Rational Expressions

Study Sheet

Rational Expressions

Excluded values, simplifying, multiplying, dividing, adding, subtracting, complex fractions, equations, and work problems

What Is a Rational Expression?

Concept
Definition

A rational expression is a ratio of two polynomials---one polynomial on top (the numerator) divided by another polynomial on the bottom (the denominator):

polynomialpolynomial,such as  2x+1x−3,x2−9x2+5x+6,5x.\frac{\text{polynomial}}{\text{polynomial}},\qquad\text{such as }\ \frac{2x+1}{x-3},\quad \frac{x^2-9}{x^2+5x+6},\quad \frac{5}{x}.

It is just like a numerical fraction (such as 34\tfrac{3}{4}), except the top and bottom can contain variables.

Example
Example: Which are rational expressions?
  • [leftmargin=*]
  • 2x+1x−3\dfrac{2x+1}{x-3} Yes --- a polynomial over a polynomial.
  • 5x2\dfrac{5}{x^2} Yes --- the constant 55 and x2x^2 are both polynomials.
  • x+1\sqrt{x}+1 No --- x\sqrt{x} is not a polynomial (a variable under a root).
Tip

Remember: A constant like 55 is a polynomial, so 5x\dfrac{5}{x} counts. But roots of a variable, or a variable in an exponent, are not allowed in a polynomial.

Excluded Values (Domain Restrictions)

Concept
Why the Denominator Matters

Division by zero is undefined. So a rational expression is undefined for any value that makes the denominator equal to 00. Those values are called excluded values.

Method: Set the denominator equal to 00, solve, and exclude those values.

Example
Example: Find the excluded values of x+4x2−x−6\dfrac{x+4}{x^2-x-6}

Set the denominator to 00 and factor:

x2−x−6=0(x−3)(x+2)=0x=3orx=−2.\begin{aligned} x^2 - x - 6 &= 0 \\ (x-3)(x+2) &= 0 \\ x = 3 \quad&\text{or}\quad x = -2. \end{aligned}

So the excluded values are x≠3x \neq 3 and x≠−2x \neq -2.

Tip

Tip: Look at the original denominator to find excluded values, before you cancel anything. Canceling can hide a restriction, but the restriction is still there.

Simplifying Rational Expressions

Concept
Factor, Then Cancel Common Factors

To simplify a rational expression:

  • [leftmargin=*]
  • Factor the numerator and denominator completely.
  • Cancel any factor that appears in both the top and the bottom.
  • State the excluded values (from the original denominator).
Example
Example: Simplify x2−9x2+5x+6\dfrac{x^2-9}{x^2+5x+6}

Factor top and bottom:

x2−9x2+5x+6=(x−3)(x+3)(x+2)(x+3).\frac{x^2-9}{x^2+5x+6} = \frac{(x-3)(x+3)}{(x+2)(x+3)}.

Cancel the common factor (x+3)(x+3):

=x−3x+2,x≠−3, x≠−2.= \frac{x-3}{x+2},\qquad x\neq -3,\ x\neq -2.
Example
Example: Simplify x2−xx\dfrac{x^2-x}{x}

Factor the top: x2−x=x(x−1)x^2 - x = x(x-1). Then cancel the xx:

x(x−1)x=x−1,x≠0.\frac{x(x-1)}{x} = x-1,\qquad x\neq 0.
Tip

You may only cancel FACTORS, never terms!   In x+2x\dfrac{x+2}{x} you cannot cancel the xx's, because x+2x+2 is a sum, not a product. Only cancel things that are multiplied. For instance x+3x+2\dfrac{x+3}{x+2} is already simplified---nothing cancels.

Multiplying Rational Expressions

Concept
Factor, Cancel, Multiply

To multiply, factor everything, cancel any factor common to a top and a bottom, then multiply the remaining tops together and the remaining bottoms together. (You do not need a common denominator to multiply.)

Example
Example: x2−4x2+6x+9⋅x+3x−2\dfrac{x^2-4}{x^2+6x+9}\cdot\dfrac{x+3}{x-2}

Factor each part:

(x−2)(x+2)(x+3)(x+3)⋅x+3x−2.\frac{(x-2)(x+2)}{(x+3)(x+3)}\cdot\frac{x+3}{x-2}.

Cancel one (x−2)(x-2) and one (x+3)(x+3):

=x+2x+3,x≠−3, x≠2.= \frac{x+2}{x+3},\qquad x\neq -3,\ x\neq 2.
Tip

Tip: Cancel before you multiply out---it keeps the numbers small. Collect the excluded values from every original denominator.

Dividing Rational Expressions

Concept
Multiply by the Reciprocal

To divide, flip the second fraction (take its reciprocal) and multiply. Then factor and cancel as usual.

AB÷CD=AB⋅DC.\frac{A}{B}\div\frac{C}{D} = \frac{A}{B}\cdot\frac{D}{C}.
Example
Example: x2−1x2−4÷x+1x−2\dfrac{x^2-1}{x^2-4}\div\dfrac{x+1}{x-2}

Flip the divisor and multiply:

x2−1x2−4⋅x−2x+1=(x−1)(x+1)(x−2)(x+2)⋅x−2x+1.\frac{x^2-1}{x^2-4}\cdot\frac{x-2}{x+1} = \frac{(x-1)(x+1)}{(x-2)(x+2)}\cdot\frac{x-2}{x+1}.

Cancel (x+1)(x+1) and (x−2)(x-2):

=x−1x+2,x≠2, x≠−2, x≠−1.= \frac{x-1}{x+2},\qquad x\neq 2,\ x\neq -2,\ x\neq -1.
Tip

Careful with restrictions when dividing. The factor you flip (x+1x+1 here) was a denominator after flipping, so x≠−1x\neq -1 too. Track excluded values from both the original bottoms and the divisor's top.

Adding & Subtracting with LIKE Denominators

Concept
Same Bottom? Combine the Tops

When the denominators are already the same, add or subtract the numerators and keep the common denominator. Then simplify if possible.

AD±BD=A±BD.\frac{A}{D}\pm\frac{B}{D} = \frac{A\pm B}{D}.
Example
Example: x2x−2−4x−2\dfrac{x^2}{x-2}-\dfrac{4}{x-2}

Subtract the numerators over the common denominator, then factor and cancel:

x2−4x−2=(x−2)(x+2)x−2=x+2,x≠2.\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2,\qquad x\neq 2.
Tip

Subtracting? Subtract the whole numerator---use parentheses. AD−BD=A−BD\dfrac{A}{D}-\dfrac{B}{D}=\dfrac{A-B}{D}, and the minus sign hits every term of BB.

Adding & Subtracting with UNLIKE Denominators

Concept
Find the LCD

When denominators differ, build a least common denominator (LCD):

  • [leftmargin=*]
  • Factor each denominator.
  • The LCD contains each different factor the greatest number of times it appears in any one denominator.
  • Rewrite each fraction over the LCD, then combine the numerators.
Example
Example: 2x+3x+1\dfrac{2}{x}+\dfrac{3}{x+1}

The denominators xx and x+1x+1 share no factors, so the LCD is x(x+1)x(x+1). Rewrite each fraction:

2x⋅x+1x+1+3x+1⋅xx=2(x+1)x(x+1)+3xx(x+1)=2x+2+3xx(x+1)=5x+2x(x+1),x≠0, x≠−1.\begin{aligned} \frac{2}{x}\cdot\frac{x+1}{x+1} + \frac{3}{x+1}\cdot\frac{x}{x} &= \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} \\ &= \frac{2x+2+3x}{x(x+1)} = \frac{5x+2}{x(x+1)},\qquad x\neq 0,\ x\neq -1. \end{aligned}
Example
Example with factoring: 3x+2+xx2−4\dfrac{3}{x+2}+\dfrac{x}{x^2-4}

Factor x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2), so the LCD is (x−2)(x+2)(x-2)(x+2):

3(x−2)(x−2)(x+2)+x(x−2)(x+2)=3x−6+x(x−2)(x+2)=4x−6(x−2)(x+2)=2(2x−3)(x−2)(x+2),x≠2, x≠−2.\begin{aligned} \frac{3(x-2)}{(x-2)(x+2)} + \frac{x}{(x-2)(x+2)} &= \frac{3x-6+x}{(x-2)(x+2)} \\ &= \frac{4x-6}{(x-2)(x+2)} = \frac{2(2x-3)}{(x-2)(x+2)},\qquad x\neq 2,\ x\neq -2. \end{aligned}
Tip

Do not just multiply the denominators blindly. Factor first so the LCD is as small as possible, and only multiply each numerator by the factors its denominator is missing.

Complex Fractions (Intro)

Concept
A Fraction Within a Fraction

A complex fraction has fractions in its numerator, its denominator, or both. Two ways to simplify:

  • [leftmargin=*]
  • Multiply through by the LCD of every little fraction, or
  • Combine the top into one fraction and the bottom into one fraction, then divide (multiply by the reciprocal).
Example
Example: 1+1x1−1x\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}

The only small denominator is xx, so multiply the top and bottom by xx:

(1+1x)⋅x(1−1x)⋅x=x+1x−1,x≠0, x≠1.\frac{\left(1+\frac{1}{x}\right)\cdot x}{\left(1-\frac{1}{x}\right)\cdot x} = \frac{x+1}{x-1},\qquad x\neq 0,\ x\neq 1.
Tip

Tip: Multiplying every part by the LCD clears all the little fractions in one clean step. Watch the excluded values: x≠0x\neq 0 (the inner denominator) and x≠1x\neq 1 (the new bottom).

Solving Rational Equations

Concept
Clear the Denominators

To solve an equation with rational expressions:

  • [leftmargin=*]
  • Note the excluded values (denominators ≠0\neq 0).
  • Multiply every term by the LCD to clear all fractions.
  • Solve the resulting equation.
  • Check each answer---reject any that is an excluded value (an extraneous solution).
Example
Example (real solution): 3x+1=2x−1\dfrac{3}{x+1}=\dfrac{2}{x-1}

Cross-multiply (or multiply by the LCD (x+1)(x−1)(x+1)(x-1)):

3(x−1)=2(x+1)3x−3=2x+2x=5.\begin{aligned} 3(x-1) &= 2(x+1) \\ 3x-3 &= 2x+2 \\ x &= 5. \end{aligned}

Since x=5x=5 is not excluded (x≠−1,1x\neq -1,1), the solution is x=5x=5.

Example
Example (extraneous!): xx−4=4x−4+2\dfrac{x}{x-4}=\dfrac{4}{x-4}+2

Excluded value: x≠4x\neq 4. Multiply every term by (x−4)(x-4):

x=4+2(x−4)x=4+2x−8x=2x−4−x=−4 ⇒ x=4.\begin{aligned} x &= 4 + 2(x-4) \\ x &= 4 + 2x - 8 \\ x &= 2x - 4 \\ -x &= -4 \ \Rightarrow\ x = 4. \end{aligned}

But x=4x=4 is excluded, so it is extraneous. There is no solution.

Tip

Always check for extraneous solutions! An answer that makes any original denominator 00 must be thrown out---even though it came from correct algebra.

Applications: Work & Rate Problems

Concept
The Work Formula

If one worker (or pipe) finishes a job in aa units of time and another in bb units, each does a fraction of the job per unit of time: 1a\frac{1}{a} and 1b\frac{1}{b}. Working together for tt units:

1a+1b=1t.\frac{1}{a} + \frac{1}{b} = \frac{1}{t}.

(Their rates add.)

Example
Example: Two pipes filling a tank

Pipe A fills a tank in 66 hours; pipe B fills it in 33 hours. Together, how long?

16+13=1t.\frac{1}{6} + \frac{1}{3} = \frac{1}{t}.

The LCD is 66: 16+26=36=12\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}. So 1t=12\dfrac{1}{t}=\dfrac{1}{2}, giving t=2t=2 hours.

Example
Example: Solving for an unknown rate

Maria paints a room in 33 hours. With Sam helping, they finish in 22 hours. How long would Sam take alone?

13+1s=12 ⇒ 1s=12−13=3−26=16.\frac{1}{3} + \frac{1}{s} = \frac{1}{2} \ \Rightarrow\ \frac{1}{s} = \frac{1}{2}-\frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}.

So s=6s=6 hours for Sam alone.

Tip

Sanity check: Working together must be faster than either person alone. If your “together” time is bigger than one of the individual times, recheck the setup.

Going Deeper: Advanced Rational Expressions

Concept
Continued & Nested Complex Fractions

A continued fraction stacks fractions inside fractions many layers deep. Simplify it from the bottom up: fully combine the innermost fraction, then work outward one level at a time. The same two tools from before still work---multiply by an LCD, or combine-then-flip---but here you apply them repeatedly.

2+12+12  =  2+1 52   =  2+25  =  125.2 + \dfrac{1}{2 + \dfrac{1}{2}} \;=\; 2 + \dfrac{1}{\ \frac{5}{2}\ } \;=\; 2 + \dfrac{2}{5} \;=\; \dfrac{12}{5}.
Example
Worked Example: Simplify x 1+1 1+1x  \dfrac{x}{\,1+\dfrac{1}{\,1+\dfrac{1}{x}\,}\,}

Start at the deepest level and climb out. First the inner denominator:

1+1x=x+1x.1+\frac{1}{x}=\frac{x+1}{x}.

Substitute and flip that little fraction:

1+1 x+1x =1+xx+1=(x+1)+xx+1=2x+1x+1.1+\dfrac{1}{\ \frac{x+1}{x}\ }=1+\frac{x}{x+1}=\frac{(x+1)+x}{x+1}=\frac{2x+1}{x+1}.

Now the whole expression is x 2x+1x+1 \dfrac{x}{\ \frac{2x+1}{x+1}\ }, so flip once more:

x1⋅x+12x+1=x(x+1)2x+1,x≠0, x≠−1, x≠−12.\frac{x}{1}\cdot\frac{x+1}{2x+1}=\frac{x(x+1)}{2x+1},\qquad x\neq 0,\ x\neq -1,\ x\neq -\tfrac12.

Each inner denominator contributes an excluded value---track them all as you unwind.

Concept
The x+1xx+\dfrac{1}{x} Family

A classic trick: if you know the value of x+1xx+\dfrac{1}{x}, you can find x2+1x2x^2+\dfrac{1}{x^2} and x3+1x3x^3+\dfrac{1}{x^3} without ever solving for xx. The key is that squaring or cubing x+1xx+\frac1x produces a “+2+2” or “+3(⋯ )+3(\cdots)” cross-term that you subtract back off.

(x+1x)2=x2+2+1x2⇒x2+1x2=(x+1x)2−2,(x+1x)3=x3+3(x+1x)+1x3⇒x3+1x3=(x+1x)3−3(x+1x).\begin{aligned} \left(x+\tfrac{1}{x}\right)^2 &= x^2 + 2 + \tfrac{1}{x^2} &&\Rightarrow\quad x^2+\tfrac{1}{x^2}=\left(x+\tfrac{1}{x}\right)^2-2,\\[2pt] \left(x+\tfrac{1}{x}\right)^3 &= x^3 + 3\left(x+\tfrac{1}{x}\right) + \tfrac{1}{x^3} &&\Rightarrow\quad x^3+\tfrac{1}{x^3}=\left(x+\tfrac{1}{x}\right)^3-3\left(x+\tfrac{1}{x}\right). \end{aligned}
Example
Worked Example: If x+1x=5x+\dfrac{1}{x}=5, find x2+1x2x^2+\dfrac{1}{x^2} and x3+1x3x^3+\dfrac{1}{x^3}

Let a=x+1x=5a=x+\dfrac1x=5. Square first:

x2+1x2=a2−2=52−2=25−2=23.x^2+\frac{1}{x^2}=a^2-2=5^2-2=25-2=23.

Now cube, using the formula above:

x3+1x3=a3−3a=53−3(5)=125−15=110.x^3+\frac{1}{x^3}=a^3-3a=5^3-3(5)=125-15=110.

Check the pattern another way: x3+1x3=(x+1x) ⁣(x2−1+1x2)=5 (23−1)=5⋅22=110.x^3+\frac{1}{x^3}=\left(x+\frac1x\right)\!\left(x^2-1+\frac{1}{x^2}\right)=5\,(23-1)=5\cdot 22=110. ✓

Tip

Why it works: multiplying out (x+1x)2\left(x+\frac1x\right)^2 or (x+1x)3\left(x+\frac1x\right)^3, the “x⋅1x=1x\cdot\frac1x=1” cross-terms are constants. Rearranging isolates the “symmetric” power sum you want. This is the same idea as (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with b=1ab=\frac1a.

Concept
Telescoping Sums via Partial Fractions

Some long sums of rational terms collapse. Split each term into a difference of simpler fractions; then consecutive pieces cancel and only the first and last survive. The workhorse identity is

1n(n+1)=1n−1n+1.\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.

Adding from n=1n=1 to n=Nn=N, every −1n+1-\frac{1}{n+1} cancels the next term's +1n+1+\frac{1}{n+1}:

∑n=1N1n(n+1)=(1−12)+(12−13)+⋯+(1N−1N+1)=1−1N+1=NN+1.\sum_{n=1}^{N}\frac{1}{n(n+1)}=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac{1}{N+1}\right)=1-\frac{1}{N+1}=\frac{N}{N+1}.
Example
Worked Example: Evaluate 11⋅2+12⋅3+13⋅4+⋯+19⋅10\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{9\cdot10}

Each term is 1n(n+1)=1n−1n+1\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac{1}{n+1}. Write them out and watch the middle vanish:

(1−12)+(12−13)+(13−14)+⋯+(19−110).\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\left(\tfrac13-\tfrac14\right)+\cdots+\left(\tfrac19-\tfrac{1}{10}\right).

Only the very first 11 and the very last −110-\tfrac{1}{10} are left:

=1−110=910.=1-\frac{1}{10}=\frac{9}{10}.

(Matches the formula NN+1\frac{N}{N+1} with N=9N=9.)

Tip

Recognizing a telescoping split: to break 1n(n+1)\dfrac{1}{n(n+1)} into An+Bn+1\dfrac{A}{n}+\dfrac{B}{n+1}, clear denominators: 1=A(n+1)+Bn1=A(n+1)+Bn. Setting n=0n=0 gives A=1A=1; setting n=−1n=-1 gives B=−1B=-1. That “cover-up” method is partial-fraction decomposition, and it is the engine behind every telescoping sum.

Concept
Preview: Asymptotes & Holes of Rational Functions

The graph of y=P(x)Q(x)y=\dfrac{P(x)}{Q(x)} is shaped by where the function blows up or flattens. After factoring and canceling:

  • [leftmargin=*]
  • Hole at x=cx=c: a factor (x−c)(x-c) that cancels from top and bottom---the graph is undefined at cc but has no infinite spike (a single missing point).
  • Vertical asymptote at x=cx=c: a factor (x−c)(x-c) left in the denominator after canceling---the graph shoots to ±∞\pm\infty.
  • Horizontal asymptote: compare degrees. Bottom-heavy ⇒y=0\Rightarrow y=0; equal degrees ⇒y=leading coeff topleading coeff bottom\Rightarrow y=\frac{\text{leading coeff top}}{\text{leading coeff bottom}}; top-heavy ⇒\Rightarrow none (a slant asymptote instead).
Example
Example: Locate the features of y=x2−1x2−x−2y=\dfrac{x^2-1}{x^2-x-2}

Factor top and bottom:

y=(x−1)(x+1)(x−2)(x+1).y=\frac{(x-1)(x+1)}{(x-2)(x+1)}.

The factor (x+1)(x+1) cancels, so there is a hole at x=−1x=-1. What remains is x−1x−2\dfrac{x-1}{x-2}, whose leftover denominator factor (x−2)(x-2) gives a vertical asymptote at x=2x=2. Top and bottom have equal degree (11 and 11) with leading coefficients 11 and 11, so the horizontal asymptote is y=11=1y=\frac{1}{1}=1.

Tip

Cancel first, then classify. A factor that cancels makes a hole; a factor that stays in the denominator makes a vertical asymptote. Same value of xx can never be both---decide it after simplifying.

Concept
Average Speed Is a Harmonic Mean

For equal distances traveled at two speeds, the average speed is not the plain average of the speeds. Since time=distancespeed\text{time}=\frac{\text{distance}}{\text{speed}}, cover distance dd out at speed v1v_1 and back at v2v_2:

v‾=total distancetotal time=2d dv1+dv2 =2 1v1+1v2 =2v1v2v1+v2.\overline{v}=\frac{\text{total distance}}{\text{total time}}=\frac{2d}{\ \frac{d}{v_1}+\frac{d}{v_2}\ }=\frac{2}{\ \frac{1}{v_1}+\frac{1}{v_2}\ }=\frac{2v_1v_2}{v_1+v_2}.

This is the harmonic mean---the same “rates add” idea as the work formula.

Example
Worked Example: Round trip at 3030 and 6060 mph

You drive to work at 3030 mph and return along the same road at 6060 mph. The average speed is not 4545. Use the harmonic mean:

v‾=2v1v2v1+v2=2(30)(60)30+60=360090=40 mph.\overline{v}=\frac{2v_1v_2}{v_1+v_2}=\frac{2(30)(60)}{30+60}=\frac{3600}{90}=40\ \text{mph}.

Sanity check with a real distance: let each leg be 6060 miles. Out takes 6030=2\frac{60}{30}=2 h, back takes 6060=1\frac{60}{60}=1 h, total 120120 miles in 33 h =40=40 mph. ✓ The slower leg dominates because you spend more time at the low speed.

Tip

Which mean? Equal times at two speeds ⇒\Rightarrow ordinary average v1+v22\frac{v_1+v_2}{2}. Equal distances ⇒\Rightarrow harmonic mean 2v1v2v1+v2\frac{2v_1v_2}{v_1+v_2}. The harmonic mean is always the smaller of the two---slow stretches cost extra time.

Formulas, Proofs & Tips

Tip
Excluded values
P(x)Q(x) is undefined where Q(x)=0\frac{P(x)}{Q(x)} \text{ is undefined where } Q(x)=0

What it means. A rational expression exists everywhere except where its denominator vanishes.

Example. 1x−2\dfrac{1}{x-2} is undefined at x=2x=2, so x≠2x\ne 2.

Why it works. Division asks "what times the denominator gives the numerator?". If the denominator is 00, no number works (or every number does, when the numerator is 00 too), so the value cannot be defined.

Tip. Find the excluded values from the original denominator, before cancelling. Cancelling can hide a restriction that still applies.