Final Exam — Cumulative (Weeks 1–15) · Objectives 1–8

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Scope: Cumulative — all eight objectives, Weeks 1–15 (simplifying with real-number properties, order of operations & exponent rules · linear equations & inequalities, incl. absolute value · functions: notation, domain & range, operations · linear functions, graphs & systems · polynomials & factoring · quadratics: equations, functions & graphs · rational & radical expressions/equations · exponential & logarithmic functions).
Format: 20 items, 100 points (5 each) · application-skewed · auto-gradable item types only (multiple-choice, multiple-answer, matching).
Points: 100 · Assignment group: Final (30% of the course grade) · Window: opens at the start of Module 16 (finals week); due Fri Dec 18, 11:59 p.m. · No AI on the Final.

This is the human-readable exam with its vetted answer key and one-line feedback. The import-ready Classic QTI 1.2 is in L-final-week-16-qti.xml (generated by a validated Python script — parses with 20 items, every single-answer item exactly one correct). The item-bank / coverage note and the Canvas placement block are at the bottom of this file.

This is the live exam. Its paired ungraded rehearsal — O-practice-final-week-16.md — mirrors this blueprint with fresh variants and shares none of these items.

No tables or calculators are needed for these items — every value is engineered to land on a clean integer or simple expression by hand. (Students may use a basic calculator and Desmos in the room per the syllabus; nothing here requires them.)

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Blueprint (items → objective)

Coverage is proportional to teaching time and weighted toward the function/quadratic/rational-radical/exponential-log material: Obj 1 = 2 · Obj 2 = 2 · Obj 3 = 3 · Obj 4 = 3 · Obj 5 = 2 · Obj 6 = 3 · Obj 7 = 3 · Obj 8 = 2. No trick questions; all arithmetic is pre-computed; every single-answer item has exactly one correct option.

#	Type	Concept	Objective	Week
1	Multiple choice	Simplify a product of powers (exponent rules)	1	1
2	Multiple choice	Order of operations with a squared negative	1	1
3	Multiple choice	Solve a linear equation	2	2
4	Multiple choice	Solve an absolute-value inequality	2	2
5	Multiple choice	Evaluate a function	3	3
6	Multiple choice	Domain of a rational function	3	3
7	Multiple choice	Composition of functions (f∘g)	3	3
8	Multiple choice	Slope through two points	4	4
9	Multiple choice	Write a line in slope-intercept form	4	4
10	Multiple choice	Solve a 2×2 system	4	5
11	Multiple choice	Factor a trinomial	5	6
12	Matching	Expand / factor (special products & GCF)	5	6
13	Multiple choice	Solve a quadratic by factoring	6	7
14	Multiple choice	Vertex from vertex form	6	9
15	Multiple choice	Discriminant → number of real solutions	6	7
16	Multiple choice	Simplify a rational expression	7	11
17	Multiple choice	Evaluate a rational exponent	7	12
18	Multiple choice	Solve a radical equation	7	12
19	Multiple choice	Evaluate a logarithm	8	14
20	Multiple answer	Select all true log/exponential statements	8	14

Objective totals: Obj 1 = 2 (10 pts) · Obj 2 = 2 (10 pts) · Obj 3 = 3 (15 pts) · Obj 4 = 3 (15 pts) · Obj 5 = 2 (10 pts) · Obj 6 = 3 (15 pts) · Obj 7 = 3 (15 pts) · Obj 8 = 2 (10 pts) → 20 items, 100 points.

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Questions, key, and feedback

Objective 1 — Simplify with Properties, Order of Operations & Exponents (Week 1)

Q1 (MC). Simplify (3x⁴)(2x⁻²), writing the result with a positive exponent.
- A. 6x⁻⁸
- B. 6x² ✅
- C. 5x²
- D. 6x⁶
Feedback: Multiply the coefficients (3 · 2 = 6) and add the exponents (4 + (−2) = 2): 6x². (A multiplies the exponents — the power-of-a-power rule, which doesn't apply here; C adds the coefficients.)

Q2 (MC). Evaluate −4² + (−2)³.
- A. 8
- B. 24
- C. −24 ✅
- D. −8
Feedback: −4² = −16 (the exponent binds tighter than the sign, so only the 4 is squared), and (−2)³ = −8; −16 + (−8) = −24. (If you read −4² as (−4)² = +16 you'd get +8 — the classic Week-1 trap.)

Objective 2 — Linear Equations & Inequalities (Week 2)

Q3 (MC). Solve 5x − 3 = 2x + 9.
- A. x = 2
- B. x = 6
- C. x = 4 ✅
- D. x = −4
Feedback: Subtract 2x from both sides → 3x − 3 = 9; add 3 → 3x = 12; divide → x = 4. Check: 5(4) − 3 = 17 = 2(4) + 9. ✓

Q4 (MC). Solve the inequality |x − 2| < 5. Which interval is the solution set?
- A. x < 7
- B. −3 < x < 7 ✅
- C. x < −3 or x > 7
- D. −7 < x < 3
Feedback: |expr| < c means −c < expr < c: −5 < x − 2 < 5; add 2 throughout → −3 < x < 7. (C is the pattern for |expr| > c; "less than" gives a single "and" interval, not an "or".)

Objective 3 — Functions: Notation, Domain & Operations (Week 3)

Q5 (MC). For f(x) = x² + 1, find f(−2).
- A. −3
- B. 5 ✅
- C. 3
- D. −5
Feedback: Substitute −2 for x: (−2)² + 1 = 4 + 1 = 5. (Note (−2)² = +4 — the parentheses square the whole −2; reading it as −(2²) = −4 gives the wrong −3.)

Q6 (MC). What is the domain of f(x) = (x + 1)/(x − 3)?
- A. All real numbers
- B. All real numbers except x = −1
- C. All real numbers except x = 3 ✅
- D. All real numbers except x = −3
Feedback: You can't divide by zero, so exclude where the denominator is 0: x − 3 = 0 → x = 3. The domain is all reals except x = 3. (B uses the numerator's zero, which is allowed.)

Q7 (MC). Let f(x) = x − 1 and g(x) = 2x. Find (f∘g)(3).
- A. 3
- B. 6
- C. 5 ✅
- D. 4
Feedback: (f∘g)(3) = f(g(3)). Inside first: g(3) = 2(3) = 6; then f(6) = 6 − 1 = 5. (Composition is nesting, not multiplying — do g first, then f.)

Objective 4 — Linear Functions, Graphs & Systems (Weeks 4–5)

Q8 (MC). Find the slope of the line through (2, −1) and (5, 8).
- A. −3
- B. 3 ✅
- C. 1/3
- D. 9
Feedback: m = (y₂ − y₁)/(x₂ − x₁) = (8 − (−1))/(5 − 2) = 9/3 = 3. (Watch the double negative in the numerator; C flips rise and run.)

Q9 (MC). Write the equation, in slope-intercept form, of the line with slope 2 that passes through (0, −3).
- A. y = −3x + 2
- B. y = 2x − 3 ✅
- C. y = 2x + 3
- D. y = −2x − 3
Feedback: Slope-intercept is y = mx + b; the slope is 2 and the point (0, −3) is the y-intercept, so b = −3 → y = 2x − 3. (A swaps the slope and intercept.)

Q10 (MC). Solve the system 2x + y = 5 and x − y = 1. The solution (x, y) is —
- A. (1, 2)
- B. (3, 1)
- C. (2, 1) ✅
- D. (2, −1)
Feedback: Add the equations (the y-terms cancel): 3x = 6 → x = 2; substitute → 2 + y = 5 → y = 1. Solution (2, 1); check x − y = 2 − 1 = 1. ✓ (A reverses the coordinates.)

Objective 5 — Polynomials & Factoring (Week 6)

Q11 (MC). Factor completely: x² − 5x − 14.
- A. (x − 2)(x + 7)
- B. (x − 7)(x + 2) ✅
- C. (x − 7)(x − 2)
- D. (x + 7)(x + 2)
Feedback: Find two numbers that multiply to −14 and add to −5: that's −7 and +2, so x² − 5x − 14 = (x − 7)(x + 2). Check (FOIL): x² + 2x − 7x − 14 = x² − 5x − 14. ✓ (A's factors add to +5, not −5.)

Q12 (Matching). Match each expression on the left to its equivalent form on the right (expand or factor as needed).
| Expression | Equivalent form |
|---|---|
| (x − 4)(x + 4) | x² − 16 |
| (x + 3)² | x² + 6x + 9 |
| x² − 5x − 14 | (x − 7)(x + 2) |
| 6x³ − 9x² | 3x²(2x − 3) |
Feedback: (x − 4)(x + 4) is a difference of squares → x² − 16 (no middle term). (x + 3)² is a perfect square → x² + 6x + 9 (the middle term is 2·3·x). x² − 5x − 14 factors to (x − 7)(x + 2). 6x³ − 9x² has GCF 3x² → 3x²(2x − 3).

Objective 6 — Quadratics: Equations, Functions & Graphs (Weeks 7, 9)

Q13 (MC). Solve x² − 4x − 5 = 0.
- A. x = 4, 5
- B. x = −5, 1
- C. x = 5, −1 ✅
- D. x = 5, 1
Feedback: Factor: x² − 4x − 5 = (x − 5)(x + 1) = 0; by the zero-product property, x − 5 = 0 or x + 1 = 0 → x = 5 or x = −1. (D drops the negative on the second root.)

Q14 (MC). The graph of f(x) = (x − 1)² − 4 is a parabola. What is its vertex?
- A. (−1, −4)
- B. (−1, 4)
- C. (1, 4)
- D. (1, −4) ✅
Feedback: In vertex form a(x − h)² + k the vertex is (h, k); here x − 1 gives h = +1 (the form already has the minus), and k = −4, so the vertex is (1, −4). (A flips the sign of h — the most common vertex error.)

Q15 (MC). Use the discriminant to determine the number of real solutions of x² + 2x + 5 = 0.
- A. Two real solutions
- B. One real solution
- C. No real solutions ✅
- D. Cannot be determined
Feedback: The discriminant is b² − 4ac = 2² − 4(1)(5) = 4 − 20 = −16. A negative discriminant means no real solutions (the parabola never crosses the x-axis). (A positive value would give two; zero would give one.)

Objective 7 — Rational & Radical Expressions and Equations (Weeks 11–12)

Q16 (MC). Simplify (x² − 9)/(x − 3) (for x ≠ 3).
- A. x − 3
- B. x + 3 ✅
- C. x² − 3
- D. −3
Feedback: Factor the numerator as a difference of squares: (x − 3)(x + 3)/(x − 3); cancel the common factor (x − 3) → x + 3 (with x ≠ 3). (A cancels the wrong factor; you can only cancel common factors, never loose terms.)

Q17 (MC). Evaluate 27^(2/3).
- A. 18
- B. 9 ✅
- C. 3
- D. 729
Feedback: A rational exponent means root then power: 27^(2/3) = (³√27)² = 3² = 9. (A multiplies 27 by 2/3; C stops at the cube root without squaring; D squares first and forgets the root.)

Q18 (MC). Solve √(x + 1) = 4.
- A. x = 3
- B. x = 16
- C. x = 15 ✅
- D. x = 7
Feedback: Square both sides: x + 1 = 16 → x = 15. Check: √(15 + 1) = √16 = 4 ✓ (not extraneous). (B forgets to subtract 1; A squares the 4 incorrectly.)

Objective 8 — Exponential & Logarithmic Functions (Week 14)

Q19 (MC). Evaluate log₂16.
- A. 8
- B. 4 ✅
- C. 2
- D. 32
Feedback: log₂16 asks "2 to what power equals 16?" — since 2⁴ = 16, the answer is 4. (A divides 16 by 2; the log is the exponent, not the quotient.)

Q20 (Multiple answer — select all that apply). Select all statements that are true.
- A. log₂8 = 3 ✅
- B. log₅1 = 0 ✅
- C. 2³ = 6
- D. ln e = 1 ✅
- E. log₁₀100 = 10
Feedback: A is true (2³ = 8). B is true (anything to the 0 power is 1, so log₅1 = 0). C is false — 2³ = 8, not 6. D is true (ln e = log_e e = 1). E is false — log₁₀100 = 2 (10² = 100), not 10. The three true statements are A, B, and D.

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Answer key (quick reference)

Q	Answer	Q	Answer
1	B (6x²)	11	B ((x − 7)(x + 2))
2	C (−24)	12	(x−4)(x+4)→x²−16 / (x+3)²→x²+6x+9 / x²−5x−14→(x−7)(x+2) / 6x³−9x²→3x²(2x−3)
3	C (x = 4)	13	C (x = 5, −1)
4	B (−3 < x < 7)	14	D (1, −4)
5	B (5)	15	C (no real solutions)
6	C (except x = 3)	16	B (x + 3)
7	C (5)	17	B (9)
8	B (3)	18	C (x = 15)
9	B (y = 2x − 3)	19	B (4)
10	C (2, 1)	20	A, B, D

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Quality gate (H5 — self-checked, computer-verified)

• Structure: 20 items, 5 points each, 100 points total; coverage Obj 1 = 2 · Obj 2 = 2 · Obj 3 = 3 · Obj 4 = 3 · Obj 5 = 2 · Obj 6 = 3 · Obj 7 = 3 · Obj 8 = 2 matches the blueprint exactly, weighted toward the function/quadratic/rational-radical/exp-log material.
• Single-answer integrity: every multiple-choice item (Q1–Q11, Q13–Q19) has exactly one correct option; the matching item (Q12) pairs all four rows one-to-one; the multiple-answer item keys Q20 → A, B, D (C and E left unselected).
• Arithmetic pre-computed and independently re-verified (Python verify_w16.py, sympy): Q1 (3·2)x^(4+(−2)) = 6x²; Q2 −4² = −16, (−2)³ = −8, sum −24; Q3 5x−3=2x+9 → x = 4; Q4 |x−2|=5 → x = −3, 7 (interior interval −3 < x < 7); Q5 (−2)²+1 = 5; Q6 x−3 = 0 → exclude 3; Q7 f(g(3)) = f(6) = 5; Q8 (8−(−1))/(5−2) = 3; Q9 slope 2, intercept −3; Q10 add equations → x = 2, y = 1; Q11/Q12 factor & expand checks (x²−5x−14 = (x−7)(x+2); (x−4)(x+4) = x²−16; (x+3)² = x²+6x+9; 6x³−9x² = 3x²(2x−3)); Q13 (x−5)(x+1) = 0 → x = 5, −1; Q14 vertex (1, −4); Q15 disc 4−20 = −16 → no real roots; Q16 (x²−9)/(x−3) = x+3; Q17 27^(2/3) = 9; Q18 √(x+1) = 4 → x = 15 (checks); Q19 log₂16 = 4; Q20 log₂8 = 3, log₅1 = 0, ln e = 1 (true), 2³ = 8 ≠ 6 and log₁₀100 = 2 ≠ 10 (false). All checks PASS (0 failures).
• Clean values by design: every item resolves to an integer, a simple ordered pair, or a one-term expression — no calculator or table is required, and every distractor encodes a specific, named error (exponent-rule mix-up, the −4² sign trap, the |x| > c vs |x| < c swap, the vertex sign flip, cancelling loose terms, root-vs-power confusion, the dropped second root).
• QTI parse confirmation: L-final-week-16-qti.xml parses as imsqti_xmlv1p2 with 20 items; every single-answer respcondition sets SCORE = 100 on exactly one option; the matching item's four partial-credit blocks add to 100; the multiple-answer item awards 100 only for the exact correct-set selection (A, B, D).
• Integrity vs. the practice final: 0 items are shared with O-practice-final-week-16.md (verified by full stem-plus-options comparison; the maximum overlap is a same-concept slot with different numbers and contexts).
• No content outside the Weeks 1–15 course definitions; no hallucinated facts.

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Item-bank & coverage note

All 20 items are fresh variants assembled from the Week 1–15 item banks per Prompt L (changed numbers and contexts to reduce answer-sharing with the weekly quizzes and the midterm), tagged course=MATH120 · exam=final · weeks=1–15 · objectives=1–8 and deposited back into the banks for future per-term ($39) regenerations:

Objective	Drawn from banks	Items
1	Week 1 (Real Numbers, Exponents & Expressions)	Q1–Q2
2	Week 2 (Linear Equations & Inequalities)	Q3–Q4
3	Week 3 (Functions: Notation, Domain & Range)	Q5–Q7
4	Weeks 4–5 (Linear Functions & Graphs; Systems)	Q8–Q10
5	Week 6 (Polynomials & Factoring)	Q11–Q12
6	Weeks 7, 9 (Quadratic Equations; Quadratic Functions & Graphs)	Q13–Q15
7	Weeks 11–12 (Rational Expressions; Radicals & Rational Exponents)	Q16–Q18
8	Week 14 (Logarithmic Functions)	Q19–Q20

Each term's update regenerates fresh final variants from these same banks; the paired practice final is regenerated alongside and continues to share none of the live items.

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Final Exam — Cumulative (Weeks 1–15)"
assignment_group          = "Final"
points_possible           = 100
grading_type              = points
available_from_offset_days = 0        # opens at the start of Module 16 (finals week)
due_offset_days           = 4        # 4 days after module start (Fri Dec 18)
published                 = true
allowed_attempts          = 1
shuffle_answers           = true
provenance                = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"