to view answer.A rational expression is the quotient of two polynomials. In other words, it is a fraction in which the numerator and the denominator are polynomial expressions. Some examples of rational expressions are:
$${\large\frac{{3x^2 - 1}}{{x + 1}}}$$
$${\large\frac{x}{6}}$$
$${\large\frac{{3y^2 + 9y - 5}}{{4y^2 - 3y + 10}}}$$
$${\large\frac{1}{2}}$$
Yes, even though $${\large\frac{1}{2}}$$ is a fraction with integers in the numerator and denominator, it is still a rational expression, because 1 and 2 are each polynomial expressions of degree 0 (since, for example, 1 can be written as $$1x^0 $$).
In general, we like fractions to be in simplified form. Recall when simplifying fractions, we were able to divide factors that were in both the numerator and the denominator. For instance, when simplifying:
$${\large\frac{{24}}{{48}}}$$
we rewrote the numerator and denominator as the product of their factors:
$${\large\frac{{24}}{{48}}} ={\large \frac{{2 \cdot 2 \cdot 2 \cdot 3}}{{2 \cdot 2 \cdot 2 \cdot 2\cdot 3}}}$$
When we multiply fractions, we multiply the numerators with the numerators and the denominators with the denominators, and so, we can rewrite this as:
$${\large\frac{{24}}{{48}}} = {\large\frac{{2 \cdot 2 \cdot 2 \cdot 3}}{{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3}} }= {\large\frac{2}{2}} \cdot{\large \frac{2}{2}} \cdot {\large\frac{2}{2}} \cdot{\large \frac{3}{3}} \cdot {\large\frac{1}{2}}$$
Notice that each fraction where the numerator equals the denominator has a value of 1, for instance, $${\large\frac{2}{2}} = 1$$. When an expression is multiplied by 1 the value of the expression is not changed. Thus, we can rewrite this as:
$${\large\frac{{24}}{{48}}} ={\large \frac{{2 \cdot 2 \cdot 2 \cdot 3}}{{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3}}} = {\large\frac{2}{2}} \cdot {\large\frac{2}{2}} \cdot{\large \frac{2}{2}} \cdot{\large \frac{3}
{3}} \cdot {\large\frac{1}{2}} = 1 \cdot 1 \cdot 1 \cdot 1 \cdot {\large\frac{1}{2}} = {\large\frac{1}{2}}$$
This is the same as if we divided three 2’s and one 3 from the numerator and the denominator (or divided 12 to begin with from the numerator and the denominator).
All other rational expressions can be simplified in the same manner. Basically, factor the numerator and denominator completely and see if any of the factors are on both the top and the bottom. If they are, then together, they divide to be 1, and thus can be divided out.
It is important to remember that there may be some values of the variable for which a rational expression is undefined. Any values of the variable that make the denominator zero must be specified as being not part of the domain of the expression. For example:
$${\large\frac{{3x^2 }}{{4x}}}$$
simplifies to: $${\large\frac{{3x}}{4} }\cdot{\large \frac{x}{x}} = {\large\frac{{3x}}{4}} \cdot 1 ={\large\frac{{3x}}{4}}$$
In this final expression, there are no x’s in the denominator. However, since there was one in the original expression, and we have just simplified that expression, we still must keep the same domain as the original expression. In the original expression, 4x is the denominator, and thus, $$4x \ne 0$$ and so $$x \ne 0$$, or else the denominator would be zero. Thus, our simplified expression should be written:
$${\large\frac{{3x}}{4}},\;\;x \ne 0$$
Why do we need to keep this domain even when our final expression does not have x in the denominator? Because we do not want to input a value of x = 0 into our simplified expression when the original expression would not have given an answer. The simplified expression has to be equivalent to the original for it to be useful, so we should not get values of the expression for values of x that they didn’t exist for in the original.
Simplify the following rational expression by reducing:
$${\large\frac{{12x^2 + 4x - 8}}{{2x^2 - 4x - 6}}}$$
1. First, we need to factor the top and the bottom. See factoring quadratics to review this:
$${\large\frac{{12x^2 + 4x - 8}}{{2x^2 - 4x - 6}}} = {\large\frac{{4\left( {3x^2 + x - 2} \right)}}{{2\left( {x^2 - 2x - 3} \right)}}} = {\large\frac{{4\left( {3x - 2} \right)\left( {x + 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 3} \right)}}}$$
2. At this point, we should find the values of x for which this original expression is undefined. Remember that the expression is undefined whenever the denominator is zero, so to find the values of x that will not be in the domain, we can set the denominator equal to zero and see which values make that occur:
$$2\left( {x + 1} \right)\left( {x - 3} \right) = 0$$
Just as when we solve quadratic equations by factoring, we realize that in order for 3 numbers to be multiplied together to be zero, one or more of them must be zero. Thus, we set each factor equal to zero and solve for x:
$$\eqalign{
& x + 1 = 0\;\;\;\;\;so\;\;\;\;x = - 1 \cr
& x - 3 = 0\;\;\;\;\;so\;\;\;\;x = 3 \cr} $$
Thus, our domain contains all values of x, except these two. We write that $$x \ne - 1\;\;\;\;and\;\;\;x \ne 3$$.
3. Finally, we can divide factors that are in both the numerator and the denominator. Remember that we can do this because they can be rewritten as a fraction that equals 1:
$$
\eqalign{
& \frac{{12x^2 + 4x - 8}}{{2x^2 - 4x - 6}} = \frac{{4\left( {3x - 2} \right)\left( {x + 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 3} \right)}} \cr
& = \frac{2}{2} \cdot \frac{{\left( {x + 1} \right)}}{{\left( {x + 1} \right)}} \cdot \frac{{2\left( {3x - 2} \right)}}{{\left( {x - 3} \right)}} \cr
& = 1 \cdot 1 \cdot \frac{{2\left( {3x - 2} \right)}}{{\left( {x - 3} \right)}} \cr} $$
Thus, our simplified expression is:
$${\large\frac{{2\left( {3x - 2} \right)}}{{\left( {x - 3} \right)}}},\;\;\;x \ne - 1,\;\;x \ne 3$$
When you are simplifying rational expression, be very careful, that you ONLY DIVIDE MULTIPLIED TERMS (in other words, factors). If the same thing is added or subtracted to the top as the bottom, it CANNOT be divided. For example:
$${\large\frac{{x - 2}}{{x^2 - 2}}} \ne{\large\frac{x}{{x^2 }}}$$
Why not? Because –2 is neither a factor of the numerator nor the denominator. We are dividing a number, which is $$\left( {x - 2} \right)$$ by another number, which is $$\left( {x^2 - 2} \right)$$. To verify that the two expressions are not equivalent, substitute a value for x, such as 3:
$${\large\frac{{x - 2}}{{x^2 - 2}}} = {\large\frac{{3 - 2}}{{9 - 2}}} = {\large\frac{1}{7}}$$
Had we cancelled the -2’s, we would have instead:
$${\large\frac{x}{{x^2 }}} = {\large\frac{3}{9}} = {\large\frac{1}{3}}$$
As you can see, dividing the –2’s changes the value of the expression. Thus, it is not a successful procedure.
Simplify the rational expression by reducing:
$${\large\frac{{x^2 - 5x + 6}}{{ - x^2 + 7x - 12}}}$$
1. First, we need to factor the numerator and denominator completely. See factoring quadratics to review this:
$${\large\frac{{x^2 - 5x + 6}}{{ - x^2 + 7x - 12}}} ={\large \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{\left( { - x + 3} \right)\left( {x - 4} \right)}}}$$
2. Now, as before, we should find the values of x that make this original expression undefined by setting each factor in the denominator equal to zero:
$$\eqalign{
& - x + 3 = 0\;\;\;\;implies\;\;\;x = 3 \cr
& x - 4 = 0\;\;\;\;\;implies\;\;\;\;x = 4 \cr} $$
Thus, we know that in our simplified (and original) expression:
$$x \ne 3\;\;and\;\;x \ne 4$$
3. Now, we want to cancel any multiplied terms that we can. At first glance, it doesn’t seem that we have any factors on both the top and bottom that are the same, but notice that $$\left( {x - 3} \right)$$ and $$\left( { - x + 3} \right)$$ are almost the same, but one is the negative of the other. In a case like this, factor a –1 out of one of these binomials. In other words:
$$\left( { - x + 3} \right) = - 1\left( {x - 3} \right)$$
Now, we can rewrite the expression:
$${\large\frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{ - \left( {x - 3} \right)\left( {x - 4} \right)}}}$$
So, the factor $$\left( {x - 3} \right)$$ is in the numerator and denominator, and so we can divide it since $${\large\frac{{\left( {x - 3} \right)}}{{\left( {x - 3} \right)}}} = 1$$. Thus, our simplified expression is:
$${\large\frac{{\left( {x - 2} \right)}}{{ - \left( {x - 4} \right)}}},\;\;x \ne 3,\;\;x \ne 4$$
(Remember that you CANNOT cancel the remaining –2 with the –4 because these are added terms, not factors.)
Some practice problems to check your skills:
Simplify the following rational expressions by reducing:
1. $${\large\frac{{4x^2 - 8x}}{{8x^3 - 24x^2 }}}$$
Think about it for a moment and then access this link to view answer.
2. $${\large\frac{{5x - 12}}{{15x - 24}}}$$
Think about it for a moment and then access this link to view answer.
3. $${\large\frac{{3x^2 - 3x - 60}}{{4x + 16}}}$$
Think about it for a moment and then access this link to view answer.
4. $${\large\frac{{20x^3 + 10x^2 - 60x}}{{15x^3 + 45x^2 + 30x}}}$$
Think about it for a moment and then access this link to view answer.
5. $${\large\frac{{x^2 - 6x + 8}}{{ - x^2 - x - 6}}}$$
Think about it for a moment and then access this link to view answer.