Simplifying Rational Expressions
Rational expressionsA fraction with a polynomial in the numerator and/or denominator. are fractions that have a polynomial in the numerator or the denominator or both. Although rational expressions can seem complicated because they contain variables, they can be simplified in the same way that numerical fractions are simplified.
The first step in simplifying a rational expression is to determine the domainThe set of all possible inputs of a function which allow the function to work., the set of all possible values of the variables. The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero. These values cannot be included in the domain, so they're called excluded valuesA value for a variable that is not allowed in an expression, such as a variable in a rational expression that would make the denominator equal zero.. We discard them right at the start, before we go any further.
For rational expressions, the domain will exclude values for which the value of the denominator is `0`. Two examples to illustrate finding the domain of an expression are shown below.
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Example |
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Problem |
Identify the domain of the expression `(3x+2)/(x-4)`. |
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`x - 4 = 0`
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Find any values for `x` that would make the denominator `=0`. |
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`x = 4` |
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When `x = 4`, the denominator is equal to `0`. |
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Answer |
The domain is all `x` not equal to `4`. |
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That wasn't hard. Let's try one that's a little more challenging:
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Example |
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Problem |
Identify the domain of the expression `(x+7)/(x^2+8x-9)`. |
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`x^2+8x-9=0`
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Find any values for `x` that would make the denominator `= 0` by setting the denominator `= 0` and solving the equation. |
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`(x + 9)(x - 1) = 0`
`x = -9` or `x = 1` |
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Solve the equation by factoring. The solutions are the values that are excluded from the domain. |
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Answer |
The domain is all `x` not equal to `-9` or `1`. |
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Find the domain of the rational expression `(5x)/(2x^2+8)`. A) all `x`
B) all `x` not equal to `2` or `8`
C) all `x` not equal to `0`
D) all `x` not equal to `-2, 2`
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Once we've figured out the excluded values, the next step is to simplify. To simplify a rational expression, we follow the same approach we'd use to simplify numerical fractions: find common factors in the numerator and denominator. Let’s start by looking at this method for a numerical fraction:
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Example |
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Problem |
Simplify `15/27`. |
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`15/27=(3*5)/(3*3*3)` |
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Factor the numerator and denominator. |
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`5/(3*3)*3/3` `5/(3*3)*1` |
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Pull out factors of `1`. |
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`5/9` |
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Simplify. |
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Answer |
`5/9` |
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Now, we could have done that problem in our heads. But it was worth writing it all down, because that's exactly how we simplify a rational expression.
So let's simplify a rational expression, using the same technique we applied to that fraction just now. Only this time, the numerator and denominator are both monomials with variables:
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Example |
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Problem |
Simplify `(5x^2y^5)/(25xy)` and state the domain for the expression. |
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`25xy = 0` `x = 0` or `y = 0`
The domain is all `x` not equal to `0` and all `y` not equal to `0`. |
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Find the excluded values, the values of `x` and `y` that make the numerator equal to `0`. |
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`(5x^2y^5)/(25xy)=(5*x*x*y*y*y*y*y)/(5*5*x*y)` |
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Factor the numerator and denominator. |
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`5/5*x/x*y/y*(x*y*y*y*y)/5`
`1*1*1*(x*y*y*y*y)/5` |
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Rewrite with factors of `1`. |
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`(xy^4)/5` |
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Simplify. |
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Answer
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`(xy^4)/5` The domain is all `x` not equal to `0` and all `y` not equal to `0`. |
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See—the same steps worked again. In the examples that follow, the numerator and the denominator are polynomials with more than one term, but the same principles of simplifying will once again apply. Factor the numerator and denominator to simplify the rational expression.
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Example |
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Problem |
Simplify and state the domain for the expression `(x+3)/(x^2+12x+27)`. |
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`x^2 + 12x + 27 = 0`
`(x + 3)(x + 9) = 0`
`x = -3` or `x = -9` |
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Determine the values for which the denominator is equal to `0`. Factor the quadratic to find the values. |
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`(x+3)/((x+3)(x+9)`
`(x+3)/(x+3)*1/(x+9)` `1*1/(x+9)` |
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Factor the numerator and denominator.
Rewrite with factors of `1`, and then simplify. |
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Answer |
`1/(x+9)` The domain is all `x` not `-9` or `-3`. |
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Example |
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Problem |
Simplify and state the domain for the expression `(x^2+10x+24)/(x^3-x^2-20x)`. |
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`x^3 - x^2 - 20x = 0` `x(x^2 - x - 20) = 0` `x(x - 5)(x + 4) = 0` `x = 0` or `x = 5` or `x = -4` |
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Determine the values for which the denominator is equal to `0`.
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`((x+4)(x+6))/(x(x^2-x-20))`
`((x+4)(x+6))/(x(x-5)(x+4)`
`(x+6)/(x(x-5))*(x+4)/(x+4)`
`(x+6)/(x(x-5))*1` |
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Factor the numerator and denominator of the rational expression.
Rewrite with factors of `1`, then simplify.
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`(x+6)/(x(x-5)` or `(x+6)/(x^2-5x)`
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Simplify. It is acceptable to either leave the denominator in factored form or to distribute multiplication. |
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Answer |
`(x+6)/(x(x-5)` or `(x+6)/(x^2-5x)` |
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The domain is all `x` not equal to `-4`, `0`, `5`. |
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No matter how many terms or variables there are in a rational expression, we can simplify it by following the same steps:
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The Steps for Simplifying a Rational Expression:
Determine the domain. The excluded values are those values that result in a denominator of `0`.
Find common factors for the numerator and denominator and simplify.
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Simplify the rational expression below and state the domain.
`(2x^2+13x+15)/(2x^2+23x+30)`
A) `14/25`; domain is all `x`
B) `(x+5)/(x+10)`; domain is all `x` not `-10` or `-3/2`
C) `(x+5)/(x+10)`; domain is all `x`
D) `1/2`; domain is all `x` not `-10` or `-3/2`
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Rational expressions, or fractions containing polynomials, can be simplified much like fractions can be simplified. To simplify a rational expression, first determine common factors of the numerator and denominator, and then remove them by rewriting them as expressions equal to `1`.
An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by `0` is undefined, any values of the variables that result in a denominator of `0` must be excluded. Excluded values must be identified in the original equation, not its factored form.