressions in the same way that we divide fractions. We can view the division as the multiplication of the first expression by the reciprocal of the second.

Dividing Rational Expressions

Let \(P, Q, U, V\) be polynomials with \(Q \neq 0\) and \(V \neq 0\) then

\[\frac{P}{Q} \div \frac{U}{V}=\frac{P}{Q} \cdot \frac{V}{U}=\frac{P \cdot V}{Q \cdot U}\nonumber\]

Example 22.7

Divide and write the answer in simplest form:

\[\frac{6 e^{2} f^{3} g}{5 e^{3} g^{3}} \div \frac{24 f^{6} g^{2}}{e f}\nonumber\]

We multiply the first rational expression by the reciprocal of the second:

\[\frac{6 e^{2} f^{3} g}{5 e^{3} g^{3}} \cdot \frac{e f}{24 f^{6} g^{2}}\nonumber\]

Next we can rearrange our problem into numerical and variable portions to get

\[\left(\frac{6}{5} \cdot \frac{1}{24}\right)\left(\frac{e^{2} f^{3} g}{e^{3} g^{3}} \cdot \frac{e f}{f^{6} g^{2}}\right)\nonumber\]

Now we independently and carefully simplify each portion

\[\frac{\not 6}{5 \cdot 4 \cdot \not 6}=\frac{1}{20}\nonumber\]

and

\[\frac{\not{e^{2}} \cdot \not{f^{3}} \cdot \not g}{\not{e^{2}} \cdot \not e \cdot \not g \cdot g^{2}} \cdot \frac{\not e \cdot \not f}{\not{f^{3}} \cdot \not f \cdot f^{2} \cdot g^{2}}=\frac{1}{f^{2} g^{4}}\nonumber\]

Finally we put our pieces back together to form the solution to our problem which is

\[\frac{1}{20 f^{2} g^{4}}\nonumber\]

Exit Problem

\[\text { Simplify: } \frac{35 x^{2} b^{3}}{-21 a^{2} y} \div \frac{15 x^{2} b^{2}}{14 a y}\nonumber\]