Why Divide Before Decomposing?
A rational function is a quotient of polynomials. If the numerator’s degree is at least the denominator’s degree, the rational function is called improper. In that case, the expression is not yet in the form of a polynomial plus a proper rational function. Polynomial long division produces that form.
This matters when preparing rational functions for later integration methods. The division itself is an algebraic step: it separates out the polynomial part and leaves a remainder whose degree is smaller than the denominator’s degree. The remainder, not the original numerator, is then the numerator of the proper rational part.
The polynomial division identity says that there are polynomials \(Q(x)\) and \(R(x)\) such that
Here, \(Q(x)\) is the quotient and \(R(x)\) is the remainder. Wherever \(D(x)\ne 0\), dividing the identity by \(D(x)\) gives
This is the desired form: a polynomial plus a proper rational function, unless the remainder is zero, in which case only the polynomial remains.
The Long-Division Procedure
Polynomial long division follows the same principle as numerical long division. At each stage, divide the leading term of the current expression by the leading term of the divisor. Place that result in the quotient, multiply the entire divisor by it, and subtract. Repeat with the new remainder until its degree is smaller than the divisor’s degree.
Divide the highest-degree term still to be processed by the divisor’s leading term. This gives the next term of the quotient.
Multiply every term of the divisor by the new quotient term, then subtract that whole product from the current expression.
If the new remainder has degree at least the divisor’s degree, find another quotient term and subtract again.
Stop when the remainder is zero or has smaller degree than the divisor. Then write \(P/D=Q+R/D\).
Subtraction applies to the whole product, not just its first term. Writing the terms in descending powers and including zero coefficients for missing powers helps keep the alignment clear.
Worked Division
Worked Example: A Cubic Divided by a Quadratic
Divide \(2x^3+3x^2-5x+6\) by \(x^2+x-2\), and express the resulting rational function as a polynomial plus a proper rational function.
The leading-term division is \(2x^3/x^2=2x\). Multiply the divisor by \(2x\):
Subtract this product from the numerator:
Now divide the leading term \(x^2\) by \(x^2\), giving \(1\). Multiply the divisor by \(1\) and subtract:
The remainder \(-2x+8\) has degree \(1\), less than the divisor’s degree \(2\). Thus the quotient is \(2x+1\), and the remainder is \(-2x+8\):
Check by recombination:
The numerator is recovered, and the remainder’s degree is smaller than the divisor’s degree, so the division is verified.
Missing Powers and Nonmonic Divisors
A missing power has coefficient zero; it is not a reason to skip or shift a term. For instance, \(3x^4-2x^2+x-5\) has the same coefficients as \(3x^4+0x^3-2x^2+x-5\). Keeping the zero term in view can prevent errors in multiplication and subtraction.
Worked Example: A Missing Power in the Numerator
Divide \(3x^4-2x^2+x-5\) by \(x^2-1\). First divide the leading terms: \(3x^4/x^2=3x^2\). Subtract the product \(3x^2(x^2-1)=3x^4-3x^2\):
The next quotient term is \(x^2/x^2=1\). Subtract \(1(x^2-1)=x^2-1\):
The remainder has degree \(1<2\), so the quotient and remainder are \(3x^2+1\) and \(x-4\). Therefore,
Check by recombining:
This recovers the numerator, and the remainder has lower degree than the divisor.
The divisor need not be monic—that is, its leading coefficient need not be \(1\). When it is not, divide the leading coefficients as well as the powers of \(x\). The next example shows why the coefficient arithmetic must be kept exact.
Worked Example: A Divisor with Leading Coefficient 2
Divide \(2x^4+x^3-3x+1\) by \(2x^2-x+1\). The numerator has a missing \(x^2\) term, so write it as \(2x^4+x^3+0x^2-3x+1\). The first quotient term is \(2x^4/(2x^2)=x^2\). Subtract:
Divide the new leading term by the divisor’s leading term: \(2x^3/(2x^2)=x\). Subtract the product \(x(2x^2-x+1)=2x^3-x^2+x\):
Since the remainder has degree \(1<2\), stop. The quotient is \(x^2+x\) and the remainder is \(-4x+1\):
Check the identity:
The recombination gives the original numerator, and the remainder satisfies the required degree condition.
Why the Procedure Works
Each subtraction removes the current highest-degree term without changing the division identity. Suppose at some stage the current polynomial is \(S(x)\), and its leading term is \(a x^m\). Let the divisor’s leading term be \(b x^n\), where \(a\ne0\) and \(b\ne0\). If \(m\ge n\), choose the next quotient term \((a/b)x^{m-n}\). Multiplying the divisor by this term produces leading term \(a x^m\). Subtracting cancels that term, leaving a polynomial of strictly smaller degree.
Thus the process cannot continue indefinitely: the nonnegative degree drops at each subtraction. It stops when the remaining polynomial is zero or has degree less than \(n\), the divisor’s degree. The quotient terms collected along the way form \(Q(x)\), and the final remainder is \(R(x)\). Because the subtractions account for exactly the products of the divisor and the quotient terms, they give \(P(x)=D(x)Q(x)+R(x)\).
A Common Subtraction Error
A frequent mistake is to subtract only the first term of the divisor’s product, or to forget that subtraction changes the sign of every term in that product. For example, in the first worked example, after choosing \(2x\), the product is \(2x^3+2x^2-4x\). Subtracting the full product gives
A wrong calculation would keep the \(-4x\) term’s sign unchanged during subtraction:
The correction is to distribute the minus sign: \(-(-4x)=+4x\), so the \(x\)-coefficient is \(-5+4=-1\), not \(-5-4=-9\). Recombining the quotient and remainder is an effective way to catch this kind of error.
Using the Result
Before proceeding to any later decomposition of a rational function, check that the division is complete and that the result has the correct form. The quotient is a polynomial; the remainder term is proper; and recombining them must recover the original numerator over the original denominator. If the remainder is zero, there is no proper rational part.
- Compare the degrees of the original numerator and denominator to decide whether division is needed.
- Carry out each multiplication and subtraction across every term.
- Stop only when the remainder is zero or has lower degree than the divisor.
- Verify the quotient and remainder with \(P=DQ+R\).
Check Your Understanding
For each question, use the quotient–remainder identity to check your reasoning.
- What degree condition makes a rational function improper?
- When dividing \(4x^3+0x^2-2x+1\) by \(2x-1\), why should the zero \(x^2\) term be kept visible during the calculation?
- Divide \(x^3+2x^2-1\) by \(x+1\). State the quotient and remainder, then verify by recombination.
- If a division produces a nonzero remainder of the same degree as the divisor, has the long division finished? Explain.
- What two checks confirm that an expression \(Q(x)+R(x)/D(x)\) is the result of dividing \(P(x)\) by \(D(x)\)?
