AP PRECALCULUS — UNIT 1B · POLYNOMIAL & RATIONAL FUNCTIONS
1.11B — Polynomial Long Division and Slant Asymptotes
Notes — Rewriting Rational Functions to Reveal Structure
💡 Learning Objectives
By the end of this lesson you will be able to (AP CED 1.11.A):
- Divide two polynomials using long division.
- Rewrite r(x) = N(x)/D(x) in the form q(x) + R(x)/D(x).
- Find the slant (oblique) asymptote of a rational function when deg N = deg D + 1.
- Use the quotient-remainder form to describe end behavior precisely.
1. Why Divide?
Any rational function r(x) = N(x) / D(x) can be rewritten as q(x) + R(x) / D(x), where q is the quotient and R is the remainder, with deg R < deg D. This form is useful because q(x) describes the end behavior of r, and R(x)/D(x) becomes small when |x| is large.
2. The Long Division Algorithm
The process mirrors numerical long division. Divide the leading term of the current remainder by the leading term of the divisor. Multiply that partial quotient by the divisor and subtract. Repeat until the degree of the remainder is less than the degree of the divisor.
📘 Example
Example 1 — Dividing x³ − 4x² + x + 6 by x − 3
Step 1: x³ ÷ x = x². Multiply: x² · (x − 3) = x³ − 3x². Subtract: (x³ − 4x²) − (x³ − 3x²) = −x². Bring down x: current is −x² + x.
Step 2: −x² ÷ x = −x. Multiply: −x · (x − 3) = −x² + 3x. Subtract: (−x² + x) − (−x² + 3x) = −2x. Bring down 6: current is −2x + 6.
Step 3: −2x ÷ x = −2. Multiply: −2(x − 3) = −2x + 6. Subtract: 0.
Quotient: x² − x − 2. Remainder: 0. So x³ − 4x² + x + 6 = (x − 3)(x² − x − 2).
⚠️ Common Mistake
Remember to INCLUDE ZERO COEFFICIENTS for missing powers of x. If the dividend skips a power (e.g., x³ − 2 has no x² or x terms), write it as x³ + 0x² + 0x − 2 before dividing. Skipping this step misaligns the subtractions.
3. Slant (Oblique) Asymptotes
If the degree of the numerator is exactly one more than the degree of the denominator, then long division produces a LINEAR quotient q(x) = mx + b, plus a remainder term that vanishes for large |x|. The line y = mx + b is called the slant asymptote.
A rational function whose numerator has degree one higher than the denominator has a slant asymptote. The dashed line is y = x − 2.
📘 Example
Example 2 — Finding a slant asymptote
Find the slant asymptote of r(x) = (x² − x − 2) / (x − 1).
Degrees: 2 and 1, differ by one. Slant asymptote exists.
Long division gives: r(x) = x − 2/(x − 1) × (oops, let's show it) — quotient is x, remainder is −x − 2.
More carefully: x² − x − 2 divided by x − 1. Step 1: x² ÷ x = x. Multiply: x(x − 1) = x² − x. Subtract: 0 − 2. Step 2: deg of remainder = 0 < deg of divisor = 1, stop.
So r(x) = x + (−2)/(x − 1). Slant asymptote: y = x.
💡 Quick Check
Which of these have a slant asymptote?
- r(x) = (2x² + 1)/(x + 3) (deg 2 / deg 1 → yes)
- r(x) = (x³ − 1)/(x² + x) (deg 3 / deg 2 → yes)
- r(x) = (x² + 1)/(x² − 4) (deg 2 / deg 2 → no — this has a horizontal asymptote)
- r(x) = (x + 5)/(x² − 1) (deg 1 / deg 2 → no — this has HA y = 0)
🎯 AP Tip
The AP CED excludes SYNTHETIC division — you are expected to use standard long division. You may still benefit from synthetic division for speed on linear divisors in non-AP contexts, but AP free-response solutions should show long-division work.
📘 Try This
Divide x³ + 2x² − 5 by x + 3 using long division. What are the quotient and remainder? Write the original expression in quotient-remainder form.
- Quotient: x² − x + 3. Remainder: −14.
- Form: (x³ + 2x² − 5) / (x + 3) = x² − x + 3 − 14/(x + 3).
4. Summary
- Polynomial long division writes r(x) = N/D as q(x) + R(x)/D(x), with deg R < deg D.
- If deg N = deg D + 1, the quotient is linear and is the slant asymptote.
- Include zero coefficients for missing powers before dividing.
- The remainder term R(x)/D(x) vanishes as |x| → ∞, confirming the end behavior y = q(x).