Why Divide Before Decomposing?
Some rational functions have numerators whose degree is at least as large as the degree of their denominators. Before using partial fractions on such a function, first use polynomial long division. The result separates off a polynomial and leaves a rational function whose numerator has lower degree than its denominator.
This order matters because the standard partial-fraction form is built for a proper rational function. Polynomial division does not itself decompose the remaining denominator; it makes the rational expression ready for that next step.
Proper and Improper Rational Functions
The degree of a nonzero polynomial is the highest exponent of its variable with a nonzero coefficient. For example, \(4x^3-2x+7\) has degree \(3\), while \(x^2+5\) has degree \(2\). When comparing degrees, a missing power has coefficient zero; it does not change the highest power that is actually present.
The remainder condition guarantees that \(R(x)/D(x)\) is proper whenever \(R\) is nonzero. If the numerator already has lower degree than the denominator, no division is needed: the quotient is zero and the remainder is the original numerator.
The equality is understood at values of \(x\) for which the original denominator is nonzero. If a common factor has been canceled, remember that the original expression may still have excluded values where its original denominator was zero.
How Polynomial Long Division Works
At each step, compare the leading term of the current dividend with the leading term of the divisor. Divide those leading terms to find the next quotient term. Multiply the entire divisor by that term, subtract the result, and continue with what remains. Stop when the remainder is zero or has lower degree than the divisor.
Subtraction applies to the whole product, so signs must be distributed carefully. If a power is missing, include its coefficient as zero when aligning terms. These habits make the algebra easier to track and the final result easier to check.
If the numerator degree is already lower, it is proper and no division is required.
Divide the current leading term by the divisor's leading term.
Multiply the full divisor by the new quotient term, then subtract the full product.
Continue until the remainder has lower degree than the divisor. Verify that dividend equals divisor times quotient plus remainder.
Worked Example: A Cubic Divided by a Quadratic
Worked Example: A Cubic Divided by a Quadratic
Rewrite \(\frac{2x^3+3x^2-5x+6}{x^2-x-2}\) as a polynomial plus a proper rational function.
The numerator has degree \(3\) and the denominator has degree \(2\), so the rational function is improper. First divide the leading terms:
Multiply the divisor by \(2x\), then subtract:
The new leading-term division is \(5x^2/x^2=5\). Multiply the divisor by \(5\) and subtract again:
The remainder \(4x+16\) has degree \(1\), which is less than the divisor degree \(2\). Therefore,
Check by recombining. Multiplying the divisor by the quotient gives \( (x^2-x-2)(2x+5)=2x^3+3x^2-9x-10\). Adding the remainder gives \(2x^3+3x^2-9x-10+4x+16=2x^3+3x^2-5x+6\), the original numerator.
Worked Example: Equal Degrees and an Integral
Worked Example: Equal Degrees and an Integral
Prepare the integrand \(\frac{3x^2+2x-1}{x^2+1}\) for partial fractions by polynomial division.
Both polynomials have degree \(2\), so the rational function is improper. Divide the leading terms: \(3x^2/x^2=3\). Multiplying the denominator by \(3\) gives \(3x^2+3\). Subtract:
The remainder has degree \(1\), less than the denominator degree \(2\). Thus,
For an indefinite integral, this gives the preparatory step \(\int \frac{3x^2+2x-1}{x^2+1}\,dx =\int 3\,dx+\int \frac{2x-4}{x^2+1}\,dx\). The polynomial part has antiderivative \(3x\), so the expression becomes \(3x+\int \frac{2x-4}{x^2+1}\,dx\), up to a constant. The remaining fraction is proper; its further treatment is a separate step.
Check the division: \(3(x^2+1)+(2x-4)=3x^2+3+2x-4=3x^2+2x-1\), as required.
Worked Example: Missing Powers and a Nonzero Remainder
Worked Example: Missing Powers and a Nonzero Remainder
Divide \(x^4+2x^2+3\) by \(x^2+2x+1\), then express the result as a polynomial plus a proper rational function.
The dividend has degree \(4\), and the divisor has degree \(2\). The first quotient term is \(x^4/x^2=x^2\). Include the missing \(x^3\) term as a zero coefficient while subtracting:
Divide the leading terms of the new expression: \((-2x^3)/x^2=-2x\). Multiply and subtract:
The next quotient term is \(5x^2/x^2=5\). Subtract the product:
The remainder degree \(1\) is less than the divisor degree \(2\). The decomposition is
To check, multiply the divisor by the quotient: \((x^2+2x+1)(x^2-2x+5)=x^4+2x^2+8x+5\). Adding the remainder gives \(x^4+2x^2+8x+5-8x-2=x^4+2x^2+3\), exactly the original dividend.
Why the Quotient-and-Remainder Form Works
The long-division steps produce an identity of the form \(N=DQ+R\), where \(N\) is the numerator, \(D\) the denominator, \(Q\) the quotient, and \(R\) the remainder. Dividing that identity by \(D\), at values where \(D(x)\ne 0\), gives
The process stops because each subtraction lowers the degree of the current expression. It must eventually reach a polynomial whose degree is less than the divisor's degree, or the zero polynomial.
The quotient and remainder are also unique. Suppose two divisions gave \(N=DQ+R\) and \(N=DQ_1+R_1\), with both remainders of degree below \(\deg D\). Subtracting the identities gives
If \(Q-Q_1\) were nonzero, the degree of the left side would be at least \(\deg D\). But \(R_1-R\) has degree less than \(\deg D\), unless it is zero. These cannot be equal unless both sides are zero. Therefore \(Q=Q_1\) and \(R=R_1\).
Why This Is the Necessary Preparation
The division separates two different kinds of terms: a polynomial \(Q(x)\), which can be integrated using the power rule, and a proper rational remainder \(R(x)/D(x)\), which can then be considered for partial fractions. The division step does not change the rational function's value where the original denominator is nonzero; it only rewrites it in a useful form.
Before dividing, compare degrees carefully. If the numerator degree is already smaller, it is already proper. If the degrees are equal, the quotient begins with a constant. If the numerator degree is larger, the quotient begins with a positive power of \(x\). In every case, stop only when the remainder has lower degree than the denominator.
A Common Sign Error
A frequent mistake is to subtract only the first term of the divisor's product instead of subtracting the entire product. In the equal-degree example, the product is \(3x^2+3\). A wrong subtraction might keep the constant \(-1\) unchanged and claim \((3x^2+2x-1)-(3x^2+3)=2x-1\). This is incorrect: the constant terms must also be subtracted, so \(-1-3=-4\), not \(-1\).
The correction is to align all powers, including the constant term, before subtracting. A reliable final check is to multiply the quotient by the denominator and add the remainder. If that does not reproduce the original numerator, some term or sign was lost.
Check Your Understanding
Use degree comparisons and the quotient-and-remainder condition to answer these questions.
- Is \(\frac{2x+1}{x^3-4}\) proper or improper? Explain using the degrees.
- What condition must the remainder satisfy when dividing a numerator by a denominator of degree \(3\)?
- When dividing \(\frac{5x^3-x+2}{x^3+1}\), what is the first quotient term?
- Why must the entire product of the divisor and each quotient term be subtracted at every long-division step?
- How can you check a proposed quotient and remainder without repeating the division?
