Unit 01
Polynomial Operations & Factoring
Core Concept
Polynomial Operations
A polynomial is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀.
Add/subtract by combining like terms (same degree).
Multiply using the distributive property or FOIL.
Divide using long division or synthetic division.
Key Factoring Patterns:
a² − b² = (a + b)(a − b)
a² + 2ab + b² = (a + b)²
a³ − b³ = (a − b)(a² + ab + b²)
a³ + b³ = (a + b)(a² − ab + b²)
Remainder Theorem: f(x) ÷ (x − c) → remainder = f(c)
Factor Theorem: (x − c) is a factor ⟺ f(c) = 0
a² − b² = (a + b)(a − b)
a² + 2ab + b² = (a + b)²
a³ − b³ = (a − b)(a² + ab + b²)
a³ + b³ = (a + b)(a² − ab + b²)
Remainder Theorem: f(x) ÷ (x − c) → remainder = f(c)
Factor Theorem: (x − c) is a factor ⟺ f(c) = 0
Add the polynomials: (2x² − 3x + 1) + (x² + 5x − 4)
Combine like terms:
x² terms: 2x² + x² = 3x²
x terms: −3x + 5x = 2x
constants: 1 − 4 = −3
Answer: 3x² + 2x − 3
x² terms: 2x² + x² = 3x²
x terms: −3x + 5x = 2x
constants: 1 − 4 = −3
Answer: 3x² + 2x − 3
01
Unit 1 · Polynomial Operations
★★☆
Simplify the following expression by adding the two polynomials:
(x² + 3x − 2) + (2x² − x + 5)
(x² + 3x − 2) + (2x² − x + 5)
Full Solution
Group like terms together:
(x² + 2x²) + (3x − x) + (−2 + 5) = 3x² + 2x + 3 Answer: 3x² + 2x + 3
(x² + 2x²) + (3x − x) + (−2 + 5) = 3x² + 2x + 3 Answer: 3x² + 2x + 3
02
Unit 1 · Factoring Quadratics
★★☆
Factor the following quadratic completely:
x² − 5x + 6
x² − 5x + 6
Full Solution
Find two numbers that multiply to +6 and add to −5: those are −2 and −3.
x² − 5x + 6 = (x − 2)(x − 3) Check: (x−2)(x−3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓ Answer: (x − 2)(x − 3)
x² − 5x + 6 = (x − 2)(x − 3) Check: (x−2)(x−3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓ Answer: (x − 2)(x − 3)
03
Unit 1 · Remainder Theorem
★★★
Given f(x) = x³ − 3x + 2, find the remainder when f(x) is divided by (x − 2).
Full Solution
By the Remainder Theorem, the remainder = f(2):
f(2) = (2)³ − 3(2) + 2 = 8 − 6 + 2 = 4 Answer: 4
f(2) = (2)³ − 3(2) + 2 = 8 − 6 + 2 = 4 Answer: 4
Unit 02
Quadratic Equations & Functions
Core Concept
Solving Quadratics — Three Methods
1. Factoring — rewrite as (x−p)(x−q)=0, then x=p or x=q.
2. Completing the Square — transform x²+bx+c into (x+h)²=k form.
3. Quadratic Formula — always works: x = (−b ± √(b²−4ac)) / 2a.
The discriminant Δ = b²−4ac tells the nature of roots: Δ > 0 → 2 real roots; Δ = 0 → 1 repeated root; Δ < 0 → no real roots.
2. Completing the Square — transform x²+bx+c into (x+h)²=k form.
3. Quadratic Formula — always works: x = (−b ± √(b²−4ac)) / 2a.
The discriminant Δ = b²−4ac tells the nature of roots: Δ > 0 → 2 real roots; Δ = 0 → 1 repeated root; Δ < 0 → no real roots.
Quadratic Formula: x = (−b ± √(b² − 4ac)) / 2a
Vertex Form: f(x) = a(x − h)² + k → vertex at (h, k)
Completing the square: x² + bx = (x + b/2)² − (b/2)²
Discriminant: Δ = b² − 4ac
Vertex Form: f(x) = a(x − h)² + k → vertex at (h, k)
Completing the square: x² + bx = (x + b/2)² − (b/2)²
Discriminant: Δ = b² − 4ac
Complete the square: x² + 4x + 1 = 0
x² + 4x = −1
(x + 2)² − 4 = −1
(x + 2)² = 3
x = −2 ± √3
(x + 2)² − 4 = −1
(x + 2)² = 3
x = −2 ± √3
04
Unit 2 · Quadratic Formula
★★★
Use the quadratic formula to solve:
x² − 4x + 1 = 0
Express your answer in the form x = a ± √b.
x² − 4x + 1 = 0
Express your answer in the form x = a ± √b.
Full Solution
a = 1, b = −4, c = 1
x = (4 ± √(16 − 4)) / 2 = (4 ± √12) / 2 = (4 ± 2√3) / 2 x = 2 ± √3 Answer: x = 2 ± √3
x = (4 ± √(16 − 4)) / 2 = (4 ± √12) / 2 = (4 ± 2√3) / 2 x = 2 ± √3 Answer: x = 2 ± √3
05
Unit 2 · Discriminant
★★☆
Compute the discriminant of 3x² − 2x + 1 = 0 and state the nature of its roots.
Full Solution
a = 3, b = −2, c = 1
Δ = b² − 4ac = (−2)² − 4(3)(1) = 4 − 12 = −8 Since Δ < 0, the equation has no real roots (two complex conjugate roots). Answer: Δ = −8, no real roots
Δ = b² − 4ac = (−2)² − 4(3)(1) = 4 − 12 = −8 Since Δ < 0, the equation has no real roots (two complex conjugate roots). Answer: Δ = −8, no real roots
06
Unit 2 · Completing the Square
★★★
Rewrite f(x) = x² + 6x + 5 in vertex form f(x) = (x + h)² + k and state the vertex.
Full Solution
x² + 6x + 5
= (x² + 6x + 9) − 9 + 5
= (x + 3)² − 4
Vertex form: (x + 3)² − 4 → vertex at (−3, −4)
Answer: (x + 3)² − 4, vertex (−3, −4)
Unit 03
Rational Exponents & Radicals
Core Concept
Exponent Rules & Radical Equations
Rational exponent: aᵐ/ⁿ = (ⁿ√a)ᵐ = ⁿ√(aᵐ).
Negative exponent: a⁻ⁿ = 1/aⁿ.
To solve radical equations, isolate the radical, then raise both sides to the appropriate power. Always check for extraneous solutions!
Negative exponent: a⁻ⁿ = 1/aⁿ.
To solve radical equations, isolate the radical, then raise both sides to the appropriate power. Always check for extraneous solutions!
Rational exponent: a^(m/n) = (n-th root of a)^m
Product rule: a^m · a^n = a^(m+n)
Power rule: (a^m)^n = a^(mn)
Radical equation: √(f(x)) = g(x) → [square both sides] → check answers
Product rule: a^m · a^n = a^(m+n)
Power rule: (a^m)^n = a^(mn)
Radical equation: √(f(x)) = g(x) → [square both sides] → check answers
Simplify: 8^(2/3)
8^(2/3) = (∛8)² = 2² = 4
07
Unit 3 · Rational Exponents
★★☆
Evaluate without a calculator:
27^(2/3)
27^(2/3)
Full Solution
27^(2/3) = (∛27)² = 3² = 9
(Because ∛27 = 3, since 3³ = 27)
Answer: 9
08
Unit 3 · Radical Equations
★★★
Solve for x:
√(2x + 3) = 5
√(2x + 3) = 5
Full Solution
√(2x + 3) = 5
Square both sides: 2x + 3 = 25
2x = 22 → x = 11
Check: √(2·11 + 3) = √25 = 5 ✓
Answer: x = 11
Unit 04
Complex Numbers & Absolute Value
Core Concept
Complex Number Arithmetic
Imaginary unit: i = √(−1), so i² = −1.
A complex number has the form a + bi (a = real part, b = imaginary part).
Absolute value inequality |ax + b| ≤ c means −c ≤ ax + b ≤ c.
|ax + b| ≥ c means ax + b ≤ −c OR ax + b ≥ c.
A complex number has the form a + bi (a = real part, b = imaginary part).
Absolute value inequality |ax + b| ≤ c means −c ≤ ax + b ≤ c.
|ax + b| ≥ c means ax + b ≤ −c OR ax + b ≥ c.
i² = −1, i³ = −i, i⁴ = 1 (cycle of 4)
Multiply: (a+bi)(c+di) = (ac−bd) + (ad+bc)i
Conjugate of (a+bi) = (a−bi); |a+bi| = √(a²+b²)
|x − a| < r ⟺ a − r < x < a + r
Multiply: (a+bi)(c+di) = (ac−bd) + (ad+bc)i
Conjugate of (a+bi) = (a−bi); |a+bi| = √(a²+b²)
|x − a| < r ⟺ a − r < x < a + r
Multiply: (2 + 3i)(2 − 3i)
= 4 − 6i + 6i − 9i² = 4 − 9(−1) = 4 + 9 = 13
(Product of conjugates always gives a real number.)
09
Unit 4 · Complex Numbers
★★★
Multiply the complex numbers and write your answer in the form a + bi:
(3 + 2i)(1 − i)
(3 + 2i)(1 − i)
Full Solution
(3 + 2i)(1 − i) = 3·1 + 3·(−i) + 2i·1 + 2i·(−i)
= 3 − 3i + 2i − 2i²
= 3 − i − 2(−1) = 3 − i + 2 = 5 − i
Answer: 5 − i
10
Unit 4 · Absolute Value Inequality
★★★
Solve and write the solution as an interval:
|2x − 3| ≤ 7
|2x − 3| ≤ 7
Full Solution
|2x − 3| ≤ 7 means:
−7 ≤ 2x − 3 ≤ 7 Add 3: −4 ≤ 2x ≤ 10 Divide by 2: −2 ≤ x ≤ 5 Answer: −2 ≤ x ≤ 5
−7 ≤ 2x − 3 ≤ 7 Add 3: −4 ≤ 2x ≤ 10 Divide by 2: −2 ≤ x ≤ 5 Answer: −2 ≤ x ≤ 5
Unit 05
Exponential & Logarithmic Functions
Core Concept
Exponent ↔ Logarithm Relationship
Exponential form: bˣ = y ↔ Logarithmic form: logb(y) = x.
logₐ(MN) = logₐM + logₐN (Product Rule)
logₐ(M/N) = logₐM − logₐN (Quotient Rule)
logₐ(Mⁿ) = n·logₐM (Power Rule)
To solve exponential equations: take log of both sides or rewrite bases.
logₐ(MN) = logₐM + logₐN (Product Rule)
logₐ(M/N) = logₐM − logₐN (Quotient Rule)
logₐ(Mⁿ) = n·logₐM (Power Rule)
To solve exponential equations: take log of both sides or rewrite bases.
Definition: log_b(x) = y ⟺ b^y = x
Change of Base: log_b(x) = ln(x) / ln(b)
Product: log(MN) = log M + log N
Quotient: log(M/N) = log M − log N
Power: log(Mⁿ) = n·log M
Exponential Growth: A = A₀·b^(t/k)
Change of Base: log_b(x) = ln(x) / ln(b)
Product: log(MN) = log M + log N
Quotient: log(M/N) = log M − log N
Power: log(Mⁿ) = n·log M
Exponential Growth: A = A₀·b^(t/k)
Solve: 2^x = 16
Write 16 as a power of 2: 16 = 2⁴
So 2^x = 2⁴ → x = 4
So 2^x = 2⁴ → x = 4
11
Unit 5 · Logarithms
★★☆
Evaluate without a calculator:
log₂(32)
log₂(32)
Full Solution
Ask: "2 to what power gives 32?"
2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32 log₂(32) = 5 Answer: 5
2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32 log₂(32) = 5 Answer: 5
12
Unit 5 · Log Properties
★★★
Use logarithm properties to simplify (assume base-10 log):
log(6) + log(5) − log(3)
log(6) + log(5) − log(3)
Full Solution
log(6) + log(5) − log(3) = log(6 · 5 / 3)
= log(30/3) = log(10) = 1
Answer: 1
13
Unit 5 · Exponential Equations
★★★
Solve for x (express as a fraction):
4^x = 32
4^x = 32
Full Solution
Rewrite both sides as powers of 2:
4^x = (2²)^x = 2^(2x) 32 = 2⁵ 2^(2x) = 2⁵ → 2x = 5 → x = 5/2 Answer: x = 5/2
4^x = (2²)^x = 2^(2x) 32 = 2⁵ 2^(2x) = 2⁵ → 2x = 5 → x = 5/2 Answer: x = 5/2
14
Unit 5 · Exponential Growth
★★★
A bacterial culture starts with 1000 bacteria and doubles every 3 hours.
Using the model A = 1000 · 2^(t/3), how many bacteria are there after t = 6 hours?
Full Solution
A = 1000 · 2^(6/3) = 1000 · 2² = 1000 · 4 = 4000
After 6 hours (2 complete doubling periods), the count quadruples.
Answer: 4000
Unit 06
Sequences & Series
Core Concept
Arithmetic & Geometric Sequences
Arithmetic sequence: each term increases by a constant difference d.
General term: aₙ = a₁ + (n − 1)d.
Geometric sequence: each term is multiplied by a constant ratio r.
General term: aₙ = a₁ · r^(n−1).
Partial sum (geometric): Sₙ = a₁(rⁿ − 1)/(r − 1).
General term: aₙ = a₁ + (n − 1)d.
Geometric sequence: each term is multiplied by a constant ratio r.
General term: aₙ = a₁ · r^(n−1).
Partial sum (geometric): Sₙ = a₁(rⁿ − 1)/(r − 1).
Arithmetic: aₙ = a₁ + (n−1)d
Arithmetic sum: Sₙ = n(a₁ + aₙ)/2
Geometric: aₙ = a₁ · r^(n−1)
Geometric sum: Sₙ = a₁(rⁿ − 1)/(r − 1) (r ≠ 1)
Infinite geometric sum (|r|<1): S = a₁/(1 − r)
Arithmetic sum: Sₙ = n(a₁ + aₙ)/2
Geometric: aₙ = a₁ · r^(n−1)
Geometric sum: Sₙ = a₁(rⁿ − 1)/(r − 1) (r ≠ 1)
Infinite geometric sum (|r|<1): S = a₁/(1 − r)
Find the 8th term of the arithmetic sequence: 2, 5, 8, 11, …
a₁ = 2, d = 3
a₈ = 2 + (8 − 1)·3 = 2 + 21 = 23
a₈ = 2 + (8 − 1)·3 = 2 + 21 = 23
15
Unit 6 · Arithmetic Sequences
★★☆
An arithmetic sequence has first term a₁ = 3 and common difference d = 4.
Find the 10th term, a₁₀.
Full Solution
aₙ = a₁ + (n − 1)d
a₁₀ = 3 + (10 − 1)·4 = 3 + 36 = 39
Answer: a₁₀ = 39
16
Unit 6 · Geometric Series
★★★
A geometric sequence has a₁ = 2 and common ratio r = 3.
Find the sum of the first 4 terms, S₄.
Full Solution
Sₙ = a₁(rⁿ − 1)/(r − 1)
S₄ = 2(3⁴ − 1)/(3 − 1) = 2(81 − 1)/2 = 2 · 80/2 = 80
Verify: 2 + 6 + 18 + 54 = 80 ✓
Answer: S₄ = 80
Unit 07
Systems, Binomial Theorem & Parabolas
Core Concept
Binomial Theorem & Parabola Vertex
Binomial Theorem: (a + b)ⁿ = Σ C(n,k)·aⁿ⁻ᵏ·bᵏ for k = 0,1,…,n.
C(n,k) = n! / (k!(n−k)!) — the binomial coefficient.
Parabola vertex: for f(x) = ax² + bx + c, the vertex x-coordinate is x = −b/(2a), and vertex y = f(−b/2a).
C(n,k) = n! / (k!(n−k)!) — the binomial coefficient.
Parabola vertex: for f(x) = ax² + bx + c, the vertex x-coordinate is x = −b/(2a), and vertex y = f(−b/2a).
Binomial expansion: (a + b)³ = a³ + 3a²b + 3ab² + b³
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
Vertex x-coordinate: h = −b/(2a)
Vertex y-coordinate: k = f(h) = c − b²/(4a)
Systems (substitution/elimination): solve for one variable, substitute.
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
Vertex x-coordinate: h = −b/(2a)
Vertex y-coordinate: k = f(h) = c − b²/(4a)
Systems (substitution/elimination): solve for one variable, substitute.
Expand: (x + 1)³
(x + 1)³ = x³ + 3x² + 3x + 1
17
Unit 7 · Systems of Equations
★★★
Solve the system of equations (find all solution pairs):
x + y = 5
xy = 6
x + y = 5
xy = 6
Full Solution
From the first equation: y = 5 − x. Substitute into xy = 6:
x(5 − x) = 6 → 5x − x² = 6 → x² − 5x + 6 = 0 (x − 2)(x − 3) = 0 → x = 2 or x = 3 If x = 2: y = 3. If x = 3: y = 2. Check: 2 + 3 = 5 ✓, 2 · 3 = 6 ✓ Answer: (2, 3) and (3, 2)
x(5 − x) = 6 → 5x − x² = 6 → x² − 5x + 6 = 0 (x − 2)(x − 3) = 0 → x = 2 or x = 3 If x = 2: y = 3. If x = 3: y = 2. Check: 2 + 3 = 5 ✓, 2 · 3 = 6 ✓ Answer: (2, 3) and (3, 2)
18
Unit 7 · Binomial Theorem
★★★
Expand completely:
(x + 2)³
(x + 2)³
Full Solution
Use (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = x, b = 2:
= x³ + 3x²(2) + 3x(2²) + 2³ = x³ + 6x² + 12x + 8 Answer: x³ + 6x² + 12x + 8
= x³ + 3x²(2) + 3x(2²) + 2³ = x³ + 6x² + 12x + 8 Answer: x³ + 6x² + 12x + 8
19
Unit 7 · Parabola Vertex
★★★
Find the vertex of the parabola:
f(x) = 2x² − 8x + 3
f(x) = 2x² − 8x + 3
Full Solution
a = 2, b = −8, c = 3
x-vertex: h = −b/(2a) = −(−8)/(2·2) = 8/4 = 2 y-vertex: k = f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5 Answer: Vertex = (2, −5)
x-vertex: h = −b/(2a) = −(−8)/(2·2) = 8/4 = 2 y-vertex: k = f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5 Answer: Vertex = (2, −5)
20
Unit 1 · Polynomial Division
★★★
Divide using polynomial long division:
(x³ − 2x² + x − 2) ÷ (x − 2)
Express the result as a quotient polynomial.
(x³ − 2x² + x − 2) ÷ (x − 2)
Express the result as a quotient polynomial.
Full Solution
First check f(2) = 8 − 8 + 2 − 2 = 0, so (x − 2) is a factor (no remainder).
x³ − 2x² + x − 2 = x²(x − 2) + 1(x − 2) = (x − 2)(x² + 1) Therefore the quotient is:
(x³ − 2x² + x − 2) ÷ (x − 2) = x² + 1 Answer: x² + 1
x³ − 2x² + x − 2 = x²(x − 2) + 1(x − 2) = (x − 2)(x² + 1) Therefore the quotient is:
(x³ − 2x² + x − 2) ÷ (x − 2) = x² + 1 Answer: x² + 1
FINAL RESULT
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