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USTET MathematicsAlgebra — Sets, Exponents, Radicals, Polynomials & EquationsStudy Notes

Study notes for Algebra — Sets, Exponents, Radicals, Polynomials & Equations that match the USTET 2026 syllabus. Built to mirror how University of Santo Tomas structures USTET Mathematics questions, these notes walk through each concept with examples, formulas, and practice questions designed for time-pressured exam conditions.

Exam context

The University of Santo Tomas Entrance Test is conducted by University of Santo Tomas and is scheduled for Early Q4 2026. The Mathematics subtest is marked as "Core section" in the official pattern, and Algebra — Sets, Exponents, Radicals, Polynomials & Equations appears in position 3rd of 9 in the USTET Mathematics review rotation. Passing mark: Competitive overall score. Recent USTET 2026 papers have drawn roughly a meaningful share of questions from this subject.

Algebra — Sets, Exponents, Radicals, Polynomials & Equations - Study notes

Algebra is the foundation of advanced mathematics and is crucial for success in college entrance exams like UPCAT. This comprehensive guide covers essential algebraic concepts including sets, exponents, radicals, polynomials, and equations. You'll learn problem-solving strategies, step-by-step solution methods, and real-world applications that will help you excel in your exams and future mathematical studies.

Summary

This chapter covers the essential algebraic concepts needed for college entrance exams. Master set theory for logical reasoning, exponent laws for simplifying expressions, radical properties for solving equations, polynomial operations for factoring, linear equations for modeling relationships, quadratic equations for complex problem-solving, and functions for understanding mathematical relationships. Practice each concept with step-by-step solutions, and always verify your answers. Remember that algebra is a language for describing mathematical relationships - the more fluent you become, the easier advanced mathematics will be. Focus on understanding the 'why' behind each rule, not just the 'how' to apply it.

Sections

A set is a well-defined collection of distinct objects called elements. Understanding sets is fundamental for algebraic reasoning and logical thinking. **Basic Set Definitions:** - **Set**: A collection of distinct objects - **Element**: An object belonging to a set, denoted by ∈ - **Empty/Null Set**: A set with no elements, denoted by ∅ or {} - **Universal Set**: The set containing all elements under consideration - **Subset**: Set A is a subset of B if every element of A is also in B (A ⊆ B) **Set Operations:** 1. **Union (A ∪ B)**: All elements in A or B or both 2. **Intersection (A ∩ B)**: Elements common to both A and B 3. **Complement (A')**: Elements in the universal set but not in A **Problem-Solving Strategy for Set Problems:** Step 1: Identify what each set represents Step 2: Draw a Venn diagram if helpful Step 3: Apply the appropriate set operation Step 4: Count elements or find the resulting set Step 5: Verify your answer makes logical sense

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Set Theory Fundamentals

Examples

  • **Example 1: Set Operations** Given: A = {1, 2, 3, 4} and B = {3, 4, 5, 6} Find: A ∪ B, A ∩ B, and A' **Step 1:** Identify the sets A = {1, 2, 3, 4}, B = {3, 4, 5, 6} **Step 2:** Find the union A ∪ B = {1, 2, 3, 4, 5, 6} (all elements from both sets) **Step 3:** Find the intersection A ∩ B = {3, 4} (common elements only) **Step 4:** Find A' (assuming U = {1, 2, 3, 4, 5, 6, 7, 8}) A' = {5, 6, 7, 8} (elements in U but not in A)
  • **Example 2: Word Problem** In a class of 50 students: 30 like Math, 25 like Science, and 15 like both subjects. How many students like neither? **Step 1:** Use the inclusion-exclusion principle |M ∪ S| = |M| + |S| - |M ∩ S| |M ∪ S| = 30 + 25 - 15 = 40 **Step 2:** Find students who like neither Neither = Total - (Like at least one) Neither = 50 - 40 = 10 students

Key Points

  • Sets contain distinct elements with no repetition
  • Use proper notation: ∈ for belongs to, ⊆ for subset
  • Venn diagrams help visualize set relationships
  • Universal set contains all elements being considered
  • Empty set is a subset of every set

Exponents represent repeated multiplication and follow specific rules that make algebraic manipulation easier. Mastering these laws is essential for solving complex algebraic problems. **Definition:** For any real number a and positive integer n: a^n = a × a × a × ... × a (n factors) **Laws of Exponents:** 1. **Product Rule**: a^m × a^n = a^(m+n) 2. **Quotient Rule**: a^m ÷ a^n = a^(m-n) (a ≠ 0) 3. **Power Rule**: (a^m)^n = a^(mn) 4. **Power of Product**: (ab)^n = a^n × b^n 5. **Power of Quotient**: (a/b)^n = a^n/b^n (b ≠ 0) 6. **Zero Exponent**: a^0 = 1 (a ≠ 0) 7. **Negative Exponent**: a^(-n) = 1/a^n (a ≠ 0) **Problem-Solving Strategy for Exponents:** Step 1: Identify which law applies Step 2: Apply the appropriate rule Step 3: Simplify the expression Step 4: Check if further simplification is possible **Common Mistakes to Avoid:** - Don't add exponents when multiplying different bases: 2^3 × 3^2 ≠ 6^5 - Remember: (a + b)^n ≠ a^n + b^n - Zero exponent always equals 1, not 0

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Exponents and Laws of Exponents

Examples

  • **Example 1: Simplifying with Laws of Exponents** Simplify: (2x^3y^2)^2 × (3x^2/y)^2 **Step 1:** Apply power rules (2x^3y^2)^2 = 2^2 × (x^3)^2 × (y^2)^2 = 4x^6y^4 (3x^2/y)^2 = 3^2 × (x^2)^2 / y^2 = 9x^4/y^2 **Step 2:** Multiply the results 4x^6y^4 × 9x^4/y^2 = 36x^6y^4 × x^4/y^2 **Step 3:** Apply product and quotient rules = 36x^(6+4)y^(4-2) = 36x^10y^2
  • **Example 2: Negative Exponents** Simplify: (2^(-3) × 3^2) / (2^(-1) × 3^(-1)) **Step 1:** Convert negative exponents 2^(-3) = 1/2^3 = 1/8 3^(-1) = 1/3 2^(-1) = 1/2 **Step 2:** Substitute and simplify = (1/8 × 9) / (1/2 × 1/3) = 9/8 / (1/6) = 9/8 × 6/1 = 54/8 = 27/4

Key Points

  • Exponents represent repeated multiplication
  • Laws of exponents make complex calculations manageable
  • Negative exponents create reciprocals
  • Zero exponent always equals 1 (base ≠ 0)
  • Apply laws step by step for complex expressions

Radicals are the inverse operation of exponents. Understanding how to manipulate radical expressions is crucial for solving equations and simplifying complex expressions. **Definition:** The nth root of a number a is written as ⁿ√a, where n is the index and a is the radicand. - If n is even: ⁿ√a is defined only when a ≥ 0 - If n is odd: ⁿ√a is defined for all real numbers **Properties of Radicals:** 1. **Product Property**: ⁿ√(ab) = ⁿ√a × ⁿ√b 2. **Quotient Property**: ⁿ√(a/b) = ⁿ√a / ⁿ√b (b ≠ 0) 3. **Power Property**: ⁿ√(a^m) = (ⁿ√a)^m 4. **Index Rule**: ᵐ√(ⁿ√a) = ᵐⁿ√a **Simplifying Radicals:** Step 1: Factor the radicand into perfect powers Step 2: Extract perfect nth powers from under the radical Step 3: Simplify the remaining expression Step 4: Rationalize denominators if necessary **Rationalizing Denominators:** - For √a: multiply by √a/√a - For (a + √b): multiply by (a - √b)/(a - √b) (conjugate) **Relationship to Exponents:** ⁿ√a = a^(1/n) This connection allows us to use exponent laws with radicals.

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Radicals and Radical Expressions

Examples

  • **Example 1: Simplifying Radicals** Simplify: √(72x^5y^3) **Step 1:** Factor the radicand 72 = 36 × 2 = 6^2 × 2 x^5 = x^4 × x = (x^2)^2 × x y^3 = y^2 × y **Step 2:** Apply the product property √(72x^5y^3) = √(6^2 × 2 × (x^2)^2 × x × y^2 × y) **Step 3:** Extract perfect squares = 6x^2y√(2xy) **Final Answer:** 6x^2y√(2xy)
  • **Example 2: Rationalizing Denominators** Rationalize: 5/(2 + √3) **Step 1:** Identify the conjugate of the denominator Conjugate of (2 + √3) is (2 - √3) **Step 2:** Multiply by the conjugate fraction 5/(2 + √3) × (2 - √3)/(2 - √3) **Step 3:** Simplify numerator and denominator Numerator: 5(2 - √3) = 10 - 5√3 Denominator: (2 + √3)(2 - √3) = 4 - 3 = 1 **Final Answer:** 10 - 5√3

Key Points

  • Radicals are inverse operations of exponents
  • Even index requires non-negative radicand
  • Product and quotient properties help simplify expressions
  • Always rationalize denominators in final answers
  • Convert to fractional exponents when helpful

Polynomials are algebraic expressions consisting of variables and coefficients combined using addition, subtraction, and multiplication. Understanding polynomial operations and factoring is essential for solving equations. **Polynomial Terminology:** - **Degree**: Highest power of the variable - **Leading Coefficient**: Coefficient of the highest degree term - **Constant Term**: Term with no variable - **Like Terms**: Terms with identical variable parts **Types of Polynomials by Degree:** - Constant (degree 0): 5 - Linear (degree 1): 3x + 2 - Quadratic (degree 2): x² + 5x + 6 - Cubic (degree 3): 2x³ + x² - 4x + 1 **Polynomial Operations:** **Addition/Subtraction:** Step 1: Remove parentheses (distribute negative signs) Step 2: Group like terms together Step 3: Combine coefficients of like terms **Multiplication:** - Monomial × Polynomial: Distribute the monomial - Binomial × Binomial: Use FOIL (First, Outer, Inner, Last) - General case: Distribute each term in the first polynomial to every term in the second **Factoring Strategies:** 1. **Greatest Common Factor (GCF)**: Factor out common factors 2. **Difference of Squares**: a² - b² = (a + b)(a - b) 3. **Perfect Square Trinomials**: a² ± 2ab + b² = (a ± b)² 4. **Quadratic Trinomials**: ax² + bx + c 5. **Sum/Difference of Cubes**: a³ ± b³ = (a ± b)(a² ∓ ab + b²)

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Polynomials: Operations and Factoring

Examples

  • **Example 1: Polynomial Multiplication** Multiply: (3x + 2)(x² - 4x + 5) **Step 1:** Distribute first term of first polynomial 3x(x² - 4x + 5) = 3x³ - 12x² + 15x **Step 2:** Distribute second term of first polynomial 2(x² - 4x + 5) = 2x² - 8x + 10 **Step 3:** Add the results 3x³ - 12x² + 15x + 2x² - 8x + 10 **Step 4:** Combine like terms 3x³ + (-12x² + 2x²) + (15x - 8x) + 10 = 3x³ - 10x² + 7x + 10
  • **Example 2: Factoring Quadratic Trinomials** Factor: 6x² + 7x - 3 **Step 1:** Find factors of ac = 6(-3) = -18 Pairs: (1,-18), (2,-9), (3,-6), (6,-3), (9,-2), (18,-1) **Step 2:** Find pair that adds to b = 7 9 + (-2) = 7 ✓ **Step 3:** Rewrite middle term 6x² + 9x - 2x - 3 **Step 4:** Factor by grouping 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3) **Verification:** (3x - 1)(2x + 3) = 6x² + 9x - 2x - 3 = 6x² + 7x - 3 ✓

Key Points

  • Combine only like terms when adding/subtracting
  • Use FOIL for multiplying binomials
  • Always check for GCF first when factoring
  • Recognize special patterns for quick factoring
  • Verify factoring by multiplying back

Linear equations are fundamental tools for modeling real-world situations. They have the form ax + b = c where a ≠ 0, and their graphs are straight lines. **Standard Form of Linear Equations:** - **One variable**: ax + b = 0 - **Two variables**: Ax + By = C - **Slope-intercept form**: y = mx + b (where m is slope, b is y-intercept) - **Point-slope form**: y - y₁ = m(x - x₁) **Solving Linear Equations - Step-by-Step Method:** Step 1: Simplify both sides (combine like terms, distribute) Step 2: Move all variable terms to one side Step 3: Move all constant terms to the other side Step 4: Divide by the coefficient of the variable Step 5: Check your solution by substituting back **Systems of Linear Equations:** Three main solution methods: 1. **Graphing**: Find intersection point 2. **Substitution**: Solve one equation for a variable, substitute into the other 3. **Elimination**: Add/subtract equations to eliminate a variable **Word Problem Strategy:** Step 1: Read carefully and identify what you're looking for Step 2: Define variables for unknown quantities Step 3: Write equations based on given information Step 4: Solve the equation(s) Step 5: Answer the original question Step 6: Check if your answer makes sense in context

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Linear Equations and Problem Solving

Examples

  • **Example 1: Solving Linear Equations** Solve: 3(2x - 4) + 5 = 2x - 7 **Step 1:** Distribute and simplify 6x - 12 + 5 = 2x - 7 6x - 7 = 2x - 7 **Step 2:** Move variable terms to left side 6x - 2x = -7 + 7 4x = 0 **Step 3:** Solve for x x = 0 **Step 4:** Check the solution 3(2(0) - 4) + 5 = 3(-4) + 5 = -12 + 5 = -7 2(0) - 7 = -7 ✓
  • **Example 2: Word Problem** Maria has ₱5 and ₱10 bills totaling ₱140. If she has 3 more ₱5 bills than ₱10 bills, how many of each bill does she have? **Step 1:** Define variables Let x = number of ₱10 bills Then x + 3 = number of ₱5 bills **Step 2:** Set up equation based on total value 10x + 5(x + 3) = 140 **Step 3:** Solve the equation 10x + 5x + 15 = 140 15x = 125 x = 25/3 ≈ 8.33 **Step 4:** Since we need whole bills, let's reconsider Let's try x = 8: 10(8) + 5(11) = 80 + 55 = 135 (too low) Let's try x = 9: 10(9) + 5(12) = 90 + 60 = 150 (too high) Re-examining: Let y = ₱5 bills, x = ₱10 bills y = x + 3 and 5y + 10x = 140 Substituting: 5(x + 3) + 10x = 140 15x + 15 = 140, so x = 125/15 = 25/3 This suggests there might be an error in the problem setup. Let's verify: if x = 8, then y = 11, total = 5(11) + 10(8) = 55 + 80 = 135

Key Points

  • Linear equations represent constant rates of change
  • Always perform same operation on both sides
  • Check solutions by substituting back into original equation
  • Word problems require careful translation to mathematical language
  • Systems can have one solution, no solution, or infinite solutions

Quadratic equations have the form ax² + bx + c = 0 where a ≠ 0. These equations appear frequently in physics, economics, and engineering applications. **Methods for Solving Quadratic Equations:** **1. Factoring Method:** Step 1: Write in standard form (ax² + bx + c = 0) Step 2: Factor the quadratic expression Step 3: Set each factor equal to zero Step 4: Solve each linear equation **2. Square Root Method (for equations like x² = k):** Step 1: Isolate the squared term Step 2: Take square root of both sides Step 3: Remember both positive and negative roots **3. Completing the Square:** Step 1: Ensure coefficient of x² is 1 Step 2: Move constant to right side Step 3: Add (b/2)² to both sides Step 4: Factor left side as perfect square Step 5: Apply square root method **4. Quadratic Formula:** For ax² + bx + c = 0: x = (-b ± √(b² - 4ac))/(2a) **The Discriminant (b² - 4ac):** - If b² - 4ac > 0: Two real solutions - If b² - 4ac = 0: One real solution (repeated root) - If b² - 4ac < 0: No real solutions (complex solutions) **Applications:** Quadratic equations model many real situations: - Projectile motion: h = -16t² + v₀t + h₀ - Area problems: A = lw where l and w are related - Profit maximization: P = -ax² + bx + c

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Quadratic Equations and Applications

Examples

  • **Example 1: Solving by Factoring** Solve: 2x² - 8x + 6 = 0 **Step 1:** Factor out GCF 2(x² - 4x + 3) = 0 x² - 4x + 3 = 0 **Step 2:** Factor the quadratic Need factors of 3 that add to -4: -3 and -1 (x - 3)(x - 1) = 0 **Step 3:** Set each factor to zero x - 3 = 0 or x - 1 = 0 x = 3 or x = 1 **Step 4:** Check both solutions For x = 3: 2(9) - 8(3) + 6 = 18 - 24 + 6 = 0 ✓ For x = 1: 2(1) - 8(1) + 6 = 2 - 8 + 6 = 0 ✓
  • **Example 2: Using Quadratic Formula** Solve: x² + 3x - 2 = 0 **Step 1:** Identify a, b, c a = 1, b = 3, c = -2 **Step 2:** Calculate discriminant b² - 4ac = 9 - 4(1)(-2) = 9 + 8 = 17 Since 17 > 0, there are two real solutions **Step 3:** Apply quadratic formula x = (-3 ± √17)/(2(1)) x = (-3 ± √17)/2 **Step 4:** Write both solutions x = (-3 + √17)/2 ≈ 0.56 x = (-3 - √17)/2 ≈ -2.56

Key Points

  • Quadratic equations can have 0, 1, or 2 real solutions
  • Choose solution method based on the form of the equation
  • Discriminant tells you about nature of solutions without solving
  • Always check solutions in original equation
  • Graph is a parabola opening up (a > 0) or down (a < 0)

A function is a special relationship between inputs and outputs where each input has exactly one output. Functions are fundamental to understanding mathematical relationships and modeling real-world phenomena. **Function Definition and Notation:** - A function f assigns exactly one output value f(x) to each input value x - Domain: Set of all possible input values (x-values) - Range: Set of all possible output values (y-values or f(x)-values) - Function notation: y = f(x) (read as "f of x") **Types of Functions:** 1. **Linear Functions**: f(x) = mx + b (straight line) 2. **Quadratic Functions**: f(x) = ax² + bx + c (parabola) 3. **Polynomial Functions**: f(x) = aₙxⁿ + ... + a₁x + a₀ 4. **Rational Functions**: f(x) = P(x)/Q(x) where P and Q are polynomials 5. **Radical Functions**: f(x) = √(g(x)) **Testing for Functions:** **Vertical Line Test**: A graph represents a function if every vertical line intersects it at most once. **Finding Domain:** Step 1: Identify restrictions - Division by zero: denominator ≠ 0 - Even roots: radicand ≥ 0 - Logarithms: argument > 0 Step 2: Express domain in interval notation or set-builder notation **Function Operations:** - Addition: (f + g)(x) = f(x) + g(x) - Subtraction: (f - g)(x) = f(x) - g(x) - Multiplication: (f · g)(x) = f(x) · g(x) - Division: (f/g)(x) = f(x)/g(x), g(x) ≠ 0 - Composition: (f ∘ g)(x) = f(g(x))

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Functions and Their Applications

Examples

  • **Example 1: Finding Domain** Find the domain of f(x) = √(x - 3)/(x + 2) **Step 1:** Identify restrictions - Radical: x - 3 ≥ 0, so x ≥ 3 - Denominator: x + 2 ≠ 0, so x ≠ -2 **Step 2:** Combine restrictions Since x ≥ 3 and x ≠ -2, and 3 > -2, the condition x ≠ -2 is automatically satisfied. **Domain:** x ≥ 3 or [3, ∞) in interval notation
  • **Example 2: Function Composition** Given f(x) = 2x + 1 and g(x) = x² - 3, find (f ∘ g)(x) and (g ∘ f)(x) **Finding (f ∘ g)(x) = f(g(x)):** Step 1: Substitute g(x) into f(x) f(g(x)) = f(x² - 3) = 2(x² - 3) + 1 Step 2: Simplify = 2x² - 6 + 1 = 2x² - 5 **Finding (g ∘ f)(x) = g(f(x)):** Step 1: Substitute f(x) into g(x) g(f(x)) = g(2x + 1) = (2x + 1)² - 3 Step 2: Simplify = 4x² + 4x + 1 - 3 = 4x² + 4x - 2 **Note:** (f ∘ g)(x) ≠ (g ∘ f)(x) in general

Key Points

  • Functions have exactly one output for each input
  • Domain restrictions come from mathematical limitations
  • Vertical line test determines if a graph is a function
  • Function composition means applying one function to the result of another
  • Real-world relationships are often modeled with functions
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