Solutions to Questions on Reducing Rational Expressions

Solutions with detailed explanations to questions on reducing rational expressions are presented.
Rational expressions are reduced through the following steps:
1) factor the numerator and denominator of the given rational expression completely
2) divide the numerator and denominator by all common factors AND put conditions on the variable so that this common factor(s) is/are not equal to zero.
3) simplify (reduce) the given rational expression

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1. For all $x \ne 1$, which of the following is equivalent to the rational expression $\dfrac{x^2 + 5x - 6}{x - 1}$ ?
Solution
Factor the numerator
$\dfrac{x^2 + 5x - 6}{x - 1} = \dfrac{(x - 1)(x + 6)}{x - 1}$
The common factor $x - 1$ is equal to zero for $x = 1$
For $x \ne 1$, divide the numerator and denominator by the common factor $(x - 1)$ .
$= \dfrac{ \cancel{(x - 1)}(x + 6)}{\cancel{x - 1}}$
and reduce the given rational function to
$= x + 6$

2. Which of the following is a simplified expression equal to Which of the following is a simplified expression equal to $\dfrac{5 - x}{2x - 10}$ for all $x \ne 5$?
Solution
Factor $-2$ in denominator $2x - 10$ of the given expression as follows
$\dfrac{5 - x}{2x - 10} = \dfrac{5 - x}{-2(5-x)}$
The common factor $5 - x$ is equal to zero for $x = 5$
For $x \ne 5$, we divide numerator and denominator by the common factor $5 - x$
$= \dfrac {\cancel{5 - x}}{-2\cancel{(5-x)}}$
and reduce the expression to
$= - \dfrac{1}{2}$

3. For all $x \ne -4$, which of the given expressions is equivalent to the expression $\dfrac{16 - x^2}{x + 4}$ ?
Solution
Factor the numerator in the given expression as follows:
$\dfrac{16 - x^2}{x + 4} = \dfrac{(4-x)(4+x)}{x + 4}$
The common factor $x + 4$ is equal to zero for $x = - 4$
For $x \ne - 4$ , divide numerator and denominator by the common factor $4 + x$
$= \dfrac{(4-x) \cancel{(4+x)}}{\cancel{x + 4}}$
and reduce the rational expression to
$= 4 - x$

4. Simplify the following rational expression $\dfrac{x + 2}{x^2 + 2x}$.
Solution
Factor the denominator in the given expression as follows
$\dfrac{x + 2}{x^2 + 2x} = \dfrac{x+2}{x(x+2)}$
The common factor $x + 2$ is equal to zero for $x = - 2$
For $x \ne -2$, divide the numerator and denominator by the common factor $x + 2$ in the above rational expression as follows:
$= \dfrac{\cancel{x+2}}{x\cancel{(x+2)}}$
and reduce the given expression to
$= \dfrac{1}{x}$

5. For all $x \ne 3$, which of the given expressions is equivalent to the expression $\dfrac{3-x}{x^2 - x - 6}$ ?
Solution
Factor the numerator and denominator in the given expression so that they have a common factor as follows:
$\dfrac{3-x}{x^2 - x - 6} = - \dfrac{x - 3}{(x + 2)(x - 3)}$
The common factor $x - 3$ is equal to zero for $x = 3$
For $x \ne 3$, divide the numerator and denominator and denominator by the common factor $x - 3$
$= - \dfrac{\cancel{x - 3}}{(x + 2) \cancel{(x - 3)}}$
and reduce the given expression to
$= - \dfrac{1}{x+2}$

6. $\dfrac{x^3 - x}{x^2 - 1} =$
Solution
Factor the numerator and denominator in the given expression as follows
$\dfrac{x^3 - x}{x^2 - 1} = \dfrac{x(x^2-1)}{x^2-1}$
The common factor $x^2 - 1$ is equal to zero for $x = 1$ or $x = - 1$
For $x$ not equal to $1$ and $-1$, the numerator and denominator in the above could be divided by the common factor $x^2-1$ as follows
$= \dfrac{x \cancel{(x^2-1)}} {\cancel{x^2-1}}$
and the expression reduced to
$= x$

7. $\dfrac{x^2 - 4}{x^2 + 4x - 12} =$

Solution
Factor the numerator and denominator in the given expression as follows
$\dfrac{x^2 - 4}{x^2 + 4x - 12} = \dfrac{(x-2)(x+2)}{(x - 2)(x + 6)}$
The common factor $x - 2$ is equal to zero for $x = 2$
For $x \ne 2$, we divide the numerator and denominator by the common factor $x - 2$
$= \dfrac{ \cancel{(x-2)}(x+2)}{\cancel{(x-2)}(x + 6)}$
and reduce the given expression to
$= \dfrac{x+2}{x+6}$

8. Simplify the rational expression $\dfrac{x^2 + 1}{x^3 + x}$.

Solution
Factor the denominator in the given expression as follows:
$\dfrac{x^2 + 1}{x^3 + x} = \dfrac{x^2 + 1}{x(x^2+1)}$
The common factor $x^2 + 1$ is never zero.
Divide the numerator and denominator by the common factor $x^2+1$.
$= \dfrac{\cancel{x^2 + 1}}{x\cancel{(x^2+1)}}$
and reduce the given expression to
$= \dfrac{1}{x}$ , for all x

9. $\dfrac{x^2 + 2x - 3}{2x^2 + 3x - 5} =$

Solution
Factor the numerator and denominator in the given expression as follows.
$\dfrac{x^2 + 2x - 3}{2x^2 + 3x - 5} = \dfrac{(x - 1)(x + 3)}{(x - 1)(2x + 5) }$
The common factor $x - 1$ is equal to zero for $x = 1$
For $x \ne 1$, divide the numerator and denominator by the common factor $x - 1$.
$= \dfrac{\cancel{(x - 1)}(x + 3)}{\cancel{(x - 1)}(2x + 5) }$
and reduce the expression to
$= \dfrac{x + 3}{2x + 5}$

10. For all $x \ne 1$, which of the following is equivalent to the rational expression $\dfrac{x-1}{(x^2 - 1)(x + 3)}$ ?

Solution
Factor the term $x^2 - 1$ in the denominator as follows:
$\dfrac{x-1}{(x^2 - 1)(x + 3)} = \dfrac{x - 1}{ (x - 1)(x + 1)(x + 3)}$
The common factor $x - 1$ is equal to zero for $x = 1$
For $x \ne 1$, divide the numerator and denominator by the common factor $x - 1$ as follows:
$\dfrac{\cancel{x - 1}}{ \cancel{(x - 1)}(x + 1)(x + 3)}$
and reduce to
$= \dfrac{1}{(x + 1)(x + 3)}$

More on Simplifying Rational Expressions are included.