Introduction
A complex rational expression is a quotient in which the numerator, the denominator, or both are themselves rational expressions. The notation can look complicated, but the central operation is familiar: dividing by a nonzero quantity is the same as multiplying by its reciprocal.
The challenge is to simplify the expression without changing where it is defined. In particular, simplifying and cancelling factors does not make an original denominator that was zero suddenly permissible. We will record restrictions from the start and keep them with the final result.
Definition and Domain Restrictions
A rational expression is a quotient of polynomials, with the understanding that its denominator must not be zero. A complex rational expression is a quotient \(R(x)/S(x)\) where \(R(x)\) and \(S(x)\) may themselves be rational expressions. For the full quotient to be defined, both expressions must be defined, and \(S(x)\) must also be nonzero.
There are two kinds of restrictions to check. First, every denominator appearing in the original expression must be nonzero. Second, the entire expression being divided by must be nonzero. If that divisor is a fraction, its numerator must be nonzero as well as its denominator, since a fraction with zero numerator is zero and cannot be used as a divisor.
A useful discipline is to find the restrictions before cancelling. The simplified formula may have a larger apparent domain than the original expression. The original restrictions remain part of the answer.
Why the Reciprocal Rule Works
For numbers \(A\) and \(B\), with \(B\ne0\), division is defined by multiplication by the reciprocal of the divisor:
If the divisor is itself \(C/D\), then it is defined only when \(D\ne0\), and it is nonzero only when \(C\ne0\). Under both conditions, its reciprocal is \(D/C\). Therefore:
The identity follows because \((C/D)(D/C)=1\) when \(C\) and \(D\) are nonzero. Thus multiplication by \(D/C\) undoes division by \(C/D\). This explains both the reciprocal step and its conditions: the divisor must be defined and nonzero.
A Reliable Simplification Process
Set each denominator in the original expression unequal to zero. Also require the whole divisor to be nonzero.
Keep the first expression and flip the numerator and denominator of the expression being divided by.
Factor polynomial numerators and denominators so that common factors are visible.
Cancel common factors, then give the simplified expression together with the restrictions found in Step 1.
Cancellation is valid only where the cancelled factor is nonzero. Since the original restrictions already exclude any values that made an original denominator zero or the divisor zero, those restrictions ensure that the reciprocal and cancellations used in the calculation are valid.
Worked Examples
Worked Example: Factoring Before Cancelling
Simplify \(\displaystyle \frac{x^2-9}{x^2+2x-3}\div\frac{x-3}{x+1}\), keeping the restrictions from the original expression.
First find the restrictions. The denominator \(x^2+2x-3\) factors as \((x+3)(x-1)\), so \(x\ne-3,1\). The denominator \(x+1\) in the divisor requires \(x\ne-1\). Finally, the divisor itself must not be zero. Since its numerator is \(x-3\), this requires \(x\ne3\).
Now change division to multiplication by the reciprocal:
Factor the polynomials:
Cancel the common factors \(x-3\) and \(x+3\), which are nonzero on the restricted domain:
Check at an allowed value, \(x=2\). In the original expression, the first fraction is \((4-9)/(4+4-3)=-5/5=-1\), and the divisor is \((2-3)/(2+1)=-1/3\). Thus the quotient is \((-1)\div(-1/3)=3\). The simplified expression gives \((2+1)/(2-1)=3\), as required.
Worked Example: A Quotient of Sums
Simplify \(\displaystyle \frac{\frac{1}{x}+\frac{1}{2}}{\frac{1}{x}-\frac{1}{2}}\).
The inner fractions require \(x\ne0\). The whole denominator \(\frac{1}{x}-\frac{1}{2}\) must also be nonzero. Solving \(\frac{1}{x}-\frac{1}{2}=0\) gives \(\frac{1}{x}=\frac{1}{2}\), so \(x=2\) is excluded. The original restrictions are \(x\ne0,2\).
Combine the terms in the numerator and denominator using common denominators:
Rewrite the outer division using the reciprocal:
The factors \(2x\) cancel because \(x\ne0\). Check with \(x=1\): the original numerator is \(1+\frac12=\frac32\), and its denominator is \(1-\frac12=\frac12\). Their quotient is \(\frac32\div\frac12=3\). The simplified expression gives \((1+2)/(2-1)=3\). The check agrees.
Worked Example: Restrictions Survive Cancellation
Simplify \(\displaystyle \frac{x^2-4}{x^2-x-2}\div\frac{x+2}{x-2}\).
The denominator of the first fraction is \(x^2-x-2=(x-2)(x+1)\), so \(x\ne2,-1\). The denominator \(x-2\) in the divisor also requires \(x\ne2\). The divisor must not be zero, so its numerator \(x+2\) must not be zero; hence \(x\ne-2\). The full set of original restrictions is \(x\ne-2,-1,2\).
Multiply by the reciprocal and factor:
Cancel the common factors \(x-2\) and \(x+2\). The remaining expression is:
Check at \(x=3\). The first fraction is \((9-4)/(9-3-2)=5/4\). The divisor is \((3+2)/(3-2)=5\). Therefore the original quotient is \((5/4)\div5=(5/4)(1/5)=1/4\). The simplified expression gives \((3-2)/(3+1)=1/4\). In particular, \(x=-2\) remains excluded even though its factor cancelled: the original divisor was zero there.
A Common Mistake: Multiplying by the Divisor
When dividing by a fraction, a frequent error is to multiply by that same fraction instead of its reciprocal. Consider:
The original denominators require \(x\ne2,-3\). The divisor must be nonzero, so \(x-1\ne0\), which adds \(x\ne1\). An incorrect step would be:
This multiplies by the divisor rather than its reciprocal. For example, at \(x=0\), the original expression is \((1/-2)\div(-1/3)=(-1/2)(-3)=3/2\), whereas the incorrect product is \((-1/2)(-1/3)=1/6\). The values do not agree.
The correct step flips the divisor:
At \(x=0\), this gives \((1)(3)/((-2)(-1))=3/2\), matching the original expression.
Why the Restrictions Matter
A simplified formula describes the same values as the original complex expression only on the original domain. For example, cancelling a factor can remove it from the simplified denominator, but it cannot erase the fact that the original expression was undefined when that factor was zero. This is why a complete answer consists of both a simplified formula and the restrictions on its variable.
Before finishing a problem, check each of the following:
- Have all denominators in the original expression been required to be nonzero?
- Have you required the entire divisor to be nonzero?
- Did you flip the divisor, rather than multiply by it unchanged?
- Did you factor before cancelling?
- Are all original restrictions still stated after simplification?
Check Your Understanding
For each question, keep the domain of the original expression in view.
- In a quotient \(R(x)/S(x)\), what two conditions must hold for the quotient to be defined?
- Why must the numerator of a rational expression being divided by be nonzero?
- Simplify \(\displaystyle \frac{x^2-1}{x^2+3x+2}\div\frac{x-1}{x+2}\), stating all original restrictions.
- For \(\displaystyle \frac{\frac{1}{x}+\frac{1}{3}}{\frac{1}{x}-\frac{1}{3}}\), which values are excluded before simplification?
- Why does cancelling a factor not restore a value excluded by an original denominator?
