Simplifying a rational expression and where it is undefined
A fraction with x in it can often be shortened by canceling. What does the short form keep, and what does it quietly lose?
Where you are 3 in Simplifying rational expressions
Why this exists
Divide one polynomial by another and you get a fraction with a variable in it, such as \displaystyle \frac{x^2 - 9}{x^2 + x - 6}. Fractions like this turn up whenever a quantity is a share, a rate or an average: the cost per item, the time per job, the part of a paint mix that is one color.
Two questions come with every one of them. Which values of x are forbidden? A fraction with 0 underneath has no value, and here the bottom changes as x changes, so some values of x break it. And can it be written shorter? Number fractions shorten by canceling a common factor, and so do these. This lesson answers both, and shows they are tied together: canceling can hide a forbidden value, so the forbidden values have to be found first and carried along.
Definition
A rational expression is a fraction whose numerator and denominator are both polynomials, with a denominator that is not the zero polynomial. \displaystyle \frac{3}{x}, \displaystyle \frac{x + 1}{x^2 - 4} and \displaystyle \frac{2x^2 - 8x}{6x} are rational expressions. So is any polynomial, such as x^2 + 1, since it is itself over 1.
An excluded value is a value of the variable that makes the denominator 0. At an excluded value the expression has no value at all. The excluded values are exactly the solutions of the equation “denominator = 0”. The numerator may be 0: \displaystyle \frac{0}{5} = 0 is a perfectly good number.
A rational expression is simplified when its numerator and denominator have no common factor other than 1 and -1. A simplified form is written together with its restrictions: the excluded values of the expression you started from, stated as x \neq \dots.
Intuition
Simplifying \displaystyle \frac{6}{8} to \displaystyle \frac{3}{4} works because \displaystyle \frac{6}{8} = \frac{3 \cdot 2}{4 \cdot 2} and \displaystyle \frac{2}{2} = 1. That is simplifying a number fraction, and it rests on Equivalent fractions. A rational expression cancels the same way. The difference is that the common factor is now something like x + 5, a number that changes with x, and at one value of x it is 0.
Look at \displaystyle \frac{x + 5}{x + 5}. At x = 1 it is \displaystyle \frac{6}{6} = 1. At x = -12 it is \displaystyle \frac{-7}{-7} = 1. At every x you try it is 1, except one: at x = -5 it is \displaystyle \frac{0}{0}, which is not a number. So \displaystyle \frac{x + 5}{x + 5} is not the expression 1. It is 1 with a single point taken out, and its graph is a level line with a hole in it.
Canceling the x + 5 gives 1, which does have a value at x = -5. The canceled form fills in a value the original never had. That is why the restriction x \neq -5 is written beside it: it is the only trace left of the hole.
Results
Excluded values. The excluded values of \displaystyle \frac{P}{Q} are the solutions of Q = 0, where Q is the denominator as given, before any canceling. Factoring Q and using the zero product property finds them.
Canceling a common factor. If A, B and C are polynomials, then
wherever B \neq 0 and C \neq 0.
Opposite factors. Since b - a = -(a - b),
wherever a \neq b.
The simplified form. A rational expression and its simplified form are equal at every value of the variable except the excluded values of the original. So the simplified form is stated with those restrictions, including any whose factor was canceled away.
Why it is true
The canceling rule is the rule for multiplying fractions, read backwards:
Each step needs a nonzero number in every denominator. In particular \displaystyle \frac{C}{C} = 1 holds only where C \neq 0, since \displaystyle \frac{0}{0} is not 1 or anything else. So the equation holds exactly where B \neq 0 and C \neq 0, which is where the left side is defined. Where C = 0 but B \neq 0, the right side has a value and the left side has none, so the two expressions differ at that point.
For opposite factors, b - a = (-1)(a - b) by the distributive property. So \displaystyle \frac{a - b}{b - a} = \frac{1 \cdot (a - b)}{(-1)(a - b)}, and canceling the common factor a - b, allowed where a \neq b, leaves \displaystyle \frac{1}{-1} = -1.
Worked examples
Finding the excluded values
Find every excluded value of (a) \displaystyle \frac{5x}{x^2 - 7x + 10} and (b) \displaystyle \frac{x - 1}{x^2 + 4}.
Solution(a) An excluded value makes the denominator 0, so solve x^2 - 7x + 10 = 0. The numbers that multiply to 10 and add to -7 are -2 and -5, so (x - 2)(x - 5) = 0. A product is 0 only when a factor is 0, so x = 2 or x = 5. Check: 4 - 14 + 10 = 0 and 25 - 35 + 10 = 0. The excluded values are 2 and 5.
The value x = 0 makes the numerator 0, and that is allowed: the expression is \displaystyle \frac{0}{10} = 0 there.
(b) Solve x^2 + 4 = 0. A square is never negative, so x^2 + 4 \geq 4 for every real x, and the denominator is never 0. There are no excluded values: the expression has a value at every real number.
Simplifying, and the restriction that disappears
Simplify \displaystyle \frac{x^2 - 9}{x^2 + x - 6} and state the restrictions.
SolutionFind the excluded values first, from the denominator as given. x^2 + x - 6 = (x + 3)(x - 2), so x \neq -3 and x \neq 2. The numerator is a difference of squares, x^2 - 9 = (x - 3)(x + 3), and now the common factor is in plain view:
Canceling x + 3 is allowed because x + 3 \neq 0 at every value that is left. The restriction x \neq 2 can still be read off the new denominator, but x \neq -3 cannot, which is why it has to be written. A table shows the difference.
| x | \displaystyle \frac{x^2 - 9}{x^2 + x - 6} | \displaystyle \frac{x - 3}{x - 2} |
|---|---|---|
| -3 | \displaystyle \frac{0}{0}, no value | \displaystyle \frac{6}{5} |
| 0 | \displaystyle \frac{-9}{-6} = \frac{3}{2} | \displaystyle \frac{-3}{-2} = \frac{3}{2} |
| 2 | \displaystyle \frac{-5}{0}, no value | \displaystyle \frac{-1}{0}, no value |
| 4 | \displaystyle \frac{7}{14} = \frac{1}{2} | \displaystyle \frac{1}{2} |
The two agree everywhere except at x = -3. On the graph that one point is a hole, while at x = 2 both of them break.
A monomial denominator
Simplify \displaystyle \frac{2x^2 - 8x}{6x} and state the restrictions.
SolutionThe denominator 6x is 0 only at x = 0, so x \neq 0. Take the greatest common factor 2x out of the numerator, and write the denominator with the same factor showing:
Canceling 2x is allowed wherever 2x \neq 0, which is exactly the restriction. The answer has no variable in its denominator, so it looks defined everywhere: at x = 0 it gives \displaystyle -\frac{4}{3}, while the original gives \displaystyle \frac{0}{0}. Only the restriction remembers the difference. Check at x = 1: \displaystyle \frac{2 - 8}{6} = -1 and \displaystyle \frac{1 - 4}{3} = -1.
Opposite factors
Simplify \displaystyle \frac{x^2 - 5x + 6}{3 - x} and state the restrictions.
Solution3 - x = 0 at x = 3, so x \neq 3. The numerator factors as (x - 2)(x - 3). The denominator 3 - x is not x - 3, but it is its opposite, 3 - x = -(x - 3). Written that way, the common factor appears:
Check at x = 0: \displaystyle \frac{6}{3} = 2 and 2 - 0 = 2.
Factor completely first
Simplify \displaystyle \frac{12 - 3x^2}{x^2 - x - 2} and state the restrictions.
SolutionThe denominator is x^2 - x - 2 = (x - 2)(x + 1), so x \neq 2 and x \neq -1. For the numerator, take out the common factor 3 first and then factor what is left, as in factoring completely:
The factor 2 - x is the opposite of x - 2 in the denominator, so write 2 - x = -(x - 2):
Nothing else cancels: x + 2 and x + 1 are different factors. Check at x = 0: the original is \displaystyle \frac{12}{-2} = -6, and \displaystyle \frac{-3 \cdot 2}{1} = -6.
Building an expression to order
Write a rational expression that simplifies to x + 2 and has exactly the excluded values 1 and -4.
SolutionWork backwards. The denominator must be 0 at x = 1 and x = -4 and nowhere else, so take (x - 1)(x + 4). For the expression to simplify to x + 2, the numerator must be x + 2 times that same product:
Its simplified form is x + 2 with x \neq 1,\ -4. A polynomial alone could never do this job, since it has no denominator and so no excluded values. Other answers work too, such as \displaystyle \frac{2(x + 2)(x - 1)(x + 4)}{2(x - 1)(x + 4)}.
A paint mix
Ilsabet tests shades of pink. At dial setting x, her mixer pours 3x liters of red paint and x^2 liters of white, where x \geq 0. Write the fraction of the batch that is red, simplify it, and say what the restriction means for the mixer.
SolutionThe batch holds 3x + x^2 liters, and 3x of them are red. Take out the common factor x below, then cancel it:
The excluded value -3 never comes up, since the dial has x \geq 0, but 0 does. At setting 0 the mixer pours nothing, and “what fraction of nothing is red” has no answer, just as the original’s \displaystyle \frac{0}{0} says. The short formula would give \displaystyle \frac{3}{3} = 1, claiming an empty batch is all red. The restriction x \neq 0 is what stops that. At setting 3 the batch is 9 liters of red and 9 of white, and \displaystyle \frac{3}{3 + 3} = \frac{1}{2} agrees.
Boundaries
Only factors cancel, never terms. The rule rests on \displaystyle \frac{C}{C} = 1 for a factor C that multiplies the whole top and the whole bottom. In \displaystyle \frac{x + 6}{6} the 6 on top is a term, added rather than multiplied, so nothing cancels. The expression is not x + 1: at x = 6 it is \displaystyle \frac{12}{6} = 2, while x + 1 = 7.
Excluded values come from the denominator you were given. Read off after canceling, some are lost, as in Example 2.
Equal only where both are defined. The statement \displaystyle \frac{x^2 - 9}{x^2 + x - 6} = \frac{x - 3}{x - 2} is true for every x except -3 and 2. It does not say the two are the same expression.
Only a difference flips sign. \displaystyle \frac{a - b}{b - a} = -1, but \displaystyle \frac{a + b}{b + a} = 1, since a sum does not change when its terms swap places.
Polynomials on top and bottom only. With a variable under a square root or in an exponent, the result is not a rational expression, and its forbidden values come from more than a denominator. Which inputs a formula accepts in general is the implied domain.
Where this goes next
Every later operation on rational expressions starts with the two moves here. Multiplying and dividing them is factoring and canceling across several fractions at once, and adding and subtracting them builds a common denominator from factored denominators. When a rational expression is set equal to something and solved, the excluded values decide which answers survive: a solution that is an excluded value is thrown out.
Read as a function, \displaystyle y = \frac{P(x)}{Q(x)} has its excluded values as the gaps in its domain. Example 2 showed both kinds of gap: a factor that cancels leaves a hole in the graph, and a factor that stays in the denominator makes a vertical asymptote. When the numerator’s degree is at least the denominator’s, polynomial long division rewrites the expression as a polynomial plus a smaller fraction. And quantities that vary inversely, such as \displaystyle y = \frac{k}{x}, are rational expressions from the start: Inverse, joint and combined variation.
Exercises
Excluded values
Find every excluded value.
- \displaystyle \frac{7}{x - 4}
- \displaystyle \frac{x + 6}{5x + 2}
- \displaystyle \frac{x - 3}{x^2 - 2x - 15}
- \displaystyle \frac{4x}{x^2 + 9}
- \displaystyle \frac{x^2 - 1}{x^3 - 4x}
Simplifying
Simplify, and state the restrictions.
- \displaystyle \frac{10x^3}{15x}
- \displaystyle \frac{4x + 12}{x + 3}
- \displaystyle \frac{x^2 - 25}{x^2 + 2x - 15}
- \displaystyle \frac{x^2 + 7x + 12}{x^2 - 16}
- \displaystyle \frac{3x^2 - 6x}{x^2 - 4x + 4}
- \displaystyle \frac{2x^2 + 5x - 3}{4x^2 - 1}
- \displaystyle \frac{x^3 + 2x^2 - 9x - 18}{x^2 + 5x + 6}
Opposite factors
Simplify, and state the restrictions.
- \displaystyle \frac{9 - x}{2x - 18}
- \displaystyle \frac{x^2 - 36}{6 - x}
- \displaystyle \frac{2x - 10}{25 - x^2}
- \displaystyle \frac{x^2 - 3x - 4}{4 + 3x - x^2}
Using the restrictions
- Simplify \displaystyle \frac{x^2 - 4x}{x^2 - 16}. Then find its value at x = 2, and at x = 4 if it has one.
- Jurek simplifies \displaystyle \frac{x^2 + 6x}{x} to x + 6 and concludes that the expression equals 6 when x = 0. Is he right?
- Find the number a for which \displaystyle \frac{x^2 + ax - 10}{x - 2} simplifies to a polynomial, and give that polynomial with its restriction.
- Write a rational expression that simplifies to \displaystyle \frac{1}{x - 3} and has exactly the excluded values 3 and -1.
- Mstislav shares x^2 - 4 liters of paint equally among x + 2 tins. Write the liters per tin as a rational expression, simplify it, and state the restriction.
Reasoning
- Is \displaystyle \frac{x^2 + 4}{x + 2} simplified? Explain.
- Show that \displaystyle \frac{(x - 1)^2}{(1 - x)^2} = 1 for every x \neq 1, even though \displaystyle \frac{x - 1}{1 - x} = -1.
- Octave finds the excluded values of \displaystyle \frac{x^2 - x}{x^2 + 2x} by simplifying first and reading the new denominator. Which excluded value does he miss, and why does it still count?
- Radegund says two rational expressions with the same simplified form must have the same excluded values. Is she right? Explain with an example.
Answers
- x = 4. The denominator x - 4 is 0 only there.
- \displaystyle x = -\frac{2}{5}. Solve 5x + 2 = 0.
- x = 5 and x = -3, since x^2 - 2x - 15 = (x - 5)(x + 3). The value x = 3 makes only the numerator 0, which is allowed.
- None. x^2 + 9 \geq 9 for every real x, so the denominator is never 0.
- x = 0, x = 2 and x = -2, since x^3 - 4x = x(x - 2)(x + 2).
- \displaystyle \frac{2x^2}{3}, x \neq 0. \displaystyle \frac{10x^3}{15x} = \frac{5x \cdot 2x^2}{5x \cdot 3}.
- 4, x \neq -3. \displaystyle \frac{4(x + 3)}{x + 3} is the constant 4 with a hole at x = -3.
- \displaystyle \frac{x - 5}{x - 3}, x \neq -5,\ 3. \displaystyle \frac{(x - 5)(x + 5)}{(x + 5)(x - 3)}.
- \displaystyle \frac{x + 3}{x - 4}, x \neq 4,\ -4. \displaystyle \frac{(x + 3)(x + 4)}{(x - 4)(x + 4)}.
- \displaystyle \frac{3x}{x - 2}, x \neq 2. \displaystyle \frac{3x(x - 2)}{(x - 2)^2}: one factor of x - 2 cancels and one stays.
- \displaystyle \frac{x + 3}{2x + 1}, \displaystyle x \neq \frac{1}{2},\ -\frac{1}{2}. \displaystyle \frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)}.
- x - 3, x \neq -2,\ -3. By grouping, the numerator is x^2(x + 2) - 9(x + 2), which is (x + 2)(x - 3)(x + 3), and the denominator is (x + 2)(x + 3).
- \displaystyle -\frac{1}{2}, x \neq 9. \displaystyle \frac{-(x - 9)}{2(x - 9)}, since 9 - x = -(x - 9).
- -x - 6, x \neq 6. \displaystyle \frac{(x - 6)(x + 6)}{-(x - 6)} = -(x + 6).
- \displaystyle -\frac{2}{x + 5}, x \neq 5,\ -5. \displaystyle \frac{2(x - 5)}{-(x - 5)(x + 5)}.
- -1, x \neq 4,\ -1. The denominator is -(x^2 - 3x - 4), the opposite of the numerator, and x^2 - 3x - 4 = (x - 4)(x + 1).
- \displaystyle \frac{x}{x + 4}, x \neq 4,\ -4. At x = 2 the value is \displaystyle \frac{2}{6} = \frac{1}{3}, and the original gives \displaystyle \frac{-4}{-12}, the same. At x = 4 there is no value: the original is \displaystyle \frac{0}{0}, even though the simplified form would say \displaystyle \frac{1}{2}.
- No. \displaystyle \frac{x(x + 6)}{x} = x + 6 only for x \neq 0. At x = 0 the expression is \displaystyle \frac{0}{0} and has no value.
- a = 3, giving x + 5, x \neq 2. For x - 2 to cancel it must be a factor of the numerator, so the numerator is 0 at x = 2: 4 + 2a - 10 = 0. Then x^2 + 3x - 10 = (x + 5)(x - 2).
- For example \displaystyle \frac{x + 1}{x^2 - 2x - 3}, which is \displaystyle \frac{x + 1}{(x - 3)(x + 1)}. The denominator must be 0 at 3 and -1, and the extra factor x + 1 must also sit on top so that it cancels.
- x - 2 liters per tin, x \neq -2. \displaystyle \frac{x^2 - 4}{x + 2} = \frac{(x - 2)(x + 2)}{x + 2}. At x = -2 there would be no tins to share among.
- Yes. If x + 2 were a factor of x^2 + 4, then by the factor theorem x^2 + 4 would be 0 at x = -2, but it is 8 there. The two share no factor, so nothing cancels.
- 1 - x = -(x - 1), so (1 - x)^2 = (-1)^2(x - 1)^2 = (x - 1)^2. Top and bottom are the same nonzero number for every x \neq 1, so the quotient is 1. Squaring removes the sign that makes the unsquared quotient -1.
- He misses x = 0. Simplifying gives \displaystyle \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}, whose denominator shows only -2. But at x = 0 the original is \displaystyle \frac{0}{0} and has no value; the canceled factor x was 0 there.
- No. \displaystyle \frac{x(x + 2)}{x} and \displaystyle \frac{(x + 2)(x - 1)}{x - 1} both simplify to x + 2, but the first excludes 0 and the second excludes 1.