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Week 8 · Midterm study guide

Midterm Exam — Study Guide

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Coverage — what the midterm covers

The midterm is cumulative over Weeks 1–7. It opens in the Week 8 module and closes at the due time your instructor sets. There is no time limit, and you get one attempt. The questions appear one at a time, and you cannot go back once you move on.

Format: 25 questions, and every question counts the same. Eighteen ask you to pick one option, three ask you to judge a statement true or false, two ask you to type a short result such as a whole number, and two ask you to pick every option that applies. Scratch paper and a basic calculator are allowed, and every question gives you what it needs.

What is in scope, with the number of midterm questions on each course objective:

Midterm coverage by course objective: the weeks that teach it, its main skills, and its number of questions out of 25
Course objective Weeks Main skills Questions
Real numbers, exponents, and expressions 1, used again in 3 and 6 The real-number sets; the order of operations; the exponent rules; distributing and evaluating 4
Linear equations and inequalities 2 Solving and checking; identities and contradictions; the flip rule; interval notation; absolute value 3
Functions 3, used again in 4 What a function is; function notation; domain and range; operations on functions; composition 5
Lines and systems 4–5 Slope; the three forms of a line; intercepts; parallel and perpendicular lines; solving and classifying systems 6
Polynomials and factoring 6 Vocabulary; adding, subtracting, and multiplying; special products; factoring 3
Quadratic equations 7 Standard form; factoring; square roots of both sides; completing the square; the quadratic formula; the discriminant 4

Lines and systems carry the most questions, and functions are close behind, so put the most review time on Weeks 3 to 5. The sixth objective also names quadratic functions and their graphs. That part starts in Week 9 and is not on the midterm.

Topic-by-topic review

Each topic gives the key idea in plain words, the course's own definitions, a course example to rehearse, the mistakes students make most often, and the cure for each. The deeper review for every topic is that week's chapter, lecture page, slide deck, and readings.

Week 1 — Real numbers, exponents, and expressions

  • Key idea: algebra is a short list of rules that never change, and every result you get can be checked.
  • The number sets: a rational number is a number that can be written as a quotient of two integers with a nonzero denominator; its decimal form either stops or repeats. An irrational number is a real number that cannot be written that way; its decimal goes on forever without repeating, as for √2 and π. The sets nest: N inside Z inside Q inside R.
  • Order of operations: grouping symbols first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. A negated power applies the exponent to the number first and the minus sign afterward.
  • Exponent rules: product rule aᵐ · aⁿ = aᵐ⁺ⁿ (add the exponents); quotient rule aᵐ ⁄ aⁿ = aᵐ⁻ⁿ; power rule (aᵐ)ⁿ = aᵐⁿ (multiply the exponents). Then power of a product (ab)ⁿ = aⁿbⁿ; zero exponent a⁰ = 1; negative exponent a⁻ⁿ = 1 ⁄ aⁿ. A negative exponent calls for a reciprocal; it never makes the value itself negative.
  • Course examples to rehearse: 2 + 3 · 4² = 2 + 3 · 16 = 2 + 48 = 50. Next, −4² = −(4²) = −16, but (−4)² = (−4)(−4) = 16. Then 5x − 2(3x − 4) = 5x − 6x + 8 = −x + 8. Finally 3(−2)² − 2(−2) + 5 = 3 · 4 + 4 + 5 = 21.
  • Predictable mistakes → cures: (1) reading −2⁴ as 16. Cure: ask whether the minus sits inside the parentheses; here it does not, so −2⁴ = −16. (2) reading x² · x³ as x⁶. Cure: multiplying powers adds exponents (x⁵); only a power of a power multiplies them. (3) reading 2⁻² as a negative number. Cure: it is the reciprocal 1 ⁄ 4. (4) reading 3x² as (3x)². Cure: the coefficient is applied after the exponent, so only x is squared.

Week 2 — Linear equations and inequalities

  • Key idea: one principle, do the same thing to both sides, and four tactics: distribute, clear fractions with the LCD, collect, and divide. Every solve ends with a check by substitution.
  • Definitions: a solution is a value of the variable that makes a statement true. An identity collapses to a true statement such as 0 = 0, so its solution set is all real numbers. A contradiction collapses to a false statement such as 3 = 7, so it has no solution.
  • The flip rule: when both sides of an inequality are multiplied or divided by a negative number, the direction of the sign reverses.
  • Interval notation: a square bracket for an included endpoint, a parenthesis for an excluded one, and always a parenthesis beside ∞. Two rays are joined with ∪.
  • Absolute value: the distance of a number from 0, so it is never negative. The split rule turns |X| = a, with a positive, into X = a or X = −a. Less than gives the bounded middle (an AND); greater than gives two outward rays (an OR).
  • Course examples to rehearse: 3(x − 2) = 2x + 5 gives x = 11, and both sides check at 27. Then −2x + 1 < 9 gives −2x < 8, so x > −4, written (−4, ∞). Next, |2x − 1| = 7 gives x = 4 or x = −3. Finally |x| < 5 is (−5, 5), and |x| ≥ 3 is (−∞, −3] ∪ [3, ∞).
  • Predictable mistakes → cures: (1) solving −3x > 9 as x > −3. Cure: circle the number you divide by; if it is negative, flip first (x < −3). (2) putting a bracket on infinity. Cure: infinity is not a number, so it always takes a parenthesis. (3) "solving" |x| = −5 as x = ±5. Cure: a distance is never negative, so there is no solution. (4) writing |x| > 3 as −3 < x < 3. Cure: greater than means farther from 0, so it is two rays.

Week 3 — Functions

  • Key idea: a function is a rule that pairs each input with exactly one output. Two different inputs may share an output, but no input may have two outputs. The vertical line test checks a graph.
  • Function notation: f(x) is one output value, read "f of x"; it is not f times x. Wrap every negative input in parentheses, and substitute a whole algebraic input wherever x appears.
  • Domain and range: the domain is every input that gives a real-number output. A polynomial takes all real numbers. A rational function leaves out the zeros of its denominator. A square root needs its radicand, the expression under the sign, to be zero or more. The range is every output the function actually produces.
  • Operations and composition: add, subtract, multiply, or divide two functions' outputs at the same input; a quotient leaves out every input where the bottom function is 0. Composition f(g(x)) applies the inner function g first and the outer function f second.
  • Course examples to rehearse: for f(x) = 3x² − 2x + 5, f(−2) = 21, f(0) = 5, and f(3) = 26. For g(x) = 3x + 4, g(a + 2) = 3a + 6 + 4 = 3a + 10. The domain of (x + 1) ⁄ (x − 3) leaves out x = 3, and the domain of √(x − 5) is [5, ∞). With p(x) = x² and q(x) = x + 1, (p ∘ q)(2) = 9 but (q ∘ p)(2) = 5, and the product p(2) · q(2) = 12 is a third number.
  • Predictable mistakes → cures: (1) evaluating 3(−2)² as 3(−4). Cure: the square of −2 is +4. (2) writing g(a + 2) = 3a + 2 + 4. Cure: the 3 multiplies every term of the input. (3) writing the domain of √(x − 5) as x > 5. Cure: √0 = 0 is allowed, so 5 is included. (4) reading (f ∘ g)(x) as a product, or applying f first. Cure: inside first, and composition is not multiplication.

Week 4 — Linear functions and their graphs

  • Key idea: equal steps in a table make a straight line, and the slope measures the step.
  • Definitions: slope is rise over run, m = (y₂ − y₁) ⁄ (x₂ − x₁), with the y-values on top; it is also a rate of change with units. Slope-intercept form is y = mx + b, where b is the y-intercept, the point (0, b). Point-slope form is y − y₁ = m(x − x₁). Standard form is Ax + By = C.
  • Special lines: a horizontal line has slope 0. A vertical line's slope is undefined, since its run is 0.
  • Comparing lines: parallel lines have equal slopes and different y-intercepts. Perpendicular lines have slopes whose product is −1; each slope is the negative reciprocal of the other.
  • Course examples to rehearse: through (0, 4) and (2, 10), m = 6 ⁄ 2 = 3. Through (−1, 4) and (3, −4), m = −8 ⁄ 4 = −2, and point-slope form leads to y = −2x + 2. The line 2x + 3y = 12 has intercepts (0, 4) and (6, 0) and slope −2⁄3. The slopes 2⁄3 and −3⁄2 multiply to −1.
  • Predictable mistakes → cures: (1) swapping m and b in y = −2x + 5. Cure: the slope multiplies x; the y-intercept stands alone. (2) computing run over rise. Cure: y-values on top. (3) giving −2⁄3 or 3⁄2 as perpendicular to 2⁄3. Cure: flip the fraction and change the sign, then check the product is −1. (4) trusting a picture. Cure: slopes 2⁄3 and −4⁄3 multiply to −8⁄9, so those lines are not perpendicular, however they look.

Week 5 — Systems of linear equations and inequalities

  • Key idea: a solution of a system is an ordered pair that makes every equation true at once; on a graph it is where the lines meet.
  • Methods: the substitution method replaces a variable with the expression another equation gives for it. The elimination method adds or subtracts the equations, after multiplying one if needed, so a variable cancels. Back-substitution then finds the other variable.
  • Classifying: a system that collapses to a false statement such as 0 = 4 is inconsistent: no solution, parallel lines. One that collapses to a true statement such as 0 = 0 is dependent: every point on the one shared line works, so it has infinitely many. Lines that cross once give exactly one solution.
  • Inequality systems: a boundary line is dashed for a strict inequality and solid when equality is allowed. A test point decides which side to shade, and the solutions are the overlap.
  • Course examples to rehearse: y = 2x with x + y = 9 gives (3, 6). Next, x + y = 10 with x − y = 4 gives (7, 3). Then 2x + 3y = 12 with x − y = 1 gives (3, 2), once you triple x − y = 1. Finally, x + y = 3 with x + y = 7 gives 0 = 4, and 2x + y = 5 with 4x + 2y = 10 gives 0 = 0.
  • Predictable mistakes → cures: (1) calling 0 = 0 "no solution" and 0 = 4 "infinitely many." Cure: a true statement means every point on the line works; a false one means none does. (2) adding equations that should be subtracted. Cure: subtract when the matching terms have the same sign. (3) checking a pair in only one equation. Cure: a pair must work in both, so check both every time.

Week 6 — Polynomials and factoring

  • Key idea: you build a polynomial by multiplying, and you take it apart by factoring. Each move undoes the other, so you can always check one with the other.
  • Vocabulary: the degree is the largest exponent on the variable. The leading coefficient multiplies the highest-power term, and the constant term has no variable. A monomial has one term, a binomial two, and a trinomial three.
  • Multiplying: every term of one factor multiplies every term of the other; FOIL tracks the four products of two binomials. The special products are (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and (a + b)(a − b) = a² − b².
  • Factoring: take out the greatest common factor first. For x² + bx + c, find two numbers whose product is c and whose sum is b. For a leading coefficient other than 1, the ac method splits the middle term and groups. A difference of squares factors as (a + b)(a − b). A sum of squares such as x² + 16 is a prime polynomial over the real numbers.
  • Course examples to rehearse: (x − 1)(x − 5) = x² − 6x + 5, and both forms give −3 at x = 2. Next, (x + 6)² = x² + 12x + 36. Then (x² − 6x + 5) − (x² − 2x − 15) = −4x + 20. Finally 2x² + 7x + 3 = (2x + 1)(x + 3), and 9x² − 25 = (3x + 5)(3x − 5).
  • Predictable mistakes → cures: (1) writing (a + b)² as a² + b². Cure: the middle term 2ab is always there; (x + 6)² has 12x. (2) "factoring" x² + 16 as (x + 4)(x + 4) or (x + 4)(x − 4). Cure: expand each; neither gives x² + 16. (3) applying the minus to only the first term when subtracting. Cure: the minus reaches every term of the second polynomial. (4) skipping the GCF. Cure: always look for a common factor first.

Week 7 — Quadratic equations

  • Key idea: when the variable is squared, move every term to one side so the equation reads ax² + bx + c = 0, then choose a method.
  • Definitions: the zero-product property says that if a product of factors equals 0, at least one factor equals 0. The square root property says that if a square equals a nonnegative number k, the quantity squared is the positive or the negative square root of k. Completing the square adds the square of half the x-coefficient to both sides.
  • The formula and the discriminant: x = (−b ± √(b² − 4ac)) ⁄ (2a). The discriminant b² − 4ac predicts the outcome. Positive means two real solutions, 0 means one repeated solution, and negative means no real solution.
  • Course examples to rehearse: x² − 6x + 5 = 0 three ways. Factoring gives (x − 1)(x − 5) = 0. Completing the square gives (x − 3)² = 4. The formula gives (6 ± 4) ⁄ 2. Every way, x = 1 or x = 5. Also, 3x² − 12x = 0 gives 3x(x − 4) = 0, so x = 0 or x = 4. Next, x² − 4x + 1 = 0 gives x = 2 ± √3, about 3.732 and 0.268. Finally, 2x² − 4x + 2 = 0 has discriminant 0, and x² + 2x + 5 = 0 has discriminant −16.
  • Predictable mistakes → cures: (1) dividing by x and losing x = 0. Cure: factor out the GCF instead. (2) writing −b as −4 when b = −4. Cure: −(−4) = 4. (3) dividing only the radical by 2a. Cure: the whole top is divided. (4) giving a real number for a negative discriminant. Cure: no real number squares to a negative, so write "no real solution."

Representative practice

These are fresh variants, not the midterm's questions. Work each worked example yourself first, then compare your steps. Every number below is computed and checked in advance.

Worked example — signs, exponents, and evaluating (Week 1)

A puzzle in a campus newsletter asks readers to tidy three expressions.

First: simplify −5² + (−2)⁴.

  1. The exponent binds before the leading minus, so −5² = −(5²) = −25.
  2. The parentheses make −2 the base, and an even power of a negative base is positive: (−2)⁴ = 16.
  3. Add: −25 + 16 = −9.

Why it matters: reading −5² as (−5)² = 25 gives 25 + 16 = 41, a different number entirely.

Second: simplify (x⁴ · x²) ⁄ x⁹ and write it without a negative exponent.

  1. Product rule on top: x⁴ · x² = x⁴⁺² = x⁶.
  2. Quotient rule: x⁶ ⁄ x⁹ = x⁶⁻⁹ = x⁻³.
  3. Negative exponent: x⁻³ = 1 ⁄ x³.
  4. Check at x = 2: the top is 16 · 4 = 64, the bottom is 2⁹ = 512, and 64 ⁄ 512 = 1 ⁄ 8. Also 1 ⁄ 2³ = 1 ⁄ 8.

Third: evaluate 2x² − 3x − 4 at x = −2, then simplify 3n − 5(n − 2).

  1. Wrap the input: 2(−2)² − 3(−2) − 4 = 2 · 4 + 6 − 4 = 10.
  2. Distribute the −5 to both terms: 3n − 5n + 10 = −2n + 10.
  3. Check at n = 4: the original gives 12 − 5 · 2 = 2, and −2 · 4 + 10 = 2.

Why it matters: writing 2(−2)² as 2(−4) gives −8 + 6 − 4 = −6. Distributing the −5 to only the first term gives −2n − 10. Both slips change the result.

Worked example — a flip and a distance (Week 2)

A phone starts at 90% battery and loses 6 percentage points for each hour of video. You want at least 18% left for the trip home. How many hours h can you watch?

  1. Write the inequality: 90 − 6h ≥ 18.
  2. Subtract 90 from both sides: −6h ≥ −72.
  3. Divide both sides by −6 and flip the sign: h ≤ 12.
  4. Check the endpoint: 90 − 6 · 12 = 18, and 18 ≥ 18 is true, so 12 is included. Check h = 13: 90 − 78 = 12, and 12 ≥ 18 is false.

In words: you can watch up to 12 hours. Hours cannot be negative, so in this setting h runs from 0 to 12, written [0, 12].

Now the distance. A thermostat holds a room within 3 degrees of 68. The temperature t satisfies |t − 68| ≤ 3.

  1. Less than or equal to means the bounded middle: −3 ≤ t − 68 ≤ 3.
  2. Add 68 to all three parts: 65 ≤ t ≤ 71, written [65, 71].
  3. Check t = 70: |70 − 68| = 2, and 2 ≤ 3 is true. Check t = 72: |72 − 68| = 4, and 4 ≤ 3 is false.

And one equation by the split rule: |4x − 2| = 10 splits into 4x − 2 = 10 or 4x − 2 = −10. So 4x = 12, x = 3, or 4x = −8, x = −2. Check: |12 − 2| = 10 and |−8 − 2| = |−10| = 10.

Worked example — a function at work (Week 3)

A campus bake sale spends C(x) = 2x + 30 dollars to bake x dozen cookies: 2 dollars a dozen plus a 30-dollar table fee. It takes in R(x) = 8x dollars.

  1. Difference of functions: (R − C)(x) = 8x − (2x + 30) = 8x − 2x − 30 = 6x − 30. The minus reaches both terms of C.
  2. Evaluate: at 10 dozen, (R − C)(10) = 6 · 10 − 30 = 30. Check: R(10) = 80 and C(10) = 50, and 80 − 50 = 30.
  3. Read it: 6x − 30 = 0 at x = 30 ⁄ 6 = 5, so the sale covers its costs at 5 dozen and earns money after that.
  4. Algebraic input: C(a + 1) = 2(a + 1) + 30 = 2a + 2 + 30 = 2a + 32.
  5. Practical domain: if the oven allows at most 20 dozen, the inputs that make sense are the whole numbers 0 to 20. The costs then run from C(0) = 30 to C(20) = 70 dollars.
  6. A domain from a formula: a square-root rule such as √(2x − 8) needs 2x − 8 ≥ 0, so x ≥ 4, and its domain is [4, ∞).

Why it matters: writing (R − C)(x) as 8x − 2x + 30 gives 6x + 30, which says the sale is ahead by 30 dollars before it sells a single cookie.

Worked example — a line from two readings (Week 4)

A candle is 12 cm tall after 1 hour of burning and 6 cm tall after 4 hours. It burns at a steady rate.

  1. Slope: m = (6 − 12) ⁄ (4 − 1) = −6 ⁄ 3 = −2. The candle loses 2 cm per hour.
  2. Point-slope form: y − 12 = −2(x − 1). Distribute: y − 12 = −2x + 2. Add 12: y = −2x + 14.
  3. Check both readings: −2 · 1 + 14 = 12 and −2 · 4 + 14 = 6.
  4. Intercepts: the y-intercept is (0, 14), so the candle started 14 cm tall. Setting y = 0 gives 2x = 14, so x = 7: the candle burns out after 7 hours.
  5. Standard form: add 2x to both sides to get 2x + y = 14.
  6. Related lines: any line parallel to this one has slope −2. A perpendicular line has slope 1⁄2, because −2 · (1⁄2) = −1.

Why it matters: run over rise gives (4 − 1) ⁄ (6 − 12) = −1⁄2, which would claim the candle shrinks only half a centimeter per hour.

Worked example — a system from a market stall (Week 5)

A farmers market stall sells berries in pints at 4 dollars and in quarts at 7 dollars. One morning it sells 45 containers and takes in 243 dollars. With x pints and y quarts:

x + y = 45 and 4x + 7y = 243

  1. Multiply first: multiply the first equation by 4 to get 4x + 4y = 180.
  2. Eliminate: subtract it from the second equation. The 4x terms cancel, and 7y − 4y = 3y, while 243 − 180 = 63. So 3y = 63 and y = 21.
  3. Back-substitute: x + 21 = 45, so x = 24.
  4. Check both equations: 24 + 21 = 45, and 4 · 24 + 7 · 21 = 96 + 147 = 243.

In words: the stall sold 24 pints and 21 quarts. The two lines cross at one point, so the system is consistent and independent.

Worked example — expanding and factoring (Week 6)

A square courtyard x meters on a side gives up a 7-meter strip along two of its edges, leaving a square x − 7 meters on a side.

  1. Square of a difference: (x − 7)² = x² − 2(x)(7) + 7² = x² − 14x + 49 square meters.
  2. Check at x = 10: (10 − 7)² = 9, and 100 − 140 + 49 = 9.
  3. The trap: x² + 49 at x = 10 gives 149, nowhere near 9. The middle term matters.

Then three factorings, each checked by expanding:

  • Trinomial, leading coefficient 1: x² − x − 20 needs product −20 and sum −1. The numbers −5 and 4 work, so x² − x − 20 = (x − 5)(x + 4).
  • The ac method: 3x² + 14x + 8 has ac = 3 · 8 = 24. The numbers 12 and 2 multiply to 24 and add to 14. Split and group: 3x² + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (3x + 2)(x + 4).
  • GCF, then a difference of squares: 5x² − 45 = 5(x² − 9) = 5(x − 3)(x + 3).

And one subtraction: (2x² + 5x − 1) − (x² − 3x + 4) = 2x² + 5x − 1 − x² + 3x − 4 = x² + 8x − 5.

Worked example — one quadratic, three methods (Week 7)

Solve x² − 2x − 15 = 0.

  1. Factoring: two numbers with product −15 and sum −2 are −5 and 3, so (x − 5)(x + 3) = 0. Then x = 5 or x = −3.
  2. Completing the square: x² − 2x = 15. Half of −2 is −1, and (−1)² = 1. Add 1 to both sides: (x − 1)² = 16, so x − 1 = ±4, and x = 5 or x = −3.
  3. The formula: a = 1, b = −2, c = −15. The discriminant is (−2)² − 4(1)(−15) = 4 + 60 = 64, and √64 = 8. So x = (2 ± 8) ⁄ 2, which is 5 or −3.
  4. Check: 25 − 10 − 15 = 0 and 9 + 6 − 15 = 0.

Now a setting. A community garden plot is 4 meters longer than it is wide, and its area is 45 square meters. With width w, w(w + 4) = 45, so w² + 4w − 45 = 0. The numbers 9 and −5 multiply to −45 and add to 4, so (w + 9)(w − 5) = 0. A width cannot be negative, so w = 5 meters and the length is 9 meters. Check: 5 × 9 = 45.

Self-check questions

Try each one, then read the check and the one-line why.

  1. A math club flyer lists √49, √10, 0.75, −2, and 2π. Which are irrational? Check: √10 and 2π. Why: √49 = 7 is a natural number, and 0.75 = 3⁄4 is rational.
  2. Compute 6⁰ + 2⁻³. Check: 1 + 1⁄8 = 9⁄8. Why: a zero exponent gives 1, and a negative exponent gives a reciprocal.
  3. Solve 4(x + 3) = 7x − 6. Check: x = 6. Why: 4x + 12 = 7x − 6 gives 18 = 3x, and both sides equal 36 at x = 6.
  4. Write |x − 3| > 2 in interval notation. Check: (−∞, 1) ∪ (5, ∞). Why: greater than means two outward rays: x − 3 < −2 or x − 3 > 2.
  5. Find the domain of √(3x + 6). Check: [−2, ∞). Why: the radicand must be zero or more, and 3x + 6 ≥ 0 gives x ≥ −2.
  6. With u(x) = x − 4 and w(x) = x², find (w ∘ u)(1) and (u ∘ w)(1). Check: 9 and −3. Why: inside first, so w(u(1)) = w(−3) = 9, while u(w(1)) = u(1) = −3.
  7. Find the slope and intercepts of 5x − 2y = 10. Check: slope 5⁄2; intercepts (2, 0) and (0, −5). Why: set each variable to 0 in turn, then take rise over run between the two points.
  8. Classify the system x − 2y = 4 and 3x − 6y = 5. Check: inconsistent, no solution. Why: tripling the first gives 3x − 6y = 12, and taking away 3x − 6y = 5 leaves 0 = 7, which is false.
  9. Does x² + 25 factor over the real numbers? Check: no; it is prime. Why: (x + 5)² = x² + 10x + 25 and (x + 5)(x − 5) = x² − 25, and neither is x² + 25.
  10. How many real solutions does x² − 2x + 6 = 0 have? Check: none. Why: the discriminant is 4 − 24 = −20, and it is negative.

Check yourself — choose one. Write two or three sentences to a friend.

  • (a) Explain what the candle line's y-intercept of 14 and its x-intercept of 7 each mean in the story.
  • (b) Explain why the garden's second solution, w = −9, is rejected even though it solves the equation.

Either one rehearses an explain-why skill the midterm checks.

Watch before the exam

These three videos come from this week's Readings & Resources page, and all three have captions. Use them to review and to connect weeks, not to learn anything new.

Study plan

A short countdown, counted back from the day you plan to sit the midterm. Week 8's chapter and its habit of choosing a method are new, so leave room for them too.

  • As soon as Week 8 opens: take the practice exam once, cold and closed-note. Note which objectives cost you the most.
  • About four days before you sit the midterm: reread the topic reviews above for your weakest two objectives. Redo the matching worked examples from scratch, without looking.
  • About two days before: run the paired exam-prep tutorial (the copy-and-paste chatbot tutor) for active recall. Tell it which areas the practice exam exposed. The midterm runs harder than the practice form, so ask the tutor for its harder drills.
  • The day before: retake the practice exam, redo every miss by hand, check every result by substitution, and work the self-check questions above out loud.
  • Right before: skim the "predictable mistakes → cures" lines for all seven weeks. Then stop studying.

This guide is for review. The exam-prep tutorial and the practice exam are where you rehearse. Retrieval practice, not rereading, is what moves scores.

Exam format

Every one of the 25 questions counts the same, and the midterm checks your own work without AI. The questions come one at a time, and there is no going back, so settle each one before you move on.

Test-taking strategies for this material:

  • Name the move before you start. Is it simplify, solve, evaluate, find a line, solve a system, factor, or solve a quadratic? The cue usually tells you in the first line.
  • Wrap every negative input in parentheses, and ask whether a minus sign sits inside or outside a power.
  • Circle the number you divide by in an inequality. If it is negative, flip the sign before you write anything else.
  • Read a, b, and c with their signs before you use the quadratic formula or the discriminant.
  • Check by substitution. Put each solution back into the original equation, and expand each factored form back out.
  • Keep exact forms. Leave a radical as a radical, such as 2 ± √3, and use a decimal only as a check.
  • Use scratch paper for every computation, even a short one. Writing each step catches most dropped signs.

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