Midterm Exam — Cumulative (Weeks 1–7) · Objectives 1–6

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Scope: Cumulative — Weeks 1–7, Objectives 1–6 (real numbers & exponents · linear equations & inequalities · functions: notation, domain & operations · linear functions, graphs & systems · polynomials & factoring · quadratic equations).
Format: 20 items, 100 points (5 each) · application-skewed · auto-gradable item types only (multiple-choice, multiple-answer, matching).
Points: 100 · Assignment group: Midterm (20% of the course grade) · Window: opens at the start of Module 8; due 6 days later. No AI on the midterm (course AI policy).

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-midterm-week-08-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-exam-week-08.md — mirrors this blueprint with fresh variants and shares none of these items.

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Blueprint (items → objective)

Coverage is proportional to teaching time: Obj 1 ≈ 3 · Obj 2 ≈ 3 · Obj 3 ≈ 3 · Obj 4 ≈ 4 · Obj 5 ≈ 3 · Obj 6 ≈ 4. No trick questions; all arithmetic is pre-computed and independently re-verified; every single-answer item has exactly one correct option; distractors target the named Week 1–7 misconceptions.

#	Type	Concept	Objective	Week
1	Multiple choice	Order of operations with negatives & exponents	1	1
2	Multiple choice	Power of a product (exponent rules)	1	1
3	Multiple choice	Simplify by distributing (incl. a negative)	1	1
4	Multiple choice	Solve a linear equation	2	2
5	Multiple choice	Solve a linear inequality (flip on ÷ by negative)	2	2
6	Multiple choice	Absolute-value equation (two cases)	2	2
7	Multiple choice	Evaluate a function f(−3)	3	3
8	Multiple choice	Domain of a radical function	3	3
9	Multiple choice	Composition of functions (f ∘ g)(2)	3	3
10	Multiple choice	Slope through two points	4	4
11	Multiple choice	Equation of a line (point-slope)	4	4
12	Multiple choice	Perpendicular slope (negative reciprocal)	4	4
13	Multiple choice	Solve a linear system (elimination)	4	5
14	Multiple choice	Multiply binomials	5	6
15	Multiple choice	Square a binomial (special product)	5	6
16	Matching	Expression ↔ factored form	5	6
17	Multiple choice	Solve a quadratic by factoring	6	7
18	Multiple choice	Quadratic with a repeated root	6	7
19	Multiple choice	Discriminant & no real solutions	6	7
20	Multiple answer	Select all solutions of a quadratic	6	7

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

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Questions, key, and feedback

Objective 1 — Real Numbers, Exponents & Expressions (Week 1)

Q1 (MC). Evaluate: −2³ + (−3)²
- A. 1 ✅
- B. 17
- C. −17
- D. −1
Feedback: The minus in −2³ is outside the cube: −2³ = −(2³) = −8. And (−3)² = 9. So −8 + 9 = 1. (B treats it as 8 + 9; C as −8 − 9; D as 8 − 9.)

Q2 (MC). Simplify: (2x²y)³
- A. 8x⁶y³ ✅
- B. 2x⁶y³
- C. 6x⁶y³
- D. 8x⁵y³
Feedback: Raise every factor to the 3rd: 2³ = 8, (x²)³ = x⁶, y³ = y³ → 8x⁶y³. (B forgets to cube the coefficient; C multiplies 2·3 instead of 2³; D adds exponents 2+3 instead of multiplying.)

Q3 (MC). Simplify: 3(2x − 4) − 2(x − 5)
- A. 4x − 2 ✅
- B. 4x − 22
- C. 6
- D. 4x + 8
Feedback: 3(2x − 4) = 6x − 12; −2(x − 5) = −2x + 10 (two negatives → +10). Combine: 6x − 12 − 2x + 10 = 4x − 2. (B writes −10 instead of +10 — the distribute-the-negative trap; C/D mis-distribute.)

Objective 2 — Linear Equations & Inequalities (Week 2)

Q4 (MC). Solve for x: 4(x − 1) = 2x + 6
- A. x = 5 ✅
- B. x = 1
- C. x = −5
- D. x = 11
Feedback: 4x − 4 = 2x + 6 → 2x = 10 → x = 5. Check: 4(5 − 1) = 16 and 2(5) + 6 = 16 ✓. (B/C/D are sign or division slips.)

Q5 (MC). Solve and give the solution set: −3x + 2 ≤ 11
- A. x ≥ −3 ✅
- B. x ≤ −3
- C. x ≥ 3
- D. x ≤ 3
Feedback: −3x ≤ 9; divide by −3 and flip the inequality: x ≥ −3. (B keeps the direction — the classic flip error; C/D drop the sign on −3.) Check x = 0: −3(0) + 2 = 2 ≤ 11 ✓ and 0 ≥ −3 ✓.

Q6 (MC). Solve: |2x − 3| = 7
- A. x = 5 or x = −2 ✅
- B. x = 5 only
- C. x = 2 or x = −5
- D. x = 5 or x = 2
Feedback: Split into two cases: 2x − 3 = 7 → x = 5; and 2x − 3 = −7 → x = −2. (B drops the second case; C/D mishandle the negative case.) Both check: |2(5) − 3| = 7 ✓, |2(−2) − 3| = |−7| = 7 ✓.

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

Q7 (MC). If f(x) = x² − 2x, find f(−3).
- A. 15 ✅
- B. 3
- C. −15
- D. −3
Feedback: f(−3) = (−3)² − 2(−3) = 9 − (−6) = 9 + 6 = 15. (B computes 9 − 6; C/D drop a sign on one term.) The −2(−3) becomes +6 — mind the double negative.

Q8 (MC). What is the domain of f(x) = √(x − 4)?
- A. x ≥ 4, i.e. [4, ∞) ✅
- B. x > 4, i.e. (4, ∞)
- C. x ≤ 4, i.e. (−∞, 4]
- D. all real numbers
Feedback: A square root needs its inside ≥ 0: x − 4 ≥ 0 → x ≥ 4, written [4, ∞). (B excludes x = 4, but √0 = 0 is defined; C reverses the inequality; D ignores the radical.)

Q9 (MC). Let f(x) = 2x + 1 and g(x) = x². Find (f ∘ g)(2).
- A. 9 ✅
- B. 25
- C. 20
- D. 5
Feedback: Composition is inside-out: do g first — g(2) = 4 — then f(4) = 2(4) + 1 = 9. (B is (g ∘ f)(2) = g(5) = 25, the wrong order; C multiplies f(2)·g(2); D is just f(2).)

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

Q10 (MC). Find the slope of the line through (−1, 2) and (3, 10).
- A. 2 ✅
- B. 1/2
- C. −2
- D. 4
Feedback: slope = (y₂ − y₁)/(x₂ − x₁) = (10 − 2)/(3 − (−1)) = 8/4 = 2. (B inverts run/rise; C flips a sign; D forgets the denominator.)

Q11 (MC). Write the equation of the line with slope −2 through (1, 5).
- A. y = −2x + 7 ✅
- B. y = −2x + 5
- C. y = −2x + 3
- D. y = 2x + 7
Feedback: Point-slope: y − 5 = −2(x − 1) → y = −2x + 2 + 5 = −2x + 7. Check (1, 5): −2(1) + 7 = 5 ✓. (B uses the y-value as the intercept without shifting; C sign-slips; D uses +2 for the slope.)

Q12 (MC). A line has equation y = (1/4)x. What is the slope of any line perpendicular to it?
- A. −4 ✅
- B. 4
- C. −1/4
- D. 1/4
Feedback: Perpendicular slope is the negative reciprocal: flip 1/4 to 4, change the sign → −4. (B forgets the sign; C negates without flipping; D is the original slope.)

Q13 (MC). Solve the system: x + y = 7 and x − y = 1.
- A. (4, 3) ✅
- B. (3, 4)
- C. (4, −3)
- D. (7, 1)
Feedback: Add the equations (the y's cancel): 2x = 8 → x = 4; then 4 + y = 7 → y = 3 → (4, 3). (B swaps x and y; C sign-slips on y; D just lists the constants.)

Objective 5 — Polynomials & Factoring (Week 6)

Q14 (MC). Multiply: (x − 5)(x + 2)
- A. x² − 3x − 10 ✅
- B. x² + 3x − 10
- C. x² − 3x + 10
- D. x² − 7x − 10
Feedback: FOIL: x² + 2x − 5x − 10 = x² − 3x − 10. (B gets the middle sign wrong; C makes the constant +10; D adds the outer/inner wrong.)

Q15 (MC). Expand: (2x + 3)²
- A. 4x² + 12x + 9 ✅
- B. 4x² + 9
- C. 4x² + 6x + 9
- D. 2x² + 12x + 9
Feedback: (a + b)² = a² + 2ab + b²: (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. (B drops the middle term — the #1 special-product error; C uses 6x not 12x; D forgets to square the 2.)

Q16 (Matching). Match each expression to its correct factored form.
| Expression | Correct factored form |
|---|---|
| x² − 9 | (x − 3)(x + 3) |
| x² + 6x + 9 | (x + 3)² |
| x² − x − 6 | (x − 3)(x + 2) |
| x² − 4x | x(x − 4) |
Feedback: x² − 9 is a difference of squares → (x − 3)(x + 3); x² + 6x + 9 is a perfect square → (x + 3)²; x² − x − 6 → two numbers multiplying to −6 and adding to −1 are −3 and +2; x² − 4x has a common factor x → x(x − 4). Each expands back to its expression.

Objective 6 — Quadratic Equations (Week 7)

Q17 (MC). Solve: x² + 2x − 8 = 0
- A. x = −4 or x = 2 ✅
- B. x = 4 or x = −2
- C. x = −8 or x = 1
- D. x = 8 or x = −1
Feedback: Factor: two numbers multiplying to −8 and adding to +2 are +4 and −2 → (x + 4)(x − 2) = 0 → x = −4 or x = 2. (B reverses both signs; C/D use the constant and middle coefficient directly.)

Q18 (MC). Solve: x² − 6x + 9 = 0
- A. x = 3 (a repeated root) ✅
- B. x = 3 or x = −3
- C. x = −3 (a repeated root)
- D. x = 6 or x = −6
Feedback: x² − 6x + 9 = (x − 3)² = 0 → x = 3, a repeated root (discriminant 36 − 36 = 0). (B treats it as a difference of squares; C gets the sign wrong; D mis-factors.)

Q19 (MC). Compute the discriminant of 2x² + 3x + 5 = 0 and state what it means.
- A. Discriminant = −31; no real solutions ✅
- B. Discriminant = 49; two real solutions
- C. Discriminant = 31; two real solutions
- D. Discriminant = 0; one repeated solution
Feedback: b² − 4ac = 3² − 4(2)(5) = 9 − 40 = −31. A negative discriminant means no real solutions. (B adds instead of subtracting; C drops the sign; D misreads it as zero.)

Q20 (Multiple answer — select all that apply). Select all values that are solutions of x² − x − 6 = 0.
- A. x = 3 ✅
- B. x = −2 ✅
- C. x = 2
- D. x = 6
- E. x = −3
Feedback: Factor: (x − 3)(x + 2) = 0 → x = 3 and x = −2. (C, D, E are not roots: e.g., 2² − 2 − 6 = −4 ≠ 0.) Key A and B; leave C, D, E unselected.

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Answer key (quick reference)

Q	Answer	Q	Answer
1	A (1)	11	A (y = −2x + 7)
2	A (8x⁶y³)	12	A (−4)
3	A (4x − 2)	13	A (4, 3)
4	A (x = 5)	14	A (x² − 3x − 10)
5	A (x ≥ −3)	15	A (4x² + 12x + 9)
6	A (x = 5 or −2)	16	x²−9→(x−3)(x+3) / x²+6x+9→(x+3)² / x²−x−6→(x−3)(x+2) / x²−4x→x(x−4)
7	A (15)	17	A (x = −4 or 2)
8	A ([4, ∞))	18	A (x = 3, repeated)
9	A (9)	19	A (−31; no real solutions)
10	A (2)	20	A and B (3 and −2)

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Quality gate (self-checked, computer-verified)

• Structure: 20 items, 5 points each, 100 points total; coverage Obj 1 = 3 · Obj 2 = 3 · Obj 3 = 3 · Obj 4 = 4 · Obj 5 = 3 · Obj 6 = 4 matches the blueprint exactly.
• Single-answer integrity: every multiple-choice item (Q1–Q15, Q17–Q19) has exactly one correct option; the matching item (Q16) pairs all four expressions one-to-one; the multiple-answer item (Q20) keys A and B (and requires C, D, E to be left unselected).
• Arithmetic pre-computed and independently re-verified (Python w08_verify.py, sympy): Q1 −8 + 9 = 1; Q2 2³ = 8 → 8x⁶y³; Q3 6x − 12 − 2x + 10 = 4x − 2; Q4 x = 5; Q5 boundary −3x + 2 = 11 → x = −3, flip → x ≥ −3; Q6 |2x − 3| = 7 → {5, −2}; Q7 9 + 6 = 15; Q8 x − 4 ≥ 0 → [4, ∞); Q9 f(g(2)) = f(4) = 9; Q10 8/4 = 2; Q11 y = −2x + 7 (passes (1, 5)); Q12 −1 ÷ (1/4) = −4; Q13 (4, 3); Q14 x² − 3x − 10; Q15 (2x + 3)² = 4x² + 12x + 9; Q16 all four factorings expand back; Q17 {−4, 2}; Q18 (x − 3)² → x = 3, disc = 0; Q19 9 − 40 = −31; Q20 {3, −2}, with 2/6/−3 confirmed non-roots. All checks PASS (0 failures).
• QTI parse confirmation: L-midterm-week-08-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 sum to 100.
• Integrity vs. the practice exam: 0 items are shared with O-practice-exam-week-08.md (verified by full stem-plus-options comparison; any same-concept slot uses different numbers/contexts).
• No content outside the Weeks 1–7 course definitions; no hallucinated facts. No AI permitted on the midterm.

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Item-bank & coverage note

All 20 items are fresh variants assembled from the Week 1–7 item banks per Prompt L (changed numbers and contexts to reduce answer-sharing with the weekly quizzes), tagged course=MATH120 · exam=midterm · weeks=1–7 · objectives=1–6 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–Q3
2	Week 2 (Linear Equations & Inequalities)	Q4–Q6
3	Week 3 (Functions: Notation, Domain & Operations)	Q7–Q9
4	Weeks 4–5 (Linear Functions, Graphs & Systems)	Q10–Q13
5	Week 6 (Polynomials & Factoring)	Q14–Q16
6	Week 7 (Quadratic Equations)	Q17–Q20

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

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Midterm Exam — Cumulative (Weeks 1–7)"
assignment_group          = "Midterm"
points_possible           = 100
grading_type              = points
available_from_offset_days = 0        # opens at the start of Module 8
due_offset_days           = 6        # 6 days after module start
published                 = true
allowed_attempts          = 1
shuffle_answers           = true
provenance                = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"