Midterm Practice Exam (ungraded) · Weeks 1–7 (Objectives 1–6)

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
What this is: a low-stakes rehearsal for the cumulative midterm. It mirrors the real exam's blueprint — same coverage, item-type mix, length, and application-skewed difficulty — but is built from fresh item-bank variants and shares none of the live midterm's questions.
Settings: ungraded (0 points) · unlimited attempts · feedback shown after submission · opens before the exam window so you can prepare. (Practice with whatever you like; the real midterm allows no AI.)

This is the human-readable practice exam with its vetted answer key and feedback (released after submission). The import-ready Classic QTI 1.2 is in O-practice-exam-week-08-qti.xml (generated by a validated Python script — parses with 20 items). The Canvas placement block is at the bottom.

Integrity note for students. Every item here is a fresh variant — new numbers and contexts — with a pre-computed, vetted answer. None of these are the live midterm questions. Working them builds the skill the midterm tests, honestly. The paired live exam is L-midterm-week-08.md.

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Blueprint (mirrors the midterm)

Coverage is proportional to teaching time, matching the real exam: Obj 1 ≈ 3 · Obj 2 ≈ 3 · Obj 3 ≈ 3 · Obj 4 ≈ 4 · Obj 5 ≈ 3 · Obj 6 ≈ 4. (The actual midterm items are not listed here — only the shared structure.)

#	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 g(−4)	3	3
8	Multiple choice	Domain of a radical function	3	3
9	Multiple choice	Composition of functions (f ∘ g)(3)	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 · Obj 2 = 3 · Obj 3 = 3 · Obj 4 = 4 · Obj 5 = 3 · Obj 6 = 4 → 20 items (ungraded; mirrors the 100-point midterm's emphasis).

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Questions, key, and feedback (feedback releases after you submit)

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

Q1 (MC). Evaluate: −3² + (−2)³
- A. −17 ✅
- B. 1
- C. −1
- D. 13
Feedback: −3² = −(3²) = −9 (the minus is outside the square); (−2)³ = −8. So −9 + (−8) = −17. (B treats it as 9 − 8; C as −9 + 8; D mis-signs.)

Q2 (MC). Simplify: (3x³y²)²
- A. 9x⁶y⁴ ✅
- B. 6x⁶y⁴
- C. 9x⁵y⁴
- D. 9x⁶y²
Feedback: Raise every factor to the 2nd: 3² = 9, (x³)² = x⁶, (y²)² = y⁴ → 9x⁶y⁴. (B multiplies 3·2 instead of 3²; C adds exponents 3+2; D forgets to square y².)

Q3 (MC). Simplify: 4(x − 3) − 3(2x − 1)
- A. −2x − 9 ✅
- B. −2x − 15
- C. 2x − 9
- D. −2x − 11
Feedback: 4(x − 3) = 4x − 12; −3(2x − 1) = −6x + 3. Combine: 4x − 12 − 6x + 3 = −2x − 9. (B writes −3 instead of +3 — the distribute-the-negative trap; C/D mis-combine.)

Objective 2 — Linear Equations & Inequalities (Week 2)

Q4 (MC). Solve for x: 5(x + 2) = 3x − 4
- A. x = −7 ✅
- B. x = 7
- C. x = −3
- D. x = 3
Feedback: 5x + 10 = 3x − 4 → 2x = −14 → x = −7. Check: 5(−7 + 2) = −25 and 3(−7) − 4 = −25 ✓. (B drops a sign; C/D are division slips.)

Q5 (MC). Solve and give the solution set: −2x + 5 < 9
- A. x > −2 ✅
- B. x < −2
- C. x > 2
- D. x < 2
Feedback: −2x < 4; divide by −2 and flip: x > −2. (B keeps the direction — the flip error; C/D drop the sign.) Check x = 0: −2(0) + 5 = 5 < 9 ✓ and 0 > −2 ✓.

Q6 (MC). Solve: |3x + 1| = 10
- A. x = 3 or x = −11/3 ✅
- B. x = 3 only
- C. x = 3 or x = 11/3
- D. x = −3 or x = 11/3
Feedback: Two cases: 3x + 1 = 10 → x = 3; and 3x + 1 = −10 → 3x = −11 → x = −11/3. (B drops the second case; C/D mishandle the negative case.) Both check: |3(3) + 1| = 10 ✓, |3(−11/3) + 1| = |−10| = 10 ✓.

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

Q7 (MC). If g(x) = x² + 3x, find g(−4).
- A. 4 ✅
- B. −28
- C. 28
- D. −4
Feedback: g(−4) = (−4)² + 3(−4) = 16 + (−12) = 4. (B treats it as −16 − 12; C/D drop or flip a sign.)

Q8 (MC). What is the domain of f(x) = √(2x − 6)?
- A. x ≥ 3, i.e. [3, ∞) ✅
- B. x > 3, i.e. (3, ∞)
- C. x ≤ 3, i.e. (−∞, 3]
- D. x ≥ 6, i.e. [6, ∞)
Feedback: Need the inside ≥ 0: 2x − 6 ≥ 0 → 2x ≥ 6 → x ≥ 3, i.e. [3, ∞). (B excludes x = 3, but √0 is defined; C reverses; D forgets to divide by 2.)

Q9 (MC). Let f(x) = 3x − 2 and g(x) = x². Find (f ∘ g)(3).
- A. 25 ✅
- B. 49
- C. 7
- D. 16
Feedback: Inside-out: g(3) = 9, then f(9) = 3(9) − 2 = 25. (B is (g ∘ f)(3) = g(7) = 49, wrong order; C is just f(3); D is just g(3) − wait, that's 9 — D is a slip toward 4².)

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

Q10 (MC). Find the slope of the line through (2, −1) and (5, 8).
- A. 3 ✅
- B. 1/3
- C. −3
- D. 9
Feedback: slope = (8 − (−1))/(5 − 2) = 9/3 = 3. (B inverts run/rise; C flips a sign; D forgets the denominator.)

Q11 (MC). Write the equation of the line with slope 3 through (2, 1).
- A. y = 3x − 5 ✅
- B. y = 3x + 1
- C. y = 3x − 6
- D. y = −3x − 5
Feedback: Point-slope: y − 1 = 3(x − 2) → y = 3x − 6 + 1 = 3x − 5. Check (2, 1): 3(2) − 5 = 1 ✓. (B uses the y-value as the intercept without shifting; C forgets the +1; D flips the slope.)

Q12 (MC). A line has equation y = (2/3)x. What is the slope of any line perpendicular to it?
- A. −3/2 ✅
- B. 3/2
- C. −2/3
- D. 2/3
Feedback: Negative reciprocal of 2/3: flip to 3/2, change the sign → −3/2. (B forgets the sign; C negates without flipping; D is the original slope.)

Q13 (MC). Solve the system: 2x + y = 8 and x − y = 1.
- A. (3, 2) ✅
- B. (2, 3)
- C. (3, −2)
- D. (1, 6)
Feedback: Add the equations (the y's cancel): 3x = 9 → x = 3; then 3 − y = 1 → y = 2 → (3, 2). (B swaps x and y; C sign-slips on y; D mis-solves.)

Objective 5 — Polynomials & Factoring (Week 6)

Q14 (MC). Multiply: (x − 4)(x + 3)
- A. x² − x − 12 ✅
- B. x² + x − 12
- C. x² − x + 12
- D. x² − 7x − 12
Feedback: FOIL: x² + 3x − 4x − 12 = x² − x − 12. (B gets the middle sign wrong; C makes the constant +12; D adds outer/inner wrong.)

Q15 (MC). Expand: (3x − 2)²
- A. 9x² − 12x + 4 ✅
- B. 9x² + 4
- C. 9x² − 6x + 4
- D. 6x² − 12x + 4
Feedback: (a − b)² = a² − 2ab + b²: (3x)² − 2(3x)(2) + 2² = 9x² − 12x + 4. (B drops the middle term; C uses 6x not 12x; D forgets to square the 3.)

Q16 (Matching). Match each expression to its correct factored form.
| Expression | Correct factored form |
|---|---|
| x² − 25 | (x − 5)(x + 5) |
| x² + 8x + 16 | (x + 4)² |
| x² + x − 6 | (x + 3)(x − 2) |
| x² − 9x | x(x − 9) |
Feedback: x² − 25 is a difference of squares → (x − 5)(x + 5); x² + 8x + 16 is a perfect square → (x + 4)²; x² + x − 6 → +3 and −2 multiply to −6 and add to +1; x² − 9x has a common factor x → x(x − 9). Each expands back to its expression.

Objective 6 — Quadratic Equations (Week 7)

Q17 (MC). Solve: x² − 2x − 15 = 0
- A. x = 5 or x = −3 ✅
- B. x = −5 or x = 3
- C. x = 15 or x = −1
- D. x = 5 or x = 3
Feedback: Two numbers multiplying to −15 and adding to −2 are −5 and +3 → (x − 5)(x + 3) = 0 → x = 5 or x = −3. (B reverses the signs; C/D mis-factor.)

Q18 (MC). Solve: x² + 10x + 25 = 0
- A. x = −5 (a repeated root) ✅
- B. x = 5 (a repeated root)
- C. x = 5 or x = −5
- D. x = −10 or x = −25
Feedback: x² + 10x + 25 = (x + 5)² = 0 → x = −5, a repeated root (discriminant 100 − 100 = 0). (B gets the sign wrong; C treats it as a difference of squares; D mis-factors.)

Q19 (MC). Compute the discriminant of 3x² + 2x + 4 = 0 and state what it means.
- A. Discriminant = −44; no real solutions ✅
- B. Discriminant = 52; two real solutions
- C. Discriminant = 44; two real solutions
- D. Discriminant = 0; one repeated solution
Feedback: b² − 4ac = 2² − 4(3)(4) = 4 − 48 = −44. Negative → 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 − 12 = 0.
- A. x = 3 ✅
- B. x = −4 ✅
- C. x = 4
- D. x = −3
- E. x = 12
Feedback: Factor: (x + 4)(x − 3) = 0 → x = 3 and x = −4. (C, D, E are not roots: e.g., 4² + 4 − 12 = 8 ≠ 0.) Key A and B; leave C, D, E unselected.

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Answer key (quick reference)

Q	Answer	Q	Answer
1	A (−17)	11	A (y = 3x − 5)
2	A (9x⁶y⁴)	12	A (−3/2)
3	A (−2x − 9)	13	A (3, 2)
4	A (x = −7)	14	A (x² − x − 12)
5	A (x > −2)	15	A (9x² − 12x + 4)
6	A (x = 3 or −11/3)	16	x²−25→(x−5)(x+5) / x²+8x+16→(x+4)² / x²+x−6→(x+3)(x−2) / x²−9x→x(x−9)
7	A (4)	17	A (x = 5 or −3)
8	A ([3, ∞))	18	A (x = −5, repeated)
9	A (25)	19	A (−44; no real solutions)
10	A (3)	20	A and B (3 and −4)

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Quality gate (self-checked, computer-verified)

• Mirror check: 20 items, coverage Obj 1 = 3 · Obj 2 = 3 · Obj 3 = 3 · Obj 4 = 4 · Obj 5 = 3 · Obj 6 = 4 — matches the midterm blueprint's emphasis and item-type mix.
• Single-answer integrity: every multiple-choice item (Q1–Q15, Q17–Q19) has exactly one correct option; the matching item (Q16) pairs all four one-to-one; the multiple-answer item (Q20) keys A and B (C, D, E left unselected).
• Arithmetic pre-computed and independently re-verified (Python w08_verify.py, sympy): Q1 −9 + (−8) = −17; Q2 3² = 9 → 9x⁶y⁴; Q3 4x − 12 − 6x + 3 = −2x − 9; Q4 x = −7; Q5 boundary −2x + 5 = 9 → x = −2, flip → x > −2; Q6 |3x + 1| = 10 → {3, −11/3}; Q7 16 + (−12) = 4; Q8 2x − 6 ≥ 0 → [3, ∞); Q9 f(g(3)) = f(9) = 25; Q10 9/3 = 3; Q11 y = 3x − 5 (passes (2, 1)); Q12 −1 ÷ (2/3) = −3/2; Q13 (3, 2); Q14 x² − x − 12; Q15 (3x − 2)² = 9x² − 12x + 4; Q16 all four factorings expand back; Q17 {5, −3}; Q18 (x + 5)² → x = −5, disc = 0; Q19 4 − 48 = −44; Q20 {3, −4}, with 4/−3/12 confirmed non-roots. All checks PASS (0 failures).
• QTI parse confirmation: O-practice-exam-week-08-qti.xml parses as imsqti_xmlv1p2 with 20 items; every single-answer respcondition sets SCORE = 100 on exactly one option. (In Canvas the placement makes it ungraded with feedback on; the engine still scores attempts so students see what they missed.)
• Integrity vs. the live exam: 0 items are shared with L-midterm-week-08.md — verified by full stem-plus-options comparison. Where a concept slot overlaps the midterm, this form uses different numbers and contexts (e.g., midterm Q1 is −2³ + (−3)²; here Q1 is −3² + (−2)³; midterm Q17 solves x² + 2x − 8 = 0 → {−4, 2}, here Q17 solves x² − 2x − 15 = 0 → {5, −3}).
• No content outside the Weeks 1–7 course definitions; no hallucinated facts.

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Item-bank & coverage note

All 20 items are fresh variants assembled from the Week 1–7 item banks per Prompt O, preferring items not used on the live midterm and authoring fresh variants where a concept overlaps. Tagged course=MATH120 · form=practice-midterm · weeks=1–7 · objectives=1–6 and deposited back into the banks for future per-term ($39) regenerations. Each term's update regenerates fresh practice variants alongside the midterm and continues to share none of the live items.

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Midterm Practice Exam (ungraded)"
assignment_group          = "Practice exercises"
points_possible           = 0
grading_type              = not_graded
allowed_attempts          = unlimited
show_feedback             = true        # released after submission
available_from_offset_days = -3        # opens 3 days before the exam window
due_offset_days           = 6         # on or before the exam due date
published                 = true
shuffle_answers           = true
provenance                = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"