Why Reduce Before Dividing?
A rational function is written as a quotient of polynomials. Before comparing the degrees of its numerator and denominator, or dividing one by the other, first look for polynomial factors that they share. Canceling such factors can reveal a much simpler expression and a simpler denominator for any later decomposition.
Cancellation does not change the rational function’s values wherever the original expression is defined. It can, however, change which denominator factors are visible and how many times they occur. It also does not change whether the rational function is proper or improper: the numerator and denominator degrees decrease by the same amount. These facts make it important to cancel correctly and then work with the reduced expression.
Common Factors and Reduced Form
Let \(P(x)\) and \(Q(x)\) be polynomials, with \(Q\) not the zero polynomial. A common factor is a polynomial that divides both \(P\) and \(Q\). If \(G(x)\) is a common factor, write \(P(x)=G(x)A(x)\) and \(Q(x)=G(x)B(x)\). Wherever \(G(x)\ne 0\) and the original denominator is nonzero,
Canceling means removing the same factor from the numerator and denominator. It does not mean removing a term that merely appears in both polynomials. A fraction is reduced when its numerator and denominator have no common nonconstant polynomial factor.
For real-coefficient polynomials, a common factor may be linear or of higher degree. For example, \(x^2-1=(x-1)(x+1)\), so \(x-1\) is a factor of \(x^2-1\). The factorization step is what makes cancellation possible; the expression \(x^2-1\) is not itself a product with a visible \(x-1\) until it is factored.
Degree Comparison After Cancellation
A nonzero rational function \(P(x)/Q(x)\), in reduced form, is called proper if \(\deg P<\deg Q\), and improper if \(\deg P\geq\deg Q\). Polynomial division is required for an improper rational function before its proper remainder can be handled separately.
Why compare after cancellation if cancellation does not change the proper/improper classification? Because the reduced form identifies the actual denominator factors that remain and avoids carrying a common factor through calculations. The classification is indeed unchanged: canceling a factor of degree \(d\) lowers both polynomial degrees by \(d\), leaving their difference fixed.
Why Cancellation Preserves the Polynomial Quotient
Polynomial division of \(A(x)\) by a nonzero polynomial \(B(x)\) gives unique polynomials \(S(x)\) and \(T(x)\) such that
Here \(S\) is the quotient and \(T\) is the remainder. The same quotient results whether we divide a rational function before or after canceling a common factor. The reduced version is usually easier to work with, and it also exposes the factors that remain in the denominator.
Express the numerator and denominator as \(P=GA\) and \(Q=GB\).
Write \(A=SB+T\), with \(T=0\) or \(\deg T<\deg B\).
Then \(P/Q=GA/(GB)=A/B=S+T/B\), wherever the original expression is defined.
This proves the claim: the polynomial quotient is \(S\) in either representation. The division identity also shows why it is useful to simplify first: the remainder is paired with the reduced denominator \(B\), not with the larger unreduced denominator \(GB\).
Worked Examples
Worked Example: Cancel a Linear Factor Before Comparing Degrees
Reduce \(\frac{x^2-1}{x-1}\), compare degrees, and identify the domain restriction.
First factor the numerator:
Cancel the common factor \(x-1\). This cancellation is valid only where \(x-1\ne0\), so \(x\ne1\):
The reduced expression is \((x+1)/1\). Its numerator has degree \(1\), and its denominator has degree \(0\), so it is still improper and is simply a polynomial. Check the result by multiplying the reduced expression by the original denominator:
The identity verifies the cancellation wherever \(x\ne1\). At \(x=1\), the original quotient is undefined, even though the reduced expression \(x+1\) has value \(2\). The two expressions agree on the original domain; the simplified formula does not restore the missing input.
Worked Example: Divide the Reduced Improper Rational Function
Reduce \(\frac{x^3-x^2+x-1}{x^2-1}\), then perform division using the reduced form.
Factor the numerator by grouping and factor the denominator:
Cancel \(x-1\), retaining the original restriction \(x\ne1\):
The reduced numerator has degree \(2\), and the denominator has degree \(1\), so division is needed. Divide \(x^2+0x+1\) by \(x+1\). The first quotient term is \(x\); subtract \(x(x+1)=x^2+x\), leaving \(-x+1\). The next term is \(-1\); subtract \(-1(x+1)=-x-1\), leaving remainder \(2\). Therefore,
Check by recombining the quotient and remainder:
Thus the division is correct. The reduced expression is valid for \(x\ne-1\), but the original expression also excludes \(x=1\), where its canceled factor was zero. This example shows that the simplified denominator used in division is \(x+1\), while the original domain restriction must still be remembered.
Worked Example: Cancellation Reduces a Repeated Denominator Factor
Reduce \(\frac{x^2-1}{(x-1)^2(x+2)}\), and identify the denominator factors that remain.
Factor the numerator and cancel one occurrence of the common factor \(x-1\):
The original expression excludes \(x=1\) and \(x=-2\). The reduced numerator has degree \(1\), while its denominator has degree \(2\), so this reduced rational function is proper. Check the cancellation by multiplying the reduced numerator and denominator by the canceled factor:
The denominator in reduced form contains one factor \(x-1\), not two. A later decomposition should be based on the reduced denominator and its factors, rather than carrying along the canceled occurrence. The value \(x=1\) remains excluded for the original function, even though the reduced formula has a finite value there.
Cancellation and Later Decomposition
When a rational function is prepared for partial-fraction decomposition, its denominator’s factorization determines which denominator factors must be represented. A common factor in the numerator and denominator is not part of the reduced denominator and should not be carried into that decomposition. If a denominator factor is repeated, cancellation can lower its multiplicity. The third example reduced a squared factor \(x-1\) to a single occurrence.
Cancellation does not itself perform the decomposition. Its role here is to establish the correct reduced rational function first. If that reduced function is improper, divide its reduced numerator by its reduced denominator. If it is proper, there is no polynomial-division step to perform. In either case, use the reduced denominator when determining what remains to be decomposed.
Common Mistake: Assuming Cancellation Changes Properness
A tempting but incorrect conclusion is that canceling a factor can change an improper rational function into a proper one. For example, \((x^2-1)/(x-1)\) begins with numerator degree \(2\) and denominator degree \(1\). After cancellation it is \((x+1)/1\), with degrees \(1\) and \(0\). It is still improper, because \(1\geq0\).
The degree difference was \(2-1=1\) before cancellation and \(1-0=1\) after. In general, if the canceled factor has degree \(d\), then
So cancellation does not change properness. It does simplify the polynomials and can remove or reduce factors that would otherwise clutter division and subsequent decomposition.
A second error is canceling terms instead of factors. For instance, \(x^2+1\) and \(x+1\) do not share the factor \(x\): the numerator \(x^2+1\) is not divisible by \(x\), since substituting \(x=0\) gives \(1\), not \(0\). Canceling the \(x\)-terms from \((x^2+1)/(x+1)\) is therefore invalid. Factorization and divisibility, not visual similarity between terms, determine whether cancellation is allowed.
Check Your Understanding
For each question, show the factor cancellation or degree comparison that supports your answer.
- Reduce \(\frac{x^2-4}{x-2}\). What value must remain excluded from the domain of the original expression?
- If numerator and denominator share a factor of degree \(3\), how does cancellation affect the difference of their degrees?
- Reduce \(\frac{x^3-x}{x^2-x}\), and state whether the reduced rational function is proper or improper.
- In \(\frac{x^2-1}{(x-1)^2(x+2)}\), how many factors of \(x-1\) remain in the reduced denominator?
- Why is it incorrect to cancel the \(x\)-terms in \(\frac{x^2+1}{x+1}\)?
