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Warm-Up

One function, three outfits:

f(x) = x² − 2x − 8          (standard form)
f(x) = (x − 4)(x + 2)       (factored form)
f(x) = (x − 1)² − 9         (vertex form)

Quick: what's the y-intercept? (Look at standard form: −8.) The zeros? (Factored: 4 and −2.) The minimum? (Vertex: −9, at x = 1.) Each question was instant in the right form and annoying in the others. Symbolic fluency on this exam isn't about grinding algebra — it's about choosing the representation that makes the question trivial, and converting when needed.

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Core Concept

Which form shows what

Conversions: expand (factored → standard), factor (standard → factored, when it cooperates), complete the square (standard → vertex, for quadratics). The AP asks "which form..." questions directly, and FRQ 3 asks you to produce an equivalent form.

Polynomial long division

Dividing p(x) by a lower-degree d(x) produces a quotient q(x) and remainder r(x) with degree of r < degree of d:

p(x)/d(x) = q(x) + r(x)/d(x)        equivalently   p(x) = d(x)·q(x) + r(x)

Worked layout for (2x³ + x² − 5x + 2) ÷ (x² + 1):

        2x  + 1
      ┌─────────────────────
x²+1  │ 2x³ +  x² − 5x + 2
        2x³       + 2x        ← 2x·(x² + 1)
        ───────────────
               x² − 7x + 2
               x²      + 1    ← 1·(x² + 1)
               ─────────
                  − 7x + 1    ← remainder (degree 1 < degree 2 ✓)

Result: (2x³ + x² − 5x + 2)/(x² + 1) = 2x + 1 + (−7x + 1)/(x² + 1).

A useful spot-check: evaluate both sides at an easy x (like x = 0): left: 2/1 = 2; right: 1 + 1/1 = 2 ✓.

One consequence worth knowing (the remainder theorem): dividing by (x − a) leaves remainder p(a), because p(x) = (x − a)q(x) + r gives p(a) = 0 + r.

Slant (oblique) asymptotes

When a rational function is top-heavy by exactly one degree (n = d + 1), the quotient from division is linear, and the graph hugs that line at both ends:

g(x) = (x² − 3x + 5)/(x − 1) = (x − 2) + 3/(x − 1)

As x → ±∞, the leftover 3/(x − 1) → 0, so the graph approaches the slant asymptote y = x − 2. In limit language: the gap between g(x) and the line shrinks to zero: lim x→±∞ [g(x) − (x − 2)] = 0.

Top-heavy by more than one degree? The quotient is a higher-degree polynomial, and the graph's ends follow that polynomial curve instead of a line (sometimes called the "end-behavior polynomial"). Same division, same logic.

The slant asymptote — like the horizontal kind — can be crossed at finite x (set the leftover fraction equal to 0: if the remainder has a real zero, the graph touches the line there).

The binomial theorem

Expanding (a + b)ⁿ without multiplying n times:

(a + b)ⁿ = Σ C(n, k) aⁿ⁻ᵏ bᵏ,   k = 0 … n

where C(n, k) = n!/(k!(n−k)!) are the numbers in row n of Pascal's triangle: row 4 is 1, 4, 6, 4, 1; row 5 is 1, 5, 10, 10, 5, 1.

Example: (x − 2)⁴ — treat b = −2 and keep its sign inside each power:

(x − 2)⁴ = x⁴ + 4x³(−2) + 6x²(−2)² + 4x(−2)³ + (−2)⁴
         = x⁴ − 8x³ + 24x² − 32x + 16

Sign pattern check: with a negative b, signs must alternate. If yours don't, you dropped a sign.

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Worked Examples

Example 1 (easy) — Form-switching on demand 🚫 No-Calc

Problem: h(x) = −2(x + 1)(x − 5). Find the y-intercept, the zeros, and the maximum value of h.

Solution: - Zeros: read off — x = −1 and x = 5. - y-intercept: h(0) = −2(1)(−5) = 10. - Maximum: vertex at the midpoint of the zeros, x = 2; h(2) = −2(3)(−3) = 18. Since a = −2 < 0 (concave down), maximum value 18.

Interpretation: Zeros from factored form, vertex from symmetry — no expansion needed anywhere.

Example 2 (medium) — Division with a missing term 🚫 No-Calc

Problem: Rewrite f(x) = (x³ − 5x + 6)/(x − 2) as quotient + remainder/divisor.

Strategy: Insert 0x² as a placeholder; divide.

Solution:

x³ + 0x² − 5x + 6 ÷ (x − 2):
x³ − 2x²                      ← x²(x − 2);  bring down: 2x² − 5x
2x² − 4x                      ← 2x(x − 2);  bring down: −x + 6
−x + 2                        ← −1(x − 2);  remainder: 4

Quotient x² + 2x − 1, remainder 4:

f(x) = x² + 2x − 1 + 4/(x − 2)

Check at x = 0: LHS 6/(−2) = −3; RHS −1 + 4/(−2) = −1 − 2 = −3 ✓

Interpretation: The placeholder 0x² prevents the most common division wreck. And the check takes ten seconds — do it every time on FRQ 3.

Example 3 (medium) — Slant asymptote, plus the crossing 🚫 No-Calc

Problem: Find the slant asymptote of g(x) = (x² − 3x + 5)/(x − 1) and determine whether the graph crosses it.

Solution: From the division above: g(x) = (x − 2) + 3/(x − 1) → slant asymptote y = x − 2. Crossing: the graph meets the line where the leftover term is zero: 3/(x − 1) = 0 has no solution (a nonzero constant over anything is never 0). The graph never crosses this slant asymptote.

Interpretation: Crossing the slant asymptote ⇔ remainder expression has a real zero. Here the remainder is the constant 3 — no zeros, no crossings.

Example 4 (AP-style) — Binomial coefficient extraction 🚫 No-Calc

Problem: What is the coefficient of x³ in the expansion of (x + 2)⁵?

Strategy: One term of the binomial theorem — don't expand everything.

Solution: The x³ term takes k = 2 (since aⁿ⁻ᵏ = x³ needs n − k = 3):

C(5, 2) · x³ · 2² = 10 · 4 · x³ = 40x³

Coefficient: 40.

Interpretation: "Find one coefficient" questions reward surgical use of the formula. Expanding all six terms is a two-minute penalty for a twenty-second task.

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Common Mistakes

1. Skipping placeholder zeros in division. Dividing x³ − 5x + 6 without 0x² misaligns every column. Fix: write every degree, always.
2. Sign errors in the subtraction step. Long division's "subtract" flips both signs of the line below. Fix: change the signs explicitly before combining, and run the x = 0 check at the end.
3. Slant asymptote = quotient + remainder. The asymptote is the quotient only (y = x − 2), not y = x − 2 + 3/(x−1). The leftover fraction is precisely the part that vanishes.
4. Binomial signs with (a − b)ⁿ. Keep b = −2 inside the powers; odd k's come out negative. Alternating signs are your checksum.
5. Using the wrong form and grinding. Asked for a vertex, students expand factored form and complete the square — when the zeros' midpoint gives it in one step. Fix: before computing, ask "which form shows this directly?"

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Practice Problems

12. (FRQ-style) 🚫 Let g(x) = (3x² − 7x + 5)/(x − 2). (i) Use polynomial long division to write g(x) in the form q(x) + r/(x − 2). (ii) State the equation of the slant asymptote and write a limit statement expressing why the graph approaches it. (iii) Does the graph of g cross its slant asymptote? Justify.

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FRQ Practice — Task Model: Symbolic Manipulations (FRQ 3 style) 🚫 No-Calc

(a) The function f is given by f(x) = (x³ − 2x² − 5x + 6)/(x − 2). (i) Rewrite f(x) in the form q(x) + r/(x − 2), where q is a polynomial and r is a constant. (ii) Verify your answer by evaluating both forms at x = 0.

(b) (i) Using your result, describe the end behavior of f with limit notation. (ii) Explain why the graph of f resembles the graph of y = q(x) for large |x|.

(c) Expand (x − 2)⁴ completely using the binomial theorem.

Model Response & Rubric (6 points)

(a) [2 pts] (i) [1 pt for quotient, 1 pt for remainder] Divide with all terms present:

x³ − 2x² − 5x + 6 ÷ (x − 2):
x³ − 2x²          ← x²(x − 2); nothing remains in x² column: bring down −5x + 6
−5x + 10          ← −5(x − 2); remainder: 6 − 10 = −4

Quotient x² − 5, remainder −4: f(x) = x² − 5 − 4/(x − 2). (ii) Original: f(0) = 6/(−2) = −3. Rewritten: 0 − 5 − 4/(−2) = −5 + 2 = −3 ✓ (the verification is part of the quotient/remainder points — an inconsistent check signals an error to hunt down)

(b) [2 pts] (i) [1 pt] As x → ±∞, f behaves like x² − 5, so lim x→−∞ f(x) = ∞ and lim x→∞ f(x) = ∞. (ii) [1 pt] The gap between f(x) and q(x) = x² − 5 is −4/(x − 2), which approaches 0 as |x| grows — so the graphs become indistinguishable for large |x|.

(c) [2 pts] Row 4 of Pascal's triangle: 1, 4, 6, 4, 1, with b = −2:

(x − 2)⁴ = x⁴ + 4x³(−2) + 6x²(4) + 4x(−8) + 16
         = x⁴ − 8x³ + 24x² − 32x + 16

[1 pt for correct coefficients/structure, 1 pt for correct alternating signs]

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Show answer key & explanations

(g) Answer Key

1. (A). x² + 5x + 9 = (x + 2)(x + 3) + 3 → remainder 3. (Or remainder theorem: p(−2) = 4 − 10 + 9 = 3.) (C) is the constant term of the dividend; (D) would mean x + 2 divides evenly.

2. (B). Division: x² − 3x + 5 = (x − 1)(x − 2) + 3 → quotient x − 2. (D) mistakes the divisor for the asymptote; (A) subtracts the wrong constant.

3. (C). Factored form displays each zero as a factor. (A) shows end behavior/y-intercept; (D) shows the vertex.

4. (D). C(5,2)·2² = 10·4 = 40. (C) forgets 2²; (B) uses 2³; (A) halves it.

5. (B). Exactly one degree of top-heaviness → linear quotient → slant line. (A) gives horizontal; (C) gives y = 0; (D) is irrelevant by itself (e.g., (x³+1)/(x−1) is top-heavy by 2 — parabolic end behavior).

6. (A). Quotient from the worked division in section (b): 2x + 1 (remainder −7x + 1 vanishes at ∞). (B) sign slip; (C) drops the constant; (D) halves the slope.

7. (B). g(0) = (−4)(1)(−2) = 8. (A) is the sign-error twin; the exam banks on it.

8. (C). Substitute x = 3: p(3) = 0·q(3) + 7 = 7 — the remainder theorem in one line. (D) forgets the zero factor.

9. (D). C(4,2) = 6 (middle of row 1, 4, 6, 4, 1). (A) is C(4,1); (C)/(B) are miscounts.

10. (A). Complete the square: x² − 6x + 4 = (x² − 6x + 9) − 9 + 4 = (x − 3)² − 5. (B) flips the shift direction; (C)/(D) botch the constant.

11. (C). (x² + 1)/x = x + 1/x; the graph meets y = x where 1/x = 0 — never. (A) is doubly wrong (x = 0 isn't even in the domain).

12. (FRQ-style, 6 points) (i) [2 pts] 3x² − 7x + 5 ÷ (x − 2): 3x(x − 2) = 3x² − 6x, leaving −x + 5; −1(x − 2) = −x + 2, leaving 3. So g(x) = 3x − 1 + 3/(x − 2). Check x = 0: LHS 5/(−2) = −2.5; RHS −1 + 3/(−2) = −2.5 ✓ (ii) [2 pts] Slant asymptote y = 3x − 1, because lim x→±∞ [g(x) − (3x − 1)] = lim x→±∞ 3/(x − 2) = 0 — the vertical gap between graph and line vanishes in both directions. (iii) [2 pts] Crossing requires 3/(x − 2) = 0, which has no solution (numerator is the nonzero constant 3). The graph never crosses its slant asymptote.

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🎯 Exam tip: FRQ 3 (Symbolic Manipulations) is the most practice-sensitive question on the exam — every part is a computation you either execute cleanly or don't. Build the habit trio now: placeholder zeros, explicit sign-flips when subtracting, and the x = 0 double-check. Students who check catch their own slips; students who don't donate the points.