2.2 Algebra, Polynomials, and Linear/Quadratic Equations

Adamson University College Entrance Test AdUCET

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

2.2 Algebra, Polynomials, and Linear/Quadratic Equations

Key Takeaways Explain This Topic AI

The quadratic formula x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} x = 2 a − b ± b 2 − 4 a c ​ ​ yields two distinct real roots when Δ > 0 \Delta > 0 Δ > 0 , one real repeated root when Δ = 0 \Delta = 0 Δ = 0 , and complex conjugate roots when Δ < 0 \Delta < 0 Δ < 0 .

Factoring identities like difference of squares a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) simplify complex polynomial expressions and speed up equation solving without full expansion.

Exponent rules require adding exponents when multiplying like bases ( x a ⋅ x b = x a + b x^a \cdot x^b = x^{a+b} x a ⋅ x b = x a + b ), multiplying exponents for powers ( ( x a ) b = x a b (x^a)^b = x^{ab} ( x a ) b = x ab ), and converting fractional powers to radicals ( x m / n = x m n x^{m/n} = \sqrt[n]{x^m} x m / n = n x m ​ ).

Logarithmic equations rely on the equivalence log ⁡ b ( x ) = y ⟺ b y = x \log_b(x) = y \iff b^y = x lo g b ​ ( x ) = y ⟺ b y = x and combining terms via product/quotient properties before converting to exponential form.

Work rate problems calculate combined completion time T T T using reciprocal sum of individual rates 1 T = 1 t 1 + 1 t 2 \frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} T 1 ​ = t 1 ​ 1 ​ + t 2 ​ 1 ​ .

Last updated: August 2026

Algebraic manipulation and equation solving form the quantitative foundation of the Adamson University College Entrance Test (AdUCET). Mastery of algebraic operations, factoring techniques, systems of equations, quadratic behavior, logarithmic rules, and multi-step word problems is required for rapid and accurate problem solving.

1. Algebraic Simplification and Exponent Rules

Simplifying algebraic expressions involves expanding products using the distributive property, combining like terms, and applying standard laws of exponents.

Exponent Rules Summary

Rule Name Mathematical Identity Operational Context & Example

Product Rule x a ⋅ x b = x a + b x^a \cdot x^b = x^{a+b} x a ⋅ x b = x a + b Add exponents when multiplying like bases: 2 3 ⋅ 2 4 = 2 7 = 128 2^3 \cdot 2^4 = 2^7 = 128 2 3 ⋅ 2 4 = 2 7 = 128

Quotient Rule x a x b = x a − b \frac{x^a}{x^b} = x^{a-b} x b x a ​ = x a − b Subtract denominator exponent from numerator exponent: y 8 y 3 = y 5 \frac{y^8}{y^3} = y^5 y 3 y 8 ​ = y 5

Power Rule ( x a ) b = x a b (x^a)^b = x^{ab} ( x a ) b = x ab Multiply inner and outer exponents: ( z 3 ) 4 = z 12 (z^3)^4 = z^{12} ( z 3 ) 4 = z 12

Negative Exponent x − a = 1 x a x^{-a} = \frac{1}{x^a} x − a = x a 1 ​ Invert the base to make exponent positive: 5 − 2 = 1 5 2 = 1 25 5^{-2} = \frac{1}{5^2} = \frac{1}{25} 5 − 2 = 5 2 1 ​ = 25 1 ​

Zero Exponent x 0 = 1 ( x ≠ 0 ) x^0 = 1 \quad (x \neq 0) x 0 = 1 ( x  = 0 ) Any non-zero base raised to power zero equals 1 1 1 : ( 17 x y 2 ) 0 = 1 (17xy^2)^0 = 1 ( 17 x y 2 ) 0 = 1

Fractional Exponent x m / n = x m n = ( x n ) m x^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m x m / n = n x m ​ = ( n x ​ ) m Denominator represents root, numerator represents power: 8 2 / 3 = ( 8 3 ) 2 = 2 2 = 4 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4 8 2/3 = ( 3 8 ​ ) 2 = 2 2 = 4

When simplifying complex expressions such as ( 3 x 2 y − 3 ) 2 9 x − 4 y 2 \frac{(3x^2 y^{-3})^2}{9x^{-4} y^2} 9 x − 4 y 2 ( 3 x 2 y − 3 ) 2 ​ , apply power rules to numerator first: 9 x 4 y − 6 9 x − 4 y 2 = 9 9 ⋅ x 4 − ( − 4 ) ⋅ y − 6 − 2 = x 8 y − 8 = x 8 y 8 \frac{9x^4 y^{-6}}{9x^{-4} y^2} = \frac{9}{9} \cdot x^{4 - (-4)} \cdot y^{-6 - 2} = x^8 y^{-8} = \frac{x^8}{y^8} 9 x − 4 y 2 9 x 4 y − 6 ​ = 9 9 ​ ⋅ x 4 − ( − 4 ) ⋅ y − 6 − 2 = x 8 y − 8 = y 8 x 8 ​ .

2. Polynomial Factoring Techniques

Factoring transforms polynomial expressions into products of simpler algebraic factors.

Key Factoring Identities

Greatest Common Factor (GCF): Factor out the largest monomial shared by all terms.

Example: 6 x 3 − 12 x 2 + 18 x = 6 x ( x 2 − 2 x + 3 ) 6x^3 - 12x^2 + 18x = 6x(x^2 - 2x + 3) 6 x 3 − 12 x 2 + 18 x = 6 x ( x 2 − 2 x + 3 ) .

Difference of Squares: Any expression of the form a 2 − b 2 a^2 - b^2 a 2 − b 2 factors directly into ( a − b ) ( a + b ) (a - b)(a + b) ( a − b ) ( a + b ) .

Example: 16 x 2 − 49 = ( 4 x − 7 ) ( 4 x + 7 ) 16x^2 - 49 = (4x - 7)(4x + 7) 16 x 2 − 49 = ( 4 x − 7 ) ( 4 x + 7 ) . Note that a sum of squares ( a 2 + b 2 a^2 + b^2 a 2 + b 2 ) does not factor over real numbers.

Factoring Quadratic Trinomials ( a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c ):

When a = 1 a = 1 a = 1 : Find two numbers r r r and s s s such that r ⋅ s = c r \cdot s = c r ⋅ s = c and r + s = b r + s = b r + s = b . Then x 2 + b x + c = ( x + r ) ( x + s ) x^2 + bx + c = (x + r)(x + s) x 2 + b x + c = ( x + r ) ( x + s ) .

When a ≠ 1 a \neq 1 a  = 1 : Use the grouping method. Find two numbers whose product is a c ac a c and whose sum is b b b , split the middle term, and factor by grouping.

3. Solving Linear Equations and Systems

A linear equation in one variable takes the form a x + b = 0 ax + b = 0 a x + b = 0 . Solving involves isolating the variable x x x through inverse arithmetic operations.

Systems of Linear Equations

When solving two simultaneous linear equations with two variables ( x x x and y y y ), AdUCET tests two primary techniques:

Substitution Method: Isolate one variable in one equation and substitute it into the second equation.

Elimination Method: Multiply equations by constants so that adding or subtracting the equations eliminates one variable.

• Worked Step-by-Step Example: System Elimination
• Solve the system: 2 x + 3 y = 18 — (Equation 1) 5 x − 2 y = 7 — (Equation 2) \begin{aligned}
• 2x + 3y &= 18 \quad \text{--- (Equation 1)} \\
• 5x - 2y &= 7 \quad \text{--- (Equation 2)}
• \end{aligned} 2 x + 3 y 5 x − 2 y ​ = 18 — (Equation 1) = 7 — (Equation 2) ​

Step 1: Multiply Equation 1 by 2 2 2 and Equation 2 by 3 3 3 to align the coefficients of y y y : 4 x + 6 y = 36 4x + 6y = 36 4 x + 6 y = 36 15 x − 6 y = 21 15x - 6y = 21 15 x − 6 y = 21

Step 2: Add the two resulting equations to eliminate y y y : ( 4 x + 15 x ) + ( 6 y − 6 y ) = 36 + 21 ⟹ 19 x = 57 ⟹ x = 3 (4x + 15x) + (6y - 6y) = 36 + 21 \implies 19x = 57 \implies x = 3 ( 4 x + 15 x ) + ( 6 y − 6 y ) = 36 + 21 ⟹ 19 x = 57 ⟹ x = 3

Step 3: Substitute x = 3 x = 3 x = 3 back into Equation 1: 2 ( 3 ) + 3 y = 18 ⟹ 6 + 3 y = 18 ⟹ 3 y = 12 ⟹ y = 4 2(3) + 3y = 18 \implies 6 + 3y = 18 \implies 3y = 12 \implies y = 4 2 ( 3 ) + 3 y = 18 ⟹ 6 + 3 y = 18 ⟹ 3 y = 12 ⟹ y = 4

Solution: ( x , y ) = ( 3 , 4 ) (x, y) = (3, 4) ( x , y ) = ( 3 , 4 ) .

4. Quadratic Equations and the Discriminant

A quadratic equation in standard form is written as a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , where a ≠ 0 a \neq 0 a  = 0 .

The Quadratic Formula

When factoring is difficult or impossible over integers, use the quadratic formula: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c ​ ​

The Discriminant ( Δ = b 2 − 4 a c \Delta = b^2 - 4ac Δ = b 2 − 4 a c )

The value under the radical, Δ = b 2 − 4 a c \Delta = b^2 - 4ac Δ = b 2 − 4 a c , dictates the nature and quantity of the roots:

• Discriminant Value Nature of Solutions Graphical Interpretation
• Δ > 0 \Delta > 0 Δ > 0 2 distinct real solutions Parabola intersects the x x x -axis at two distinct points

Δ = 0 \Delta = 0 Δ = 0 1 real solution (repeated root) Parabola vertex touches the x x x -axis at exactly one point

Δ < 0 \Delta < 0 Δ < 0 2 complex conjugate solutions (no real roots) Parabola does not intersect the x x x -axis

5. Fundamentals of Logarithms

A logarithm is the inverse operation of exponentiation. The fundamental equivalence is: log ⁡ b ( x ) = y ⟺ b y = x ( b > 0 , b ≠ 1 , x > 0 ) \log_b(x) = y \iff b^y = x \quad (b > 0, b \neq 1, x > 0) lo g b ​ ( x ) = y ⟺ b y = x ( b > 0 , b  = 1 , x > 0 )

Essential Logarithmic Properties

Product Rule: log ⁡ b ( M ⋅ N ) = log ⁡ b ( M ) + log ⁡ b ( N ) \log_b(M \cdot N) = \log_b(M) + \log_b(N) lo g b ​ ( M ⋅ N ) = lo g b ​ ( M ) + lo g b ​ ( N )

Quotient Rule: log ⁡ b ( M N ) = log ⁡ b ( M ) − log ⁡ b ( N ) \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) lo g b ​ ( N M ​ ) = lo g b ​ ( M ) − lo g b ​ ( N )

Power Rule: log ⁡ b ( M k ) = k ⋅ log ⁡ b ( M ) \log_b(M^k) = k \cdot \log_b(M) lo g b ​ ( M k ) = k ⋅ lo g b ​ ( M )

Identity & Zero Rules: log ⁡ b ( b ) = 1 \log_b(b) = 1 lo g b ​ ( b ) = 1 and log ⁡ b ( 1 ) = 0 \log_b(1) = 0 lo g b ​ ( 1 ) = 0

Change of Base Formula: log ⁡ b ( x ) = log ⁡ c ( x ) log ⁡ c ( b ) \log_b(x) = \frac{\log_c(x)}{\log_c(b)} lo g b ​ ( x ) = l o g c ​ ( b ) l o g c ​ ( x ) ​

Worked Example: Logarithmic Equation

Solve for x x x : log ⁡ 2 ( x ) + log ⁡ 2 ( x − 3 ) = 2 \log_2(x) + \log_2(x - 3) = 2 lo g 2 ​ ( x ) + lo g 2 ​ ( x − 3 ) = 2 .

Apply Product Rule: log ⁡ 2 ( x ( x − 3 ) ) = 2 \log_2(x(x - 3)) = 2 lo g 2 ​ ( x ( x − 3 )) = 2 .

Convert to exponential form: x ( x − 3 ) = 2 2 ⟹ x 2 − 3 x = 4 ⟹ x 2 − 3 x − 4 = 0 x(x - 3) = 2^2 \implies x^2 - 3x = 4 \implies x^2 - 3x - 4 = 0 x ( x − 3 ) = 2 2 ⟹ x 2 − 3 x = 4 ⟹ x 2 − 3 x − 4 = 0 .

Factor: ( x − 4 ) ( x + 1 ) = 0 ⟹ x = 4 (x - 4)(x + 1) = 0 \implies x = 4 ( x − 4 ) ( x + 1 ) = 0 ⟹ x = 4 or x = − 1 x = -1 x = − 1 .

Check domain ( x > 0 x > 0 x > 0 and x − 3 > 0 x - 3 > 0 x − 3 > 0 ): x = − 1 x = -1 x = − 1 is extraneous. Thus, the only valid solution is x = 4 x = 4 x = 4 .

6. Applied Multi-Step Word Problems

AdUCET word problems translate English scenarios into algebraic expressions.

Problem Types & Models

Cost, Revenue, and Profit:

Profit P ( x ) = R ( x ) − C ( x ) P(x) = R(x) - C(x) P ( x ) = R ( x ) − C ( x ) , where Revenue R ( x ) = p ⋅ x R(x) = p \cdot x R ( x ) = p ⋅ x and Total Cost C ( x ) = Fixed Cost + Variable Cost ⋅ x C(x) = \text{Fixed Cost} + \text{Variable Cost} \cdot x C ( x ) = Fixed Cost + Variable Cost ⋅ x . Break-even occurs when P ( x ) = 0 P(x) = 0 P ( x ) = 0 .

Work Rate Model:

If Worker A finishes a job in t 1 t_1 t 1 ​ hours and Worker B finishes in t 2 t_2 t 2 ​ hours, their combined time T T T satisfies: 1 T = 1 t 1 + 1 t 2 ⟹ T = t 1 t 2 t 1 + t 2 \frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} \implies T = \frac{t_1 t_2}{t_1 + t_2} T 1 ​ = t 1 ​ 1 ​ + t 2 ​ 1 ​ ⟹ T = t 1 ​ + t 2 ​ t 1 ​ t 2 ​ ​

Motion / Distance-Rate-Time:

d = r ⋅ t d = r \cdot t d = r ⋅ t . For round trips or opposing motion, set up equations balancing total distance or time differences.

• Step-by-Step Worked Word Problems
• Work Rate Example

Pipe A can fill a reservoir in 6 hours, while Pipe B can fill it in 12 hours. How many hours will it take to fill the reservoir if both pipes operate simultaneously?

Step 1: Identify individual hourly rates. Pipe A's rate is 1 6 \frac{1}{6} 6 1 ​ reservoir/hour; Pipe B's rate is 1 12 \frac{1}{12} 12 1 ​ reservoir/hour.

Step 2: Combine rates: 1 6 + 1 12 = 2 12 + 1 12 = 3 12 = 1 4 \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} 6 1 ​ + 12 1 ​ = 12 2 ​ + 12 1 ​ = 12 3 ​ = 4 1 ​ reservoir/hour.

Step 3: Calculate total time T = 1 Combined Rate = 1 1 / 4 = 4 T = \frac{1}{\text{Combined Rate}} = \frac{1}{1/4} = 4 T = Combined Rate 1 ​ = 1/4 1 ​ = 4 hours.

Age Problem Example

A father is currently 3 times as old as his daughter. In 12 years, the sum of their ages will be 80. How old is the daughter now?

Step 1: Define variables. Let daughter's current age be d d d . Father's current age is 3 d 3d 3 d .

Step 2: Express future ages in 12 years. Daughter will be d + 12 d + 12 d + 12 ; Father will be 3 d + 12 3d + 12 3 d + 12 .

Step 3: Formulate and solve the sum equation: ( d + 12 ) + ( 3 d + 12 ) = 80 ⟹ 4 d + 24 = 80 ⟹ 4 d = 56 ⟹ d = 14 (d + 12) + (3d + 12) = 80 \implies 4d + 24 = 80 \implies 4d = 56 \implies d = 14 ( d + 12 ) + ( 3 d + 12 ) = 80 ⟹ 4 d + 24 = 80 ⟹ 4 d = 56 ⟹ d = 14 The daughter is currently 14 years old.