📚 Concept Review
1
Polynomials & Factor Theorem
Unit 1
A polynomial p(x) has factor (x − c) if and only if p(c) = 0. Dividing p(x) by (x − c) gives zero remainder when (x − c) is a factor.
Factor Theorem: (x − c) is a factor of p(x) ⟺ p(c) = 0
Remainder Theorem: p(x) ÷ (x − c) → remainder = p(c)
Synthetic Division: use coefficients of p(x), divisor value = c
Remainder Theorem: p(x) ÷ (x − c) → remainder = p(c)
Synthetic Division: use coefficients of p(x), divisor value = c
📋 Memorize
If p(c) = 0, then (x − c) is a factor. The remainder when dividing by (x − c) equals p(c).
✍ Example
Is (x − 2) a factor of p(x) = x³ − 3x² + x + 2?
p(2) = 8 − 12 + 2 + 2 = 0 ✓ → Yes, (x − 2) is a factor.
2
Rational Expressions
Unit 2
A rational expression is a fraction where numerator and denominator are polynomials. Simplify by factoring and canceling common factors. Note excluded values (where denominator = 0).
Simplify: factor top and bottom, cancel common factors
Multiply: (a/b)·(c/d) = ac/bd, then simplify
Divide: (a/b)÷(c/d) = (a/b)·(d/c)
Multiply: (a/b)·(c/d) = ac/bd, then simplify
Divide: (a/b)÷(c/d) = (a/b)·(d/c)
📋 Memorize
Always factor FIRST before canceling. State excluded values (x ≠ values that make denominator = 0).
✍ Example
Simplify: (x² − 4) / (x + 2)
= (x+2)(x−2)/(x+2) = x − 2, where x ≠ −2
3
Complex Numbers & Radicals
Unit 3
Complex numbers have the form a + bi, where i = √(−1). Powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1.
i¹ = i i² = −1 i³ = −i i⁴ = 1 (then repeats)
|a + bi| = √(a² + b²) (modulus)
Conjugate of (a+bi) is (a−bi); product = a²+b²
|a + bi| = √(a² + b²) (modulus)
Conjugate of (a+bi) is (a−bi); product = a²+b²
📋 Memorize
To find iⁿ: divide n by 4, use the remainder. Remainder 0→1, 1→i, 2→−1, 3→−i.
✍ Example
Find i²⁵
25 ÷ 4 = 6 remainder 1 → i²⁵ = i¹ = i
4
Quadratic Functions & Discriminant
Unit 4
The discriminant b² − 4ac determines the nature of roots. Vertex form y = a(x − h)² + k gives vertex (h, k) directly.
Quadratic Formula: x = (−b ± √(b²−4ac)) / 2a
Discriminant D = b²−4ac:
D > 0 → 2 real roots | D = 0 → 1 real root | D < 0 → 2 complex roots
Vertex form: y = a(x−h)² + k, vertex = (h, k)
Discriminant D = b²−4ac:
D > 0 → 2 real roots | D = 0 → 1 real root | D < 0 → 2 complex roots
Vertex form: y = a(x−h)² + k, vertex = (h, k)
📋 Memorize
Vertex: h = −b/(2a), then k = f(h). Axis of symmetry: x = −b/(2a).
✍ Example
Discriminant of 2x² − 4x + 3 = 0?
D = (−4)² − 4(2)(3) = 16 − 24 = −8 → 2 complex roots
5
Polynomial Functions & End Behavior
Unit 5
End behavior of a polynomial is determined by its leading term (degree and sign of leading coefficient). As x → ±∞, other terms become negligible.
Even degree, + leading → both ends UP (↑↑)
Even degree, − leading → both ends DOWN (↓↓)
Odd degree, + leading → left DOWN, right UP (↓↑)
Odd degree, − leading → left UP, right DOWN (↑↓)
Even degree, − leading → both ends DOWN (↓↓)
Odd degree, + leading → left DOWN, right UP (↓↑)
Odd degree, − leading → left UP, right DOWN (↑↓)
📋 Memorize
Same direction (both up or both down) = EVEN degree. Opposite directions = ODD degree.
✍ Example
End behavior of f(x) = −3x⁴ + 5x²?
Even degree (4), negative leading → both ends go DOWN (↓↓)
6
Exponential Growth & Decay
Unit 6
Exponential functions model growth and decay. The base b determines behavior: b > 1 is growth, 0 < b < 1 is decay. The function always passes through (0, a).
Growth: f(x) = a · bˣ, b > 1, a > 0
Decay: f(x) = a · bˣ, 0 < b < 1, a > 0
Compound Interest: A = P(1 + r/n)^(nt)
Continuous: A = Pe^(rt)
Decay: f(x) = a · bˣ, 0 < b < 1, a > 0
Compound Interest: A = P(1 + r/n)^(nt)
Continuous: A = Pe^(rt)
📋 Memorize
Growth rate r: f(x) = a(1+r)ˣ. Decay rate r: f(x) = a(1−r)ˣ. Half-life: T½ = ln(2)/k.
✍ Example
Population starts at 500, grows 4% yearly. Formula?
P(t) = 500(1.04)ᵗ
7
Logarithms & Logarithmic Laws
Unit 7
Logarithms are the inverse of exponential functions. logₐ(x) = y means aʸ = x. All three log laws are essential for solving equations.
Product Rule: log(MN) = log M + log N
Quotient Rule: log(M/N) = log M − log N
Power Rule: log(Mⁿ) = n · log M
Change of Base: logₐ(x) = log(x)/log(a) = ln(x)/ln(a)
Special: log(1) = 0 | log base a of a = 1
Quotient Rule: log(M/N) = log M − log N
Power Rule: log(Mⁿ) = n · log M
Change of Base: logₐ(x) = log(x)/log(a) = ln(x)/ln(a)
Special: log(1) = 0 | log base a of a = 1
📋 Memorize
log means log₁₀. ln means logₑ (natural log). logₐ(aˣ) = x and a^(logₐx) = x (inverse properties).
✍ Example
Expand: log(x²y / z³)
= 2log(x) + log(y) − 3log(z)
8
Sequences & Series
Unit 8
Arithmetic sequences have a common difference d; geometric sequences have a common ratio r. Sums can be found with specific formulas.
Arithmetic: aₙ = a₁ + (n−1)d
Arithmetic Sum: Sₙ = n/2 · (a₁ + aₙ) = n/2 · [2a₁ + (n−1)d]
Geometric: aₙ = a₁ · r^(n−1)
Geometric Sum: Sₙ = a₁(1 − rⁿ)/(1 − r), r ≠ 1
Infinite Geo: S∞ = a₁/(1 − r), |r| < 1
Arithmetic Sum: Sₙ = n/2 · (a₁ + aₙ) = n/2 · [2a₁ + (n−1)d]
Geometric: aₙ = a₁ · r^(n−1)
Geometric Sum: Sₙ = a₁(1 − rⁿ)/(1 − r), r ≠ 1
Infinite Geo: S∞ = a₁/(1 − r), |r| < 1
📋 Memorize
Arithmetic → constant DIFFERENCE (add). Geometric → constant RATIO (multiply). Infinite sum only works if |r| < 1.
✍ Example
Find the 10th term of: 3, 7, 11, 15, …
d = 4, a₁ = 3. a₁₀ = 3 + 9(4) = 3 + 36 = 39
9
Rational Functions & Asymptotes
Unit 9
Rational functions have vertical asymptotes where the denominator = 0 (after simplifying). Horizontal asymptotes depend on the degrees of numerator and denominator.
Vertical Asymptote (VA): set denominator = 0 (after simplifying)
Horizontal Asymptote (HA):
deg(N) < deg(D) → y = 0
deg(N) = deg(D) → y = ratio of leading coefficients
deg(N) > deg(D) → no HA (oblique asymptote)
Horizontal Asymptote (HA):
deg(N) < deg(D) → y = 0
deg(N) = deg(D) → y = ratio of leading coefficients
deg(N) > deg(D) → no HA (oblique asymptote)
📋 Memorize
If numerator and denominator share a common factor, there is a HOLE (not a VA) at that x-value.
✍ Example
Find HA of f(x) = (3x² − 1)/(x² + 5)
Same degree → HA: y = 3/1 = 3
10
Combinations, Permutations & Probability
Unit 10
Combinations count selections without regard to order. Permutations count arrangements where order matters. Binomial probability uses combinations.
Combination: C(n,r) = n! / [r!(n−r)!]
Permutation: P(n,r) = n! / (n−r)!
Binomial Prob: P(X=k) = C(n,k) · pᵏ · (1−p)^(n−k)
Note: C(n,0)=C(n,n)=1 | C(n,1)=C(n,n−1)=n
Permutation: P(n,r) = n! / (n−r)!
Binomial Prob: P(X=k) = C(n,k) · pᵏ · (1−p)^(n−k)
Note: C(n,0)=C(n,n)=1 | C(n,1)=C(n,n−1)=n
📋 Memorize
Combinations: order does NOT matter (choose). Permutations: order DOES matter (arrange). C(n,r) is always ≤ P(n,r).
✍ Example
How many ways to choose 3 students from 8?
C(8,3) = 8!/(3!·5!) = (8·7·6)/(3·2·1) = 56
✎ Practice Quiz
20 Essential Problems
Select one answer per question. Immediate feedback with full explanation after each answer.
0
Correct
0
Answered
20
Remaining
Question 01
Polynomials
Which of the following is a factor of p(x) = x³ − 6x² + 11x − 6?
Select the best answer.
Explanation
Use the Factor Theorem: test x = 3 → p(3) = 27 − 54 + 33 − 6 = 0 ✓
Since p(3) = 0, (x − 3) is a factor.
Verify others: p(−1) = −1−6−11−6 = −24 ≠ 0 The polynomial factors as (x−1)(x−2)(x−3).
Since p(3) = 0, (x − 3) is a factor.
Verify others: p(−1) = −1−6−11−6 = −24 ≠ 0 The polynomial factors as (x−1)(x−2)(x−3).
Question 02
Polynomials
When p(x) = 2x³ − 3x² + x − 5 is divided by (x − 2), what is the remainder?
Select the best answer.
Explanation
By the Remainder Theorem, remainder = p(2).
p(2) = 2(8) − 3(4) + 2 − 5 = 16 − 12 + 2 − 5 = 1 The remainder is 1.
p(2) = 2(8) − 3(4) + 2 − 5 = 16 − 12 + 2 − 5 = 1 The remainder is 1.
Question 03
Rational Expressions
Simplify: (x² − 9) / (x² − x − 6)
Select the best answer.
Explanation
Factor both numerator and denominator:
Numerator: x² − 9 = (x+3)(x−3) Denominator: x² − x − 6 = (x−3)(x+2) Cancel common factor (x − 3):
Result = (x+3)/(x+2), where x ≠ 3 and x ≠ −2
Numerator: x² − 9 = (x+3)(x−3) Denominator: x² − x − 6 = (x−3)(x+2) Cancel common factor (x − 3):
Result = (x+3)/(x+2), where x ≠ 3 and x ≠ −2
Question 04
Radical Functions
What is the domain of f(x) = √(2x − 8)?
Select the best answer.
Explanation
The expression under a square root must be ≥ 0:
2x − 8 ≥ 0 2x ≥ 8 → x ≥ 4 Domain: [4, ∞), written as x ≥ 4.
2x − 8 ≥ 0 2x ≥ 8 → x ≥ 4 Domain: [4, ∞), written as x ≥ 4.
Question 05
Complex Numbers
What is the value of i³⁴?
Select the best answer.
Explanation
Divide the exponent by 4 and find the remainder:
34 ÷ 4 = 8 remainder 2 i³⁴ = i² = −1 Cycle: i¹=i, i²=−1, i³=−i, i⁴=1, then repeats.
34 ÷ 4 = 8 remainder 2 i³⁴ = i² = −1 Cycle: i¹=i, i²=−1, i³=−i, i⁴=1, then repeats.
Question 06
Quadratics
How many real solutions does 3x² − 6x + 4 = 0 have?
Select the best answer.
Explanation
Calculate the discriminant D = b² − 4ac:
a=3, b=−6, c=4 D = (−6)² − 4(3)(4) = 36 − 48 = −12 Since D < 0, there are NO real solutions — two complex (imaginary) solutions.
a=3, b=−6, c=4 D = (−6)² − 4(3)(4) = 36 − 48 = −12 Since D < 0, there are NO real solutions — two complex (imaginary) solutions.
Question 07
Quadratics
The function f(x) = 2(x − 3)² + 5 has its vertex at which point?
Select the best answer.
Explanation
Vertex form is y = a(x − h)² + k, where vertex = (h, k).
f(x) = 2(x − 3)² + 5 → h = 3, k = 5 Vertex = (3, 5). Note: the sign inside is (x − h), so h = +3 (not −3).
f(x) = 2(x − 3)² + 5 → h = 3, k = 5 Vertex = (3, 5). Note: the sign inside is (x − h), so h = +3 (not −3).
Question 08
Polynomial End Behavior
Which describes the end behavior of g(x) = −2x⁵ + 4x³ − x?
Select the best answer.
Explanation
Leading term: −2x⁵ (degree 5 = ODD, leading coefficient NEGATIVE).
Odd + negative leading → left UP, right DOWN As x→−∞: −2x⁵ → −2(−∞)⁵ = −2(−∞) = +∞
As x→+∞: −2x⁵ → −∞
x→−∞: g→+∞ ; x→+∞: g→−∞
Odd + negative leading → left UP, right DOWN As x→−∞: −2x⁵ → −2(−∞)⁵ = −2(−∞) = +∞
As x→+∞: −2x⁵ → −∞
x→−∞: g→+∞ ; x→+∞: g→−∞
Question 09
Exponential Functions
A bacteria culture starts with 200 bacteria and doubles every 3 hours. How many bacteria are present after 9 hours?
Select the best answer.
Explanation
N(t) = 200 · 2^(t/3)
After 9 hours: t = 9, so t/3 = 3
N(9) = 200 · 2³ = 200 · 8 = 1600 9 hours = 3 doubling periods: 200 → 400 → 800 → 1600.
N(9) = 200 · 2³ = 200 · 8 = 1600 9 hours = 3 doubling periods: 200 → 400 → 800 → 1600.
Question 10
Logarithms
Which expression equals log₂(32)?
Select the best answer.
Explanation
log₂(32) asks: "2 to what power gives 32?"
2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32 ∴ log₂(32) = 5
2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32 ∴ log₂(32) = 5
Question 11
Logarithms
Solve for x: log₃(x − 4) = 2
Select the best answer.
Explanation
Convert logarithm to exponential form:
log₃(x − 4) = 2 → 3² = x − 4 9 = x − 4 → x = 13 Check: log₃(13 − 4) = log₃(9) = log₃(3²) = 2 ✓
log₃(x − 4) = 2 → 3² = x − 4 9 = x − 4 → x = 13 Check: log₃(13 − 4) = log₃(9) = log₃(3²) = 2 ✓
Question 12
Sequences
Find the sum of the first 15 terms of the arithmetic sequence: 4, 9, 14, 19, …
Select the best answer.
Explanation
Identify: a₁ = 4, d = 5, n = 15.
a₁₅ = 4 + 14(5) = 4 + 70 = 74 S₁₅ = 15/2 · (a₁ + a₁₅) = 15/2 · (4 + 74) = 15/2 · 78 = 15 · 39 = 585
a₁₅ = 4 + 14(5) = 4 + 70 = 74 S₁₅ = 15/2 · (a₁ + a₁₅) = 15/2 · (4 + 74) = 15/2 · 78 = 15 · 39 = 585
Question 13
Sequences
In a geometric sequence, a₁ = 3 and the common ratio r = −2. What is the 5th term?
Select the best answer.
Explanation
aₙ = a₁ · r^(n−1)
a₅ = 3 · (−2)^(5−1) = 3 · (−2)⁴
(−2)⁴ = +16 (even power → positive)
a₅ = 3 · 16 = 48
Sequence: 3, −6, 12, −24, 48.
Question 14
Rational Functions
What is the horizontal asymptote of f(x) = (5x³ − 2x) / (2x³ + x² − 1)?
Select the best answer.
Explanation
Degrees of numerator and denominator are EQUAL (both degree 3).
When deg(N) = deg(D): HA = ratio of leading coefficients Leading coeff of numerator: 5 Leading coeff of denominator: 2 HA: y = 5/2
When deg(N) = deg(D): HA = ratio of leading coefficients Leading coeff of numerator: 5 Leading coeff of denominator: 2 HA: y = 5/2
Question 15
Polynomial Inequalities
Solve the polynomial inequality: x² − 5x + 6 < 0
Select the best answer.
Explanation
Factor the quadratic:
x² − 5x + 6 = (x − 2)(x − 3) Zeros at x = 2 and x = 3. Test intervals:
x < 2: (−)(−) = + → positive (NOT < 0)
2 < x < 3: (+)(−) = − → negative ✓ (satisfies < 0)
x > 3: (+)(+) = + → positive (NOT < 0)
Solution: 2 < x < 3
x² − 5x + 6 = (x − 2)(x − 3) Zeros at x = 2 and x = 3. Test intervals:
x < 2: (−)(−) = + → positive (NOT < 0)
2 < x < 3: (+)(−) = − → negative ✓ (satisfies < 0)
x > 3: (+)(+) = + → positive (NOT < 0)
Solution: 2 < x < 3
Question 16
Conic Sections
The parabola x² = 12y opens in which direction, and what is the focus?
Select the best answer.
Explanation
Standard form x² = 4py opens upward if p > 0.
x² = 12y → 4p = 12 → p = 3 Since p = 3 > 0: opens UPWARD.
Focus is at (0, p) = (0, 3) Directrix: y = −p = −3.
x² = 12y → 4p = 12 → p = 3 Since p = 3 > 0: opens UPWARD.
Focus is at (0, p) = (0, 3) Directrix: y = −p = −3.
Question 17
Rational Functions
Find all vertical asymptotes of f(x) = (x + 1) / (x² − 3x − 4)
Select the best answer.
Explanation
Factor the denominator:
x² − 3x − 4 = (x − 4)(x + 1) Numerator = x + 1. Common factor (x + 1) cancels → HOLE at x = −1 (not a VA).
After canceling: f(x) = 1/(x − 4)
Vertical Asymptote: x = 4 only There is a removable discontinuity (hole) at x = −1.
x² − 3x − 4 = (x − 4)(x + 1) Numerator = x + 1. Common factor (x + 1) cancels → HOLE at x = −1 (not a VA).
After canceling: f(x) = 1/(x − 4)
Vertical Asymptote: x = 4 only There is a removable discontinuity (hole) at x = −1.
Question 18
Probability
A fair coin is flipped 6 times. What is the probability of getting exactly 4 heads?
Select the best answer.
Explanation
Use Binomial Probability: n=6, k=4, p=1/2.
P(X=4) = C(6,4) · (1/2)⁴ · (1/2)² C(6,4) = 6!/(4!·2!) = 15 P(X=4) = 15 · (1/16) · (1/4) = 15/64
P(X=4) = C(6,4) · (1/2)⁴ · (1/2)² C(6,4) = 6!/(4!·2!) = 15 P(X=4) = 15 · (1/16) · (1/4) = 15/64
Question 19
Logarithms
Which expression is equivalent to log(x²y³/z)?
Select the best answer.
Explanation
Apply log rules step by step:
log(x²y³/z) = log(x²y³) − log(z) [Quotient Rule] = log(x²) + log(y³) − log(z) [Product Rule] = 2log(x) + 3log(y) − log(z) [Power Rule]
log(x²y³/z) = log(x²y³) − log(z) [Quotient Rule] = log(x²) + log(y³) − log(z) [Product Rule] = 2log(x) + 3log(y) − log(z) [Power Rule]
Question 20
Combinations
A committee of 4 must be chosen from 10 people. In how many ways can this be done?
Select the best answer.
Explanation
Order doesn't matter (committee), so use Combinations:
C(10,4) = 10! / (4! · 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 5040 / 24 = 210
C(10,4) = 10! / (4! · 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 5040 / 24 = 210
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🔑 Answer Key & Solutions
Complete Solutions
Detailed explanations for all 20 problems
01
B — (x − 3)
Factor Theorem: p(3) = 27−54+33−6 = 0. Full factorization: (x−1)(x−2)(x−3).
02
C — 1
Remainder Theorem: p(2) = 2(8)−3(4)+2−5 = 16−12+2−5 = 1.
03
A — (x+3)/(x+2)
Factor: (x+3)(x−3) ÷ (x−3)(x+2). Cancel (x−3). Result: (x+3)/(x+2), x≠3, x≠−2.
04
D — x ≥ 4
Set 2x−8 ≥ 0 → x ≥ 4. Domain: [4, ∞).
05
B — −1
34 ÷ 4 = 8 remainder 2. i³⁴ = i² = −1.
06
C — No real solutions
D = (−6)²−4(3)(4) = 36−48 = −12 < 0. Two complex roots.
07
A — (3, 5)
Vertex form y=a(x−h)²+k. Here h=3, k=5. Vertex = (3, 5).
08
D — x→−∞, g→+∞; x→+∞, g→−∞
Leading term −2x⁵: odd degree, negative → left UP, right DOWN.
09
B — 1600
N(9) = 200·2^(9/3) = 200·2³ = 200·8 = 1600. Three doublings: 200→400→800→1600.
10
C — 5
log₂(32): 2⁵ = 32. Therefore log₂(32) = 5.
11
A — x = 13
log₃(x−4)=2 → 3²=x−4 → 9=x−4 → x=13. Check: log₃(9)=2 ✓
12
C — 585
a₁=4, d=5, n=15. a₁₅=4+14(5)=74. S₁₅=15/2·(4+74)=15·39=585.
13
B — 48
a₅=3·(−2)⁴=3·16=48. Even power makes result positive. Sequence: 3,−6,12,−24,48.
14
C — y = 5/2
Equal degrees (both 3). HA = ratio of leading coefficients = 5/2.
15
A — 2 < x < 3
Factor: (x−2)(x−3)<0. Sign test: negative only between roots 2 and 3.
16
D — Upward; focus at (0, 3)
x²=4py → 4p=12 → p=3>0. Opens up. Focus=(0,p)=(0,3).
17
A — x = 4 only
Denom: (x−4)(x+1). Numerator (x+1) cancels → hole at x=−1. VA: x=4 only.
18
A — 15/64
C(6,4)·(1/2)⁴·(1/2)² = 15·(1/64) = 15/64.
19
A — 2log(x) + 3log(y) − log(z)
Quotient rule splits fraction, product rule splits multiplication, power rule brings exponents down.
20
D — 210
C(10,4) = (10·9·8·7)/(4·3·2·1) = 5040/24 = 210.