4.5: Simplifying Rational Expressions
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Reducing Rational Expressions in x
Now that we’ve discussed some fundamental ideas and techniques, let’s apply what we’ve learned to rational expressions that are functions of an independent variable (usually x). Let’s start with a simple example.
Reduce the rational expression \[\frac{2 x-6}{x^{2}-7 x+12} \nonumber \] to lowest terms. For what values of x is your result valid?
Solution
In the numerator, factor out a 2, as in 2x − 6 = 2(x − 3).
The denominator is a quadratic trinomial with ac = (1)(12) = 12. The integer pair −3 and −4 has product 12 and sum −7, so the denominator factors as shown.
\[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(x-3)}{(x-3)(x-4)} \nonumber \]
Now that both numerator and denominator are factored, we can cancel common factors.
\[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(x-3)}{(x-3)(x-4)}=\frac{2}{x-4} \nonumber \]
Thus, we have shown that \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2}{x-4} \nonumber \]
In equation (15), we are stating that the expression on the left (the original expression) is identical to the expression on the right for all values of x.
Actually, there are two notable exceptions, the first of which is x = 3. If we substitute x = 3 into the left-hand side of equation (15), we get
\[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(3)-6}{(3)^{2}-7(3)+12}=\frac{0}{0} \nonumber \]
We cannot divide by zero, so the left-hand side of equation (15) is undefined if x = 3. Therefore, the result in equation (15) is not valid if x = 3.
Similarly, if we insert x = 4 in the left-hand side of equation (15),
\[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(4)-6}{(4)^{2}-7(4)+12}=\frac{2}{0} \nonumber \]
Again, division by zero is undefined. The left-hand side of equation (15) is undefined if x = 4, so the result in equation (15) is not valid if x = 4. Note that the right-hand side of equation (15) is also undefined at x = 4.
However, the algebraic work we did above guarantees that the left-hand side of equation (15) will be identical to the right-hand side of equation (15) for all other values of x. For example, if we substitute x = 5 into the left-hand side of equation (15),
\[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(5)-6}{(5)^{2}-7(5)+12}=\frac{4}{2}=2 \nonumber \]
On the other hand, if we substitute x = 5 into the right-hand side of equation (15),
\[\frac{2}{x-4}=\frac{2}{5-4}=2 \nonumber \]
Hence, both sides of equation (15) are identical when x = 5. In a similar manner, we could check the validity of the identity in equation (15) for all other values of x
You can use the graphing calculator to verify the identity in equation (15). Load the left- and right-hand sides of equation (15) in Y= menu, as shown in Figure \(\PageIndex{1}\)(a). Press 2nd TBLSET and adjust settings as shown in Figure \(\PageIndex{1}\)(b). Be sure that you highlight AUTO for both independent and dependent variables and press ENTER on each to make the selection permanent. In Figure \(\PageIndex{1}\)(b), note that we’ve set TblStart = 0 and ∆Tbl = 1. Press 2nd TABLE to produce the tabular results shown in Figure \(\PageIndex{1}\)(c).
Remember that we placed the left- and right-hand sides of equation (15) in Y1 and Y2, respectively.
- In the tabular results of Figure \(\PageIndex{1}\)(c), note the ERR (error) message in Y1 when x = 3 and x = 4. This agrees with our findings above, where the left-hand side of equation (15) was undefined because of the presence of zero in the denominator when x = 3 or x = 4.
- In the tabular results of Figure \(\PageIndex{1}\)(c), note that the value of Y1 and Y2 agree for all other values of x.
We are led to the following key result.
In general, when you reduce a rational expression to lowest terms, the expression obtained should be identical to the original expression for all values of the variables in each expression, save those values of the variables that make any denominator equal to zero. This applies to the denominator in the original expression, all intermediate expressions in your work, and the final result. We will refer to any values of the variable that make any denominator equal to zero as restrictions.
Let’s look at another example.
Reduce the expression \[\frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36} \nonumber \] to lowest terms. State all restrictions.
Solution
The numerator is a quadratic trinomial with ac = (2)(−12) = −24. The integer pair −3 and 8 have product −24 and sum 5. Break the middle term of the polynomial in the numerator into a sum using this integer pair, then factor by grouping.
\[\begin{aligned} 2 x^{2}+5 x-12 &=2 x^{2}-3 x+8 x-12 \\ &=x(2 x-3)+4(2 x-3) \\ &=(x+4)(2 x-3) \end{aligned} \nonumber \]
Factor the denominator by grouping.
\[\begin{aligned} 4 x^{3}+16 x^{2}-9 x-36 &=4 x^{2}(x+4)-9(x+4) \\ &=\left(4 x^{2}-9\right)(x+4) \\ &=(2 x+3)(2 x-3)(x+4) \end{aligned} \nonumber \]
Note how the difference of two squares pattern was used to factor \(4x^2 − 9 = (2x + 3)(2x − 3)\) in the last step.
Now that we’ve factored both numerator and denominator, we cancel common factors.
\[\begin{aligned} \frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36} &=\frac{(x+4)(2 x-3)}{(2 x+3)(2 x-3)(x+4)} \\ &=\frac{(x+4)(2 x-3)}{(2 x+3)(2 x-3)(x+4)} \\ &=\frac{1}{2 x+3} \end{aligned} \nonumber \]
We must now determine the restrictions. This means that we must find those values of x that make any denominator equal to zero.
- In the body of our work, we have the denominator (2x + 3)(2x − 3)(x + 4). If we set this equal to zero, the zero product property implies that \[2 x+3=0 \quad \text { or } \quad 2 x-3=0 \quad \text { or } \quad x+4=0 \nonumber \]
Each of these linear factors can be solved independently \[x=-3 / 2 \quad \text { or } \quad x=3 / 2 \quad \text { or } \quad x=-4 \nonumber \]
Each of these x-values is a restriction.
- In the final rational expression, the denominator is 2x + 3. This expression equals zero when x = −3/2 and provides no new restrictions.
- Because the denominator of the original expression, namely \(4 x^{3}+16 x^{2}-9 x-36\), is identical to its factored form in the body our work, this denominator will produce no new restrictions.
Thus, for all values of x \[\frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36}=\frac{1}{2 x+3} \nonumber \]
provided \(x \neq −3/2, 3/2, or −4\). These are the restrictions. The two expressions are identical for all other values of x.
Finally, let’s check this result with our graphing calculator. Load each side of equation (18) into the Y= menu, as shown in Figure \(\PageIndex{2}\)(a). We know that we have a restriction at x = −3/2, so let’s set TblStart = −2 and \(\Delta \mathrm{Tbl}=0.5\), as shown in Figure \(\PageIndex{2}\)(b). Be sure that you have AUTO set for both independent and dependent variables. Push the TABLE button to produce the tabular display shown in Figure \(\PageIndex{2}\)(c).
Remember that we placed the left- and right-hand sides of equation (18) in Y1 and Y2, respectively.
- In Figure \(\PageIndex{2}\)(c), note that the expressions Y1 and Y2 agree at all values of x except x = −1.5. This is the restriction −3/2 we found above.
- Use the down arrow key to scroll down in the table shown in Figure \(\PageIndex{2}\)(c) to produce the tabular view shown in Figure \(\PageIndex{2}\)(d). Note that Y1 and Y2 agree for all values of x except x = 1.5. This is the restriction 3/2 we found above.
- We leave it to our readers to uncover the restriction at x = −4 by using the uparrow to scroll up in the table until you reach an x-value of −4. You should uncover another ERR (error) message at this x-value because it is a restriction. You get the ERR message due to the fact that the denominator of the left-hand side of equation (18) is zero at x = −4.