Why separate the quotient?
Partial fractions are set up for a proper rational function: one whose numerator has degree less than the degree of its denominator. An improper rational function does not meet that condition. Polynomial division resolves this first, expressing the original function as a polynomial plus a proper rational remainder.
The polynomial part and the remainder have different roles. The quotient is already a polynomial; only the proper remainder needs a partial-fraction decomposition. Keeping them visibly separate helps prevent a common error: decomposing the remainder correctly, then accidentally omitting the quotient from the original function.
For a nonzero remainder, the fraction \(R(x)/D(x)\) is proper and is the part to decompose. The quotient \(Q(x)\) stays outside the partial fractions. If division leaves a zero remainder, there is no proper rational part to decompose.
The setup after division
First factor the denominator of the proper remainder over the real numbers. The standard partial-fraction templates from earlier material determine the form of the rational terms. For distinct linear factors, use a constant numerator over each factor. For an irreducible quadratic factor, use a linear numerator over that factor. The examples below use distinct factors; repeated factors require additional terms, covered in the material on repeated factors.
Here the quadratic is assumed irreducible over the real numbers, and all constants in the numerators are to be determined. To find them, multiply through by the denominator. The result is a polynomial identity: it must hold for every \(x\), so its coefficients on the two sides must match.
Worked examples
Worked Example: A quotient and two distinct linear factors
Decompose the rational part of \(\frac{x^3+2x^2-2x-1}{x^2-x-2}\), keeping the polynomial quotient separate.
First factor the denominator: \(x^2-x-2=(x-2)(x+1)\). Divide the numerator by \(x^2-x-2\). The leading terms give \(x^3/x^2=x\), so subtract \(x(x^2-x-2)=x^3-x^2-2x\). This leaves \(3x^2-1\). Next, \(3x^2/x^2=3\), so subtract \(3(x^2-x-2)=3x^2-3x-6\). The remainder is \(3x+5\), whose degree is less than 2. Equivalently, check the division by multiplying back:
Therefore the quotient is \(x+3\), and the proper remainder is \(3x+5\). Set up the decomposition of that remainder:
Multiply the second identity by \((x-2)(x+1)\):
Set \(x=2\). Then \(11=3A\), so \(A=11/3\). Set \(x=-1\). Then \(2=-3B\), so \(B=-2/3\). Thus the complete expression is:
Check the decomposition by recombining the fractions. Their numerator is \(\frac{11}{3}(x+1)-\frac{2}{3}(x-2)=3x+5\), as required. The quotient check above then confirms the original numerator as well.
Worked Example: A linear factor and an irreducible quadratic
Decompose \(\frac{2x^4-3x^3+6x^2-5x+4}{(x-1)(x^2+1)}\), keeping the quotient separate.
The denominator is \(x^3-x^2+x-1\). Division gives quotient \(2x-1\) and remainder \(3x^2-2x+3\). To verify, multiply the quotient by the denominator and add the stated remainder:
The remainder has degree 2, less than the denominator's degree 3. Use one constant numerator for \(x-1\) and a linear numerator for the irreducible quadratic \(x^2+1\):
Clear the denominators in the second identity:
Expanding the right side gives \((A+B)x^2+(C-B)x+(A-C)\). Matching coefficients yields \(A+B=3\), \(C-B=-2\), and \(A-C=3\). The third equation gives \(C=A-3\); the second then gives \(B=A-1\). Substituting into the first gives \(2A-1=3\), so \(A=2\), \(B=1\), and \(C=-1\). Hence:
To check, recombine the rational terms. Their numerator is \(2(x^2+1)+(x-1)(x-1)=3x^2-2x+3\). Multiplying the quotient by the denominator and adding this remainder reproduces the original numerator, as shown by the division check.
Worked Example: Two distinct irreducible quadratics
Decompose \(\frac{2x^4+3x^3+10x^2+6x+11}{(x^2+1)(x^2+4)}\), keeping the quotient separate.
The denominator is \(x^4+5x^2+4\). Division gives quotient 2 and remainder \(3x^3+6x+3\). Check by multiplication and addition:
The remainder has degree 3, less than the denominator's degree 4. Set up a linear numerator over each irreducible quadratic:
One way to obtain the coefficients is to expand the proposed numerator identity. In this example, the values \(A=1\), \(B=1\), \(C=2\), and \(D=-1\) give:
Thus the decomposition is:
Recombining the two fractions gives the displayed remainder numerator \(3x^3+6x+3\). Adding the quotient contribution \(2(x^4+5x^2+4)\) reproduces the original numerator, verifying both parts of the expression.
Why the setup works
The quotient–remainder identity is the reason the separation is exact, not an approximation or a change to the original function. Dividing gives \(P=QD+R\). Wherever \(D(x)\ne0\), division by \(D(x)\) gives \(P/D=Q+R/D\). The partial-fraction step then rewrites \(R/D\) without changing its value.
Find \(Q\) and \(R\) such that \(P=QD+R\), with \(R=0\) or \(\deg R<\deg D\).
Write \(P/D=Q+R/D\). Do not absorb or discard the polynomial term.
Factor \(D\), write the appropriate terms for its factors, and determine their coefficients.
Recombine the rational terms to recover \(R/D\), then use \(P=QD+R\) to check the original function.
Multiplying a proposed partial-fraction identity by its denominator clears the fractions and produces a polynomial identity. This is why coefficient matching is valid: if two polynomials are equal for every \(x\), their corresponding coefficients must agree. For a concrete decomposition, recombining the terms checks that the proposed coefficients give exactly the proper remainder.
A common mistake: dropping the quotient
For the first example, the proper remainder decomposes as \(\frac{3x+5}{(x-2)(x+1)}=\frac{11}{3(x-2)}-\frac{2}{3(x+1)}\). It would be wrong to conclude that the original rational function equals only those two fractions. That expression represents the remainder fraction, not the original improper function. The division identity shows why: the original expression also contains the quotient \(x+3\).
Another tempting shortcut is to begin writing fractions before dividing, then leave an unaccounted polynomial part in the coefficient equations. Division first gives a clear target: the terms on the right of the partial-fraction identity must recombine to the proper remainder numerator, not to the full original numerator.
Check Your Understanding
For each question, keep the quotient and proper remainder distinct.
- What condition on the degrees of the numerator and denominator makes a rational function proper?
- After dividing \(P(x)\) by \(D(x)\), which part is decomposed into partial fractions?
- For a denominator with distinct linear factors \(x-a\) and \(x-b\), what form does the partial-fraction setup take?
- Why is it valid to match coefficients after multiplying a decomposition by its denominator?
- What must you do to verify that the decomposition represents the original improper rational function, rather than just its remainder?
