Topic 1.11B · CED: Equivalent Representations of Polynomial and Rational Expressions

Polynomial Long Division and Slant Asymptotes

This note is a review — you've already seen slant asymptotes (1.7B) and the full rational function checklist (1.8 through 1.10). Here we practice those skills heavily, add synthetic division as a shortcut, and learn to read end behavior straight from quotient+remainder form.

7 MIN READ8 IDEAS23 PROBLEMS21 flashcards

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01
Slant asymptote exists when the numerator's degree is exactly one more than the denominator's. Find it by dividing — the quotient is the asymptote.

Practice 23 problems Next: Chapter 1.12A Translations of Functions

Remember: What Is a Slant Asymptote?

KEY RULE

Slant asymptote exists when the numerator's degree is EXACTLY ONE more than the denominator's. Find it by dividing — the QUOTIENT is the asymptote.

Remember: The Full Checklist

CONCEPT

From 1.8 through 1.10

Domain → Simplify (check for holes) → Vertical asymptotes → Zeros → Horizontal/slant asymptote.

We'll use this same checklist in the comprehensive problem later in this note.

New Skill: Synthetic Division (a Faster Shortcut)

When dividing by a LINEAR factor $(x - c)$ , synthetic division gives the same answer as long division, using only the coefficients — much faster to write out.

Worked example

Divide $(x^{3} - 2x^{2} - 5x + 7)$ by $(x - 4)$ , two ways.

f(x) = \frac{x^3 - 2x^2 - 5x + 7}{x - 4}

Method 1: Long Division

\begin{array}{r} x^2+2x+3\phantom{{}+{}7} \\ x-4\,\overline{\smash{\big)}\,x^3-2x^2-5x+7} \\ \underline{-\;(x^3-4x^2)\phantom{{}-{}5x+7}} \\ 2x^2-5x\phantom{{}+{}7} \\ \underline{-\;(2x^2-8x)\phantom{{}+{}7}} \\ 3x+7 \\ \underline{-\;(3x-12)} \\ 19 \end{array}

Method 2: Synthetic Division

\begin{array}{r|rrrr} 4 & 1 & -2 & -5 & 7 \\ & & 4 & 8 & 12 \\ \hline & 1 & 2 & 3 & \boxed{19} \end{array}

01
Both methods agree:

f(x) = x^2 + 2x + 3 + \frac{19}{x - 4}

COMMON MISTAKE

Synthetic division ONLY works when dividing by a linear factor $(x - c)$ — not by anything degree 2 or higher.

Use $c = 4$ for dividing by $(x - 4)$ , but $c = -4$ for dividing by $(x + 4)$ — a very common sign slip.

Quick check

To divide by $x + 3$ with synthetic division, what number goes in the box?

New Skill: Reading End Behavior from Quotient + Remainder Form

If a rational function is already given in quotient-plus-remainder form, you don't need to do any division — the remainder fraction vanishes as $x \to \pm \infty$ , so end behavior follows the quotient alone.

Worked example

Find the end behavior of $f(x) = -3x + 2 + (5x - 1)/(x^{2} + 2x - 8)$ , using limit notation.

f(x) = -3x + 2 + \frac{5x - 1}{x^2 + 2x - 8}

01
As $x \to \pm \infty$ , the fraction $(5x-1)/(x^{2}+2x-8)$ shrinks toward 0 (numerator degree 1 < denominator degree 2).
02
So the end behavior matches the quotient, −3x+2, alone — a simple line with negative slope:
$\lim_{x \to -\infty} f(x) = +\infty \qquad \lim_{x \to \infty} f(x) = -\infty$

Practice: Horizontal Asymptote, Slant Asymptote, or Neither?

Worked example

A. $f(x) = (5x^{4} - 2x^{2} + 1) / (3x^{4} + x - 2)$

f(x) = \frac{5x^4 - 2x^2 + 1}{3x^4 + x - 2}

01
Degrees are EQUAL (both 4) → horizontal asymptote, $y = 5 / 3$ .

Worked example

B. $f(x) = (2x^{5} + x^{3} + 1) / (4x^{3} - 2x + 3)$

f(x) = \frac{2x^5 + x^3 + 1}{4x^3 - 2x + 3}

01
Numerator degree 5, denominator degree 3 — a gap of 2 (not 1) → NEITHER a horizontal nor a slant asymptote.

Worked example

C. $f(x) = (x^{3} + 4x^{2} - 1) / (2x^{2} + 3)$

f(x) = \frac{x^3 + 4x^2 - 1}{2x^2 + 3}

01
Numerator degree 3, denominator degree 2 — a gap of exactly 1 → SLANT asymptote (divide to find its equation).

Practice: Finding the Slant Asymptote Equation

Worked example

A. Find the slant asymptote of $f(x) = (2x^{2} + 5x - 3) / (x + 4)$ .

f(x) = \frac{2x^2 + 5x - 3}{x + 4}

01
Divide (long division or synthetic, $c=-4$ ):
$f(x) = 2x - 3 + \frac{9}{x + 4}$
02
Slant asymptote: $y = 2x - 3$ .

Worked example

B. Find the slant asymptote of $g(x) = (x^{3} - 3x^{2} + 2) / (x^{2} - 1)$ .

g(x) = \frac{x^3 - 3x^2 + 2}{x^2 - 1}

01
Run the checklist first — factor and look for a hole before dividing. The numerator has a root at $x = 1$ , so $(x-1)$ comes out of it:
$x^3 - 3x^2 + 2 = (x - 1)(x^2 - 2x - 2) \qquad x^2 - 1 = (x - 1)(x + 1)$
02
$(x-1)$ cancels, so there is a HOLE at $x = 1$ . Its height comes from the simplified form:
$\frac{x^2 - 2x - 2}{x + 1} \bigg|_{x = 1} = \frac{1 - 2 - 2}{2} = -\frac{3}{2}$

Hole at $\left(1, -\dfrac{3}{2}\right)$ , and the only vertical asymptote is $x = -1$ .

03
Now divide for the end behavior:

g(x) = x - 3 + \frac{x - 1}{x^2 - 1}

04
That remainder simplifies too — $\dfrac{x-1}{x^2-1} = \dfrac{1}{x+1}$ , which shrinks to $0$ far from the origin.

05
Slant asymptote: $y = x - 3$ .

COMMON MISTAKE

Dividing straight away and reporting only the slant asymptote. The division is correct, but it hides the hole at $x=1$ and makes $x^2-1$ look like two vertical asymptotes when there is only one.

Quick check

Before dividing $\frac{x^{3} + x^{2} - 9x - 9}{x^{2} - 1}$ for its slant asymptote, what should you check first?

Comprehensive Practice: Everything at Once

Worked example

For $f(x) = (2x^{3} - 8x) / (x^{2} - x - 6)$ , find: domain, hole(s), zero(s), vertical asymptote(s), slant asymptote, and y-intercept.

f(x) = \frac{2x^3 - 8x}{x^2 - x - 6}

01
Factor everything:
$2x^3 - 8x = 2x(x - 2)(x + 2)$ $x^2 - x - 6 = (x - 3)(x + 2)$
02
The factor $(x+2)$ appears in both — it cancels completely:
$f(x) = \frac{2x(x - 2)}{x - 3} \quad (x \neq -2)$
03
Domain: all reals except $x=-2$ (hole) and $x=3$ (vertical asymptote).
04
Hole location — plug $x=-2$ into the simplified form:
$\lim_{x \to -2} f(x) = \frac{2(-2)(-4)}{-5} = \frac{16}{-5} = -\frac{16}{5}$
05
Zeros (from simplified numerator 2x(x−2)): $x=0$ and $x=2$ . Vertical asymptote: $x=3$ .
06
Numerator degree 2, denominator degree 1 (after simplifying) — a gap of 1 → SLANT asymptote. Divide:
$2x^2 - 4x = (x - 3)(2x + 2) + 6$ $f(x) = 2x + 2 + \frac{6}{x - 3} \;\Rightarrow\; \text{slant asymptote: } y = 2x + 2$
07
y-intercept: plug $x=0$ into the simplified form → $f(0) = 0$ (matches the zero at $x=0$ ).

CONCEPT

Final Answers

Domain: x≠−2, 3

Hole: (−2, −16/5)

Zeros: $x=0$ , $x=2$

Vertical asymptote: $x=3$

Slant asymptote: $y=2x+2$

y-intercept: $(0, 0)$

Common slips

Synthetic division only works when dividing by a linear factor $(x - c)$ — not by anything degree 2 or higher.

Use $c = 4$ for dividing by $(x - 4)$ , but $c = -4$ for dividing by $(x + 4)$ — a very common sign slip.
Dividing straight away and reporting only the slant asymptote. The division is correct, but it hides the hole at $x=1$ and makes $x^2-1$ look like two vertical asymptotes when there is only one.

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21 cards · Rational functions, Horizontal asymptote or end behavior?, Equivalent forms and division

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Recap card

4 lines to re-read the night before.

01
Synthetic division is a faster shortcut for long division, but only works for linear divisors $(x-c)$ .
02
If a function is already in quotient+remainder form, end behavior follows the quotient alone — the remainder fraction vanishes at infinity.
03
Compare degrees first: equal → horizontal asymptote; gap of 1 → slant; gap of 2+ → neither.
04
A comprehensive problem often combines holes, vertical asymptotes, zeros, and a slant asymptote all in one function — work through the checklist in order.

Practice 23 problems Next: Chapter 1.12A Translations of Functions

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Polynomial Long Division and Slant Asymptotes

Remember: What Is a Slant Asymptote?

Remember: The Full Checklist

New Skill: Synthetic Division (a Faster Shortcut)

Method 1: Long Division

Method 2: Synthetic Division

New Skill: Reading End Behavior from Quotient + Remainder Form

Practice: Horizontal Asymptote, Slant Asymptote, or Neither?

Practice: Finding the Slant Asymptote Equation

Comprehensive Practice: Everything at Once

Recap card

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