Why Zeros, Poles, and Multiplicities Matter
Factored rational functions show important information at a glance. A factor in the numerator can indicate where the function is zero; a factor in the denominator can indicate a pole. The exponent on a factor records its multiplicity, which helps determine whether the function's sign changes as the input passes that value.
There is one qualification that must not be overlooked: a rational expression has an original domain restriction wherever its original denominator is zero. Cancelling a common factor can simplify the formula, but it does not put an excluded input back into the domain. That distinction is essential when deciding whether a value is a zero, a pole, or neither.
Definitions for Factored Rational Functions
Consider a rational function written as \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial. Its original domain consists of the real inputs \(x\) for which \(Q(x)\ne0\). A real number \(c\) is a zero of \(f\) if \(c\) belongs to this original domain and \(f(c)=0\).
To classify factors reliably, first cancel all common polynomial factors, but keep a record of the original domain restrictions. Write the resulting expression as \(A(x)/B(x)\), where \(A\) and \(B\) have no common factor. An input at which \(B(c)=0\) is a pole of the rational function. Since \(A\) and \(B\) have no common factor, \(A(c)\ne0\) at such a real pole.
For example, if the reduced numerator contains \((x-2)^3\), then \(2\) is a zero of multiplicity \(3\), provided \(2\) was in the original domain. If the reduced denominator contains \((x-5)^2\), then \(5\) is a pole of multiplicity \(2\). The original domain condition matters even after reduction: if a common factor was cancelled at \(c\), then \(c\) remains excluded from the domain.
Worked Example: Classifying Factors Without Cancellation
Worked Example: Classifying Factors Without Cancellation
Find the real zeros and poles, with their multiplicities, of
The denominator is zero at \(x=4\) and \(x=-3\), so these inputs are excluded from the domain. There are no common factors to cancel.
The numerator is zero at \(x=2\) and \(x=-1\). Neither input makes the denominator zero. At \(x=2\), the denominator is \((2-4)^2(2+3)=4\cdot5=20\ne0\), so \(2\) is a zero of multiplicity \(3\). At \(x=-1\), the denominator is \((-1-4)^2(-1+3)=25\cdot2=50\ne0\), so \(-1\) is a zero of multiplicity \(2\).
At \(x=4\), the numerator is \((4-2)^3(4+1)^2=2^3\cdot5^2=200\ne0\), so \(4\) is a pole of multiplicity \(2\). At \(x=-3\), the numerator is \((-3-2)^3(-3+1)^2=(-5)^3(-2)^2=-125\cdot4=-500\ne0\), so \(-3\) is a pole of multiplicity \(1\).
Check: The numerator factors account for the two zeros, and the denominator factors account for the two poles. Each listed zero is in the domain, and each listed pole has a nonzero numerator.
Cancellation and the Original Domain
Common factors require extra care. Suppose an expression is given as \(P(x)/Q(x)\), and both polynomials contain \(x-c\). The original expression is undefined at \(c\), because \(Q(c)=0\). Cancelling the common factor gives an expression that agrees with the original wherever the original is defined, but does not change the original domain.
After cancellation, a factor may remain in the numerator at an input that the original denominator excluded. That input is not a zero: a zero must be a value at which the function is defined and equals zero. Likewise, if all denominator factors at an excluded input cancel, that input is not a pole of the reduced rational function. It is still excluded from the original domain.
Worked Example: A Cancelled Factor Does Not Automatically Give a Zero
Classify the zeros and poles of
The original denominator vanishes at \(x=1\) and \(x=4\). Thus the original domain excludes both values. Cancelling one factor of \(x-1\) gives
The reduced numerator is zero at \(x=1\) and \(x=-2\). But \(x=1\) is not in the original domain, so it is not a zero of \(g\). At \(x=-2\), the original denominator is \((-2-1)(-2-4)^2=(-3)\cdot36=-108\ne0\), so \(-2\) is a zero of multiplicity \(1\).
The reduced denominator has a factor \((x-4)^2\). The original numerator at \(4\) is \((4-1)^2(4+2)=9\cdot6=54\ne0\); therefore \(4\) is a pole of multiplicity \(2\). The point \(1\) is excluded but is neither a zero nor a pole of the reduced function.
Check: At \(-2\), the numerator vanishes and the denominator does not, as required for a zero. At \(4\), the reduced denominator vanishes while its reduced numerator is \((4-1)(4+2)=18\ne0\), as required for a pole. The domain still excludes \(1\) and \(4\).
Why Multiplicity Predicts a Sign Change
When moving across a real number \(c\), the factor \(x-c\) changes sign: it is negative to the left of \(c\) and positive to the right. Its power \((x-c)^m\) changes sign across \(c\) exactly when \(m\) is odd. If \(m\) is even, the factor has the same sign on both sides.
For a zero, the reduced function near \(c\) can be written as \((x-c)^m h(x)\), where \(h(c)\ne0\). For a pole of multiplicity \(m\), it can be written as \(h(x)/(x-c)^m\), again with \(h(c)\ne0\). In either case, the nonvanishing factor \(h\) has a fixed sign sufficiently close to \(c\), so the parity of \(m\) determines whether the signs on the two sides differ.
Here is the factor argument in full. Near \(c\), all factors other than the relevant power of \(x-c\) are nonzero. Their signs therefore remain fixed on a sufficiently small interval on each side of \(c\). Moving from left to right changes the sign of \(x-c\) once. Raising it to an even power gives a positive factor on both sides; raising it to an odd power gives a negative factor on the left and a positive factor on the right. Taking a reciprocal does not change whether two signs agree or differ. Thus the parity rule applies to poles as well as zeros.
Worked Example: Sign Changes and an Excluded Input
Worked Example: Sign Changes and an Excluded Input
Determine the signs on the intervals separated by the critical values for
The original domain excludes \(x=-1\) and \(x=3\). Cancelling the common factor gives
The reduced numerator has a factor \(x-2\), so \(2\) is a zero of multiplicity \(1\). The reduced denominator has \((x-3)^3\), so \(3\) is a pole of multiplicity \(3\). Although the reduced numerator vanishes at \(-1\), that input is excluded from the original domain, so \(-1\) is not a zero. It remains a critical value for a sign chart because it separates intervals of the domain.
Choose one test value in each interval. At \(x=-2\),
At \(x=0\),
At \(x=5/2\),
At \(x=4\),
The sign chart is therefore negative on \((-\infty,-1)\), positive on \((-1,2)\), negative on \((2,3)\), and positive on \((3,\infty)\). The sign changes across \(-1\) because the reduced expression has a remaining factor \(x+1\) of odd multiplicity, even though \(-1\) is excluded and is not a zero. It changes across \(2\) because the zero has odd multiplicity, and across \(3\) because the pole has odd multiplicity.
Check: The four test values give signs \(-,+,-,+\), matching the chart. The chart marks \(-1\) and \(3\) as excluded, and \(2\) as a zero; only \(3\) is a pole.
A Common Mistake
A frequent error is to cancel a factor and then treat every zero of the reduced numerator as a zero of the original function. For example, consider
The original domain is \(x\ne1\). Cancelling gives \(q(x)=x-1\) for \(x\ne1\). The incorrect conclusion is: “The reduced numerator \(x-1\) vanishes at \(1\), so \(1\) is a zero.” That conclusion ignores the domain condition. Since \(q(1)\) is undefined, \(1\) is not a zero. The reduced formula is negative for \(x<1\) and positive for \(x>1\), but the excluded input is still not a zero or a pole.
A Reliable Classification Process
Use the following procedure when a rational function is given in factored form.
Set the original denominator equal to zero and exclude each real solution from the domain.
Reduce the expression, but retain every restriction recorded in Step 1.
A real root of the reduced numerator is a zero only if it belongs to the original domain. Its exponent gives the multiplicity.
A real root of the reduced denominator is a pole. Its exponent gives the multiplicity.
Odd multiplicity changes the sign across the value; even multiplicity does not. Keep excluded inputs marked in the sign chart.
Why These Distinctions Matter
Zeros, poles, and excluded inputs play different roles in solving rational inequalities and organizing sign charts. A zero is a defined input where the function equals zero. A pole is a point where the reduced denominator vanishes and the reduced numerator does not. An excluded input created by cancellation is still absent from the original domain, even when the reduced formula has a finite value there.
Multiplicity adds useful sign information without requiring a fresh test calculation on every interval. Odd multiplicities indicate a sign change; even multiplicities indicate no sign change. This rule applies to zeros and poles, and sign charts must also respect all original domain restrictions. Keeping these ideas separate prevents both incorrect classifications and incorrect inequality solutions.
Check Your Understanding
For each question, keep the original domain in view as well as the reduced expression.
- For \(\frac{(x-3)^2(x+1)}{(x-3)(x-5)^3}\), identify the original domain restrictions, zeros, poles, and their multiplicities.
- Why is a root of the reduced numerator not automatically a zero of the original rational function?
- Does a pole of multiplicity \(4\) produce a sign change across the pole? Explain using the factor's parity.
- For \(\frac{(x+2)^3}{(x-1)^2}\), determine whether the sign changes across each zero or pole.
- If a common factor is cancelled, should its original denominator root still be marked as excluded in a sign chart? Why?
