Why Divide Polynomials in a Rational Function?
A rational function is a quotient of polynomials. When the numerator has degree at least as large as the denominator, polynomial long division rewrites that quotient as a polynomial plus a fraction whose numerator has lower degree than its denominator. The rewritten form can make the algebraic structure easier to read and work with.
For instance, the expression \(\frac{x^2+3x+1}{x+1}\) can be written as a polynomial and a remainder fraction. The goal is not merely to obtain a shorter expression: the quotient and remainder must satisfy an exact polynomial identity, and the rational expression is still defined only where its denominator is nonzero.
The degree of a nonzero polynomial is the highest exponent with a nonzero coefficient. A rational expression \(\frac{R(x)}{D(x)}\) is called proper when the degree of its numerator is less than the degree of its denominator. If the numerator already has lower degree than the denominator, no division steps are needed: the quotient is zero and the remainder is the original numerator.
How Long Division Works
Write both polynomials in descending powers of \(x\), aligning matching powers. If a power is missing, write its coefficient as zero. This keeps every subtraction in the correct column.
At each step, divide the leading term of the current polynomial by the leading term of the divisor. That gives the next term of the quotient. Multiply the entire divisor by this term, subtract the product, and continue with the new polynomial. Stop when the remainder is zero or has degree smaller than the divisor.
Write terms in descending powers and include zero-coefficient terms for missing powers.
Divide the leading term of the current dividend by the leading term of the divisor.
Multiply the divisor by the new quotient term, then subtract that product from the current dividend.
Stop when the remainder is zero or has degree less than the divisor.
Worked Examples
Worked Example: Divide a Cubic by a Quadratic
Rewrite \(\frac{2x^3+3x^2-5x+6}{x^2-x+2}\) as a polynomial plus a proper fraction.
First divide the leading terms: \(2x^3\div x^2=2x\). Multiply the divisor by \(2x\), then subtract:
The new leading-term division is \(5x^2\div x^2=5\). Multiply the divisor by \(5\) and subtract:
The remainder has degree \(1\), which is less than the divisor's degree \(2\), so division stops. The quotient is \(2x+5\) and the remainder is \(-4x-4\). Thus, wherever \(x^2-x+2\ne0\),
Check: Multiply the divisor by the quotient and add the remainder:
This recovers the original numerator, and the remainder degree is less than the divisor degree.
Worked Example: Divide When the Divisor Is Not Monic
Divide \(3x^4-2x^3+0x^2+5x-7\) by \(2x^2+x-1\). The explicit zero coefficient keeps the missing \(x^2\) term visible.
Divide leading terms: \(3x^4\div 2x^2=\frac{3}{2}x^2\). Multiply and subtract:
Now \(-\frac{7}{2}x^3\div 2x^2=-\frac{7}{4}x\). Multiply and subtract again:
The next quotient term is \(\frac{13}{4}x^2\div 2x^2=\frac{13}{8}\). Multiply and subtract:
The remainder has degree \(1\), less than the divisor's degree \(2\). Therefore the quotient is \(\frac{3}{2}x^2-\frac{7}{4}x+\frac{13}{8}\), and the remainder is \(\frac{13}{8}x-\frac{43}{8}\). The rational expression is
Check: Multiplying the divisor by the quotient gives \(3x^4-2x^3+0x^2+\frac{27}{8}x-\frac{13}{8}\). Adding the remainder gives \(3x^4-2x^3+5x-7\), since \(\frac{27}{8}+\frac{13}{8}=5\) and \(-\frac{13}{8}-\frac{43}{8}=-7\). The identity checks.
Worked Example: An Exact Division
Divide \(x^3+2x^2-5x-6\) by \(x+3\). The leading-term division gives \(x^3\div x=x^2\). After multiplying and subtracting, continue with the new leading term:
The quotient is \(x^2-x-2\), and the remainder is zero. Thus, for \(x\ne-3\),
Check: Multiplying back gives
The original rational expression is still undefined at \(x=-3\), even though the quotient polynomial is defined there. The division identity does not change the original domain.
Why the Procedure Must Finish Correctly
Each subtraction is chosen to cancel the current leading term. If the current polynomial has leading term \(a x^m\) and the divisor has leading term \(b x^n\), with \(a\ne0\) and \(b\ne0\), their quotient term is \(\frac{a}{b}x^{m-n}\). Multiplying the divisor by this term produces leading term \(a x^m\). Subtracting therefore removes the highest power currently present.
As long as the current polynomial's degree is at least the divisor's degree, this step lowers its degree. A nonzero polynomial has a nonnegative integer degree, so the degree cannot keep decreasing indefinitely. The process stops with zero or a polynomial of lower degree than the divisor. At every step, the current expression equals the divisor times the quotient terms found so far, plus the current remainder. Consequently the final polynomials satisfy \(N(x)=D(x)Q(x)+R(x)\).
The construction above establishes existence. To see uniqueness, suppose there are two such pairs, \(Q_1,R_1\) and \(Q_2,R_2\). Subtract their identities:
If \(Q_1-Q_2\) were nonzero, the degree of the left side would be \(\deg D+\deg(Q_1-Q_2)\), which is at least \(\deg D\). But the right side has degree less than \(\deg D\), unless it is zero. A nonzero polynomial cannot have two different degrees in this identity, so both sides must be zero. Since \(D\) is nonzero, \(Q_1=Q_2\); then the original identity gives \(R_1=R_2\). The quotient and remainder are therefore unique.
A Common Subtraction Error
When subtracting the product of the divisor and a quotient term, the subtraction applies to every term in that product. For example, divide \(x^2+3x+1\) by \(x+1\). The first quotient term is \(x\), and \(x(x+1)=x^2+x\). A common incorrect step is to write
The sign in front of the product changes both terms: \(-(x^2+x)=-x^2-x\), not \(-x^2+x\). The correct subtraction is
Dividing \(2x\) by \(x\) gives \(2\). Subtract \(2(x+1)=2x+2\), leaving \(-1\). Thus the quotient is \(x+2\) and the remainder is \(-1\). Checking gives \((x+1)(x+2)-1=x^2+3x+1\).
Keep the Domain in View
Polynomial division gives an identity between polynomials. Turning that identity into a rational-expression equality requires the denominator to be nonzero. For an original expression \(\frac{N(x)}{D(x)}\), the division form \(Q(x)+\frac{R(x)}{D(x)}\) has the same restriction \(D(x)\ne0\). Do not use the quotient polynomial alone to claim that the original expression is defined at values excluded by its denominator.
The main practical checks are simple: every subtraction must be performed across all terms; the final remainder must have lower degree than the divisor; and multiplying the divisor by the quotient, then adding the remainder, must recover the original numerator.
Check Your Understanding
Use the quotient-and-remainder identity to answer these questions.
- When is a rational expression already proper, and what are its quotient and remainder in that case?
- Why must a missing power be represented with a zero coefficient during long division?
- Divide \(x^2+4x+1\) by \(x+2\). State the quotient and remainder, and verify by multiplying back.
- What condition on the remainder tells you that the division is complete?
- If a quotient polynomial is defined at a value where the original denominator is zero, does that make the original rational expression defined there? Explain.
