Why Cancellation Needs a Domain Condition
A common factor can make a rational expression easier to read and evaluate. But cancelling that factor does not erase the restrictions imposed by the original denominator. The simplified formula may be defined at inputs where the original expression was not.
The key distinction is between an algebraic formula and a function with a specified domain. Two formulas can give the same values wherever both are defined without defining the same function, because their domains may differ.
Domains Before and After Cancellation
For a quotient to be defined at an input, its numerator and denominator must be defined there, and its denominator must be nonzero. In the polynomial case, the factors are defined for every real input, so the denominator being nonzero is the only restriction.
Suppose \(G\), \(U\), and \(V\) are polynomials. Consider the original quotient \(\frac{G(x)U(x)}{G(x)V(x)}\) and the reduced quotient \(\frac{U(x)}{V(x)}\). The original denominator is zero if \(G(x)=0\) or \(V(x)=0\), while the reduced denominator is zero only if \(V(x)=0\). Thus their domains are:
Every input in \(D_{\mathrm{original}}\) is also in \(D_{\mathrm{reduced}}\), but the reduced expression may have a larger domain. If \(G\) has a real zero that is not also a zero of \(V\), that input is excluded from the original quotient and allowed by the reduced one.
For more general expressions, it is not enough to check that the displayed denominator is nonzero. Each factor must itself be defined. If \(D_G,D_U,D_V\) are the domains of \(G,U,V\), respectively, then the original quotient is defined precisely at inputs \(x\) satisfying
The reduced quotient is defined at inputs satisfying \(x\in D_U\cap D_V\) and \(V(x)\ne 0\). In particular, cancelling \(G\) does not remove restrictions coming from \(U\) or \(V\).
Why the Values Agree on the Original Domain
Here is the precise cancellation result for polynomials. The same pointwise argument works for other expressions whenever all the stated expressions are defined at the input being considered.
The reason is the ordinary rule for division: if \(a\ne 0\) and \(b\ne 0\), multiplying both numerator and denominator of a quotient by the same nonzero number does not change its value. At a fixed input \(x\), the common factor is the number \(G(x)\). It can be cancelled only when that number is nonzero.
Assume \(G(x)\ne 0\) and \(V(x)\ne 0\). Then \(G(x)V(x)\ne 0\), so the original quotient is defined.
Since \(G(x)\ne 0\), the factor \(G(x)\) can be cancelled from the numerator and denominator.
The original quotient equals \(U(x)/V(x)\), which is defined because \(V(x)\ne 0\). Thus the two expressions agree at this input.
In symbols, each step of the pointwise calculation is justified by the nonzero conditions:
The cancellation is not a way to evaluate the original quotient at a zero of \(G\). At such an input, the original denominator is zero. The reduced formula may produce a number there, but that number belongs to the reduced formula, not to the original quotient.
Worked Examples
Worked Example: Cancelling a Linear Factor
Consider \(f(x)=\frac{x^2-9}{x-3}\). Find its domain, simplify it, and compare the original and reduced expressions at an allowed input and at the excluded input.
First, factor the numerator: \(x^2-9=(x-3)(x+3)\). The original denominator \(x-3\) is zero when \(x=3\), so the original domain is \(\mathbb{R}\setminus\{3\}\).
For \(x\ne 3\), the common factor \(x-3\) is nonzero and can be cancelled:
The reduced formula \(x+3\) is defined for every real number, but the original quotient remains undefined at \(x=3\). Check an allowed input, \(x=5\): the original gives \(\frac{25-9}{5-3}=\frac{16}{2}=8\), and the reduced formula gives \(5+3=8\). The values agree. At \(x=3\), the original gives \(\frac{9-9}{3-3}=\frac{0}{0}\), which is undefined; the reduced formula gives \(3+3=6\). This is not a value of the original expression. The comparison confirms equality where the original is defined and does not assign it a value at the excluded input.
Worked Example: Two Denominator Restrictions
Simplify \(q(x)=\frac{x^2-4}{x^2-2x}\), state both domains, and check the equality at \(x=1\).
Factor the numerator and denominator: \(x^2-4=(x-2)(x+2)\) and \(x^2-2x=x(x-2)\). The original denominator is zero when \(x=0\) or \(x=2\), so the original domain is \(\mathbb{R}\setminus\{0,2\}\).
When \(x\ne 2\), cancel the common factor \(x-2\). The reduced expression is \(\frac{x+2}{x}\), whose domain is \(\mathbb{R}\setminus\{0\}\). Therefore:
At \(x=1\), the original value is \(\frac{1-4}{1-2}=\frac{-3}{-1}=3\), and the reduced value is \(\frac{1+2}{1}=\frac{3}{1}=3\). At \(x=2\), the original denominator is \(2^2-2(2)=4-4=0\), so the original quotient is undefined. The reduced expression gives \(\frac{2+2}{2}=\frac{4}{2}=2\), but that value is not a value of the original quotient. The remaining restriction \(x\ne 0\) also matters: the reduced denominator is still zero at \(x=0\).
Worked Example: Restrictions from Component Expressions
Consider \(r(x)=\frac{(x+1)\left(\frac{1}{x-2}\right)}{(x+1)(x-3)}\). Cancel the common factor and compare the domains carefully.
The factor \(x+1\) is defined for all real \(x\). The component \(\frac{1}{x-2}\) is undefined at \(x=2\), so \(x=2\) must be excluded. The original denominator \((x+1)(x-3)\) is zero at \(x=-1\) and \(x=3\), so these inputs must also be excluded. Hence the original domain is \(\mathbb{R}\setminus\{-1,2,3\}\).
For inputs in that domain, \(x+1\ne 0\), so cancellation gives:
The final formula’s own domain excludes \(x=2\) and \(x=3\), but not \(x=-1\). At \(x=0\), the original value is \(\frac{1(-1/2)}{1(-3)}=\frac{-1/2}{-3}=\frac{1}{6}\), and the reduced value is \(\frac{1}{(-2)(-3)}=\frac{1}{6}\). At \(x=-1\), the original denominator is zero, so the original is undefined; the reduced formula gives \(\frac{1}{(-3)(-4)}=\frac{1}{12}\). At \(x=2\), both expressions are undefined because the component \(\frac{1}{x-2}\) is undefined. This example shows why domains of component expressions must be included, not just zeros of the displayed denominator.
A Common Mistake: Giving the Original a New Value
A tempting but incorrect conclusion is that cancellation makes an excluded input acceptable in the original expression. For example, one might write:
The last conclusion is wrong. The original denominator at \(x=1\) is \(1-1=0\), and the original numerator is \(1^2-1=0\). Thus the original quotient would require division by zero and has no value there. Cancellation is valid only under the condition \(x-1\ne 0\).
The corrected statement is:
The reduced formula gives \(2\) at \(x=1\), but this is a value of the reduced formula alone. A formula obtained by cancellation does not change the original expression’s domain.
Why the Distinction Matters
Keeping the domain attached to an expression prevents incorrect substitutions and preserves the meaning of the original function. Whenever a common factor is cancelled, record any excluded inputs from the original denominator and from any component expressions. Then check that the input you plan to use belongs to the original domain.
For polynomial factors, a reliable procedure is:
- Find where the original denominator is zero and exclude those inputs.
- Factor the numerator and denominator.
- Cancel a common factor only with its zero inputs excluded.
- Keep the original exclusions when stating where the expressions agree.
- Check any requested input in the original domain before evaluating.
The reduced formula can be useful because it is shorter, and it agrees with the original at every input where the original is defined. But the agreement of values on that domain does not mean the two expressions have identical domains. This is the precise sense in which cancellation preserves values only on the shared domain.
Check Your Understanding
For each question, keep separate the values of the expressions and the inputs where each expression is defined.
- For \(\frac{x^2-16}{x-4}\), what is the original domain, and for which inputs does the cancelled formula agree with the original?
- Why is it necessary to require \(G(x)\ne 0\) before cancelling \(G(x)\) from \(\frac{G(x)U(x)}{G(x)V(x)}\)?
- Can a reduced quotient be defined at an input where the original quotient is undefined? Give a reason.
- If \(U\) itself is undefined at an input, does cancelling a factor elsewhere make the quotient defined there? Explain.
- State the domain of \(\frac{(x-1)(x+2)}{(x-1)(x-3)}\) before cancellation, and the domain of the reduced formula.
