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1.2 Algebraic Expressions, Polynomials & Linear Equations
Ateneo College Entrance Test ACET
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1 Chapter 1: Mathematics
1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics
2 Chapter 2: English Proficiency
2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words
3 Chapter 3: Reading Comprehension
3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments
4 Chapter 4: Logical and Abstract Reasoning
4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic
5 Chapter 5: General Knowledge
5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia
6 Chapter 6: ACET Essay Writing & Exam Strategies
6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness
1.2 Algebraic Expressions, Polynomials & Linear Equations
Key Takeaways Explain This Topic AI
Recognize special polynomial products and algebraic identities instantly to simplify complex rational expressions without expanding.
Solve linear systems rapidly using elimination or determinant-based cross-multiplication strategies.
Always reverse inequality signs when multiplying or dividing by negative quantities in single or compound inequalities.
Validate radical and rational equation solutions against domain boundaries to eliminate extraneous roots.
Last updated: July 2026
Algebraic Expressions, Polynomials & Linear Equations
Algebra serves as the mathematical backbone of the ACET. A substantial portion of the exam tests your ability to manipulate expressions, factor high-degree polynomials, resolve systems of linear equations under strict time limits, and solve inequalities without making sign errors.
1. Core Algebraic Identities & Factoring Patterns
Instant recognition of factoring structures enables you to reduce rational expressions within seconds:
Difference of Two 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 )
Perfect Square Trinomials: a 2 ± 2 a b + b 2 = ( a ± b ) 2 a^2 \pm 2ab + b^2 = (a \pm b)^2 a 2 ± 2 ab + b 2 = ( a ± b ) 2
Sum and Difference of Two Cubes: a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3 + b^3 = (a + b)(a^2 - ab + b^2) a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 )
Square of a Trinomial: ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
Cube of a Binomial: ( a ± b ) 3 = a 3 ± 3 a 2 b + 3 a b 2 ± b 3 (a \pm b)^3 = a^3 \pm 3a^2 b + 3ab^2 \pm b^3 ( a ± b ) 3 = a 3 ± 3 a 2 b + 3 a b 2 ± b 3
ACET Speed Tactic (Expression Simplification): Simplify x 3 − 8 x 2 − 4 \frac{x^3 - 8}{x^2 - 4} x 2 − 4 x 3 − 8 .
Recognize x 3 − 8 x^3 - 8 x 3 − 8 as a difference of cubes ( x − 2 ) ( x 2 + 2 x + 4 ) (x - 2)(x^2 + 2x + 4) ( x − 2 ) ( x 2 + 2 x + 4 ) and x 2 − 4 x^2 - 4 x 2 − 4 as a difference of squares ( x − 2 ) ( x + 2 ) (x - 2)(x + 2) ( x − 2 ) ( x + 2 ) . Canceling ( x − 2 ) (x - 2) ( x − 2 ) yields x 2 + 2 x + 4 x + 2 \frac{x^2 + 2x + 4}{x + 2} x + 2 x 2 + 2 x + 4 immediately.
- 2. Polynomial Arithmetic & Division
- Polynomial Long Division vs. Synthetic Division
Synthetic division provides a streamlined algorithm when dividing a polynomial P ( x ) P(x) P ( x ) by a linear factor of the form ( x − c ) (x - c) ( x − c ) :
To divide P ( x ) = 2 x 3 − 5 x 2 + 4 x − 7 P(x) = 2x^3 - 5x^2 + 4x - 7 P ( x ) = 2 x 3 − 5 x 2 + 4 x − 7 by ( x − 3 ) (x - 3) ( x − 3 ) :
Write the root c = 3 c = 3 c = 3 on the left.
List the coefficients: 2, -5, 4, -7 .
Bring down the leading coefficient 2 .
- Multiply by c = 3 c=3 c = 3 and add down iteratively: 3 2 − 5 4 − 7 6 3 21 2 1 7 14 \begin{array}{r|rrrr}
- 3 & 2 & -5 & 4 & -7 \\
- & & 6 & 3 & 21 \\
- \hline
- & 2 & 1 & 7 & 14
- \end{array} 3 2 2 − 5 6 1 4 3 7 − 7 21 14
Quotient: 2 x 2 + x + 7 2x^2 + x + 7 2 x 2 + x + 7 , Remainder: 14 14 14 .
- 3. Systems of Linear Equations in Two and Three Variables
- For a 2 × 2 2 \times 2 2 × 2 system: a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 \begin{aligned}
- a_1 x + b_1 y &= c_1 \\
- a_2 x + b_2 y &= c_2
- \end{aligned} a 1 x + b 1 y a 2 x + b 2 y = c 1 = c 2
- System Classifications & Geometric Interpretations
Consistent & Independent: Unique solution exists. Lines intersect at one point. a 1 a 2 ≠ b 1 b 2 \frac{a_1}{a_2} \neq \frac{b_1}{b_2} a 2 a 1 = b 2 b 1 .
Inconsistent: No solution exists. Lines are parallel and distinct. a 1 a 2 = b 1 b 2 ≠ c 1 c 2 \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} a 2 a 1 = b 2 b 1 = c 2 c 1 .
Consistent & Dependent: Infinitely many solutions. Lines are coincident. a 1 a 2 = b 1 b 2 = c 1 c 2 \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} a 2 a 1 = b 2 b 1 = c 2 c 1 .
Determinants & Cramer's Rule ( 2 × 2 2 \times 2 2 × 2 Matrix) D = ∣ a 1 b 1 a 2 b 2 ∣ = a 1 b 2 − a 2 b 1 D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1 D = a 1 a 2 b 1 b 2 = a 1 b 2 − a 2 b 1 x = D x D = c 1 b 2 − c 2 b 1 a 1 b 2 − a 2 b 1 , y = D y D = a 1 c 2 − a 2 c 1 a 1 b 2 − a 2 b 1 x = \frac{D_x}{D} = \frac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{D_y}{D} = \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1} x = D D x = a 1 b 2 − a 2 b 1 c 1 b 2 − c 2 b 1 , y = D D y = a 1 b 2 − a 2 b 1 a 1 c 2 − a 2 c 1
4. Absolute Value Equations & Inequalities
Absolute value represents distance on a real number line: ∣ x ∣ = x 2 ≥ 0 |x| = \sqrt{x^2} \ge 0 ∣ x ∣ = x 2 ≥ 0 .
Key Analytical Rules
Equality: ∣ a x + b ∣ = c ⟺ a x + b = c or a x + b = − c ( c ≥ 0 ) |ax + b| = c \iff ax + b = c \quad \text{or} \quad ax + b = -c \quad (c \ge 0) ∣ a x + b ∣ = c ⟺ a x + b = c or a x + b = − c ( c ≥ 0 ) .
Bounded Inequality ("Less Than"): ∣ a x + b ∣ ≤ c ⟺ − c ≤ a x + b ≤ c |ax + b| \le c \iff -c \le ax + b \le c ∣ a x + b ∣ ≤ c ⟺ − c ≤ a x + b ≤ c .
Unbounded Inequality ("Greater Than"): ∣ a x + b ∣ ≥ c ⟺ a x + b ≥ c or a x + b ≤ − c |ax + b| \ge c \iff ax + b \ge c \quad \text{or} \quad ax + b \le -c ∣ a x + b ∣ ≥ c ⟺ a x + b ≥ c or a x + b ≤ − c .
Triangle Inequality: ∣ a + b ∣ ≤ ∣ a ∣ + ∣ b ∣ |a + b| \le |a| + |b| ∣ a + b ∣ ≤ ∣ a ∣ + ∣ b ∣ .
CRITICAL ACET TRAP: When solving − 4 ∣ 2 x − 3 ∣ ≥ − 12 -4 |2x - 3| \ge -12 − 4∣2 x − 3∣ ≥ − 12 , you MUST divide by − 4 -4 − 4 first and flip the inequality sign before removing absolute value bars: ∣ 2 x − 3 ∣ ≤ 3 |2x - 3| \le 3 ∣2 x − 3∣ ≤ 3 .
5. Linear & Compound Inequalities
When multiplying or dividing both sides of an inequality by a negative quantity, the inequality symbol must be reversed .
- Interval Notation
- Open interval: ( a , b ) ⟺ a < x < b (a, b) \iff a < x < b ( a , b ) ⟺ a < x < b
- Closed interval: [ a , b ] ⟺ a ≤ x ≤ b [a, b] \iff a \le x \le b [ a , b ] ⟺ a ≤ x ≤ b
- Half-open interval: [ a , b ) ⟺ a ≤ x < b [a, b) \iff a \le x < b [ a , b ) ⟺ a ≤ x < b
- 6. Rational & Radical Equations: Spotting Extraneous Roots
Extraneous roots occur when mathematical transformations (such as squaring both sides or multiplying by a variable denominator) introduce invalid solutions that violate domain constraints.
Radical Equation Rule
Squaring both sides of f ( x ) = g ( x ) \sqrt{f(x)} = g(x) f ( x ) = g ( x ) requires g ( x ) ≥ 0 g(x) \ge 0 g ( x ) ≥ 0 . Always substitute candidate roots back into the original un-squared equation .
- Worked Step-by-Step ACET Exam Problems
- Problem 1: System of Equations with Dependent Parameters
Find the value of the constant k k k such that the following system has infinitely many solutions: 3 x − 5 y = 12 6 x + ( 2 k − 1 ) y = 24 \begin{aligned}
- 3x - 5y &= 12 \\
- 6x + (2k - 1)y &= 24
- \end{aligned} 3 x − 5 y 6 x + ( 2 k − 1 ) y = 12 = 24
Step 1: Identify dependent system conditions. For infinitely many solutions, the ratio of coefficients must be equal: a 1 a 2 = b 1 b 2 = c 1 c 2 \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} a 2 a 1 = b 2 b 1 = c 2 c 1
Step 2: Set up the ratio equality. 3 6 = − 5 2 k − 1 = 12 24 \frac{3}{6} = \frac{-5}{2k - 1} = \frac{12}{24} 6 3 = 2 k − 1 − 5 = 24 12
Notice 3 6 = 1 2 \frac{3}{6} = \frac{1}{2} 6 3 = 2 1 and 12 24 = 1 2 \frac{12}{24} = \frac{1}{2} 24 12 = 2 1 .
Step 3: Solve for k k k . − 5 2 k − 1 = 1 2 ⟹ 1 ⋅ ( 2 k − 1 ) = 2 ⋅ ( − 5 ) ⟹ 2 k − 1 = − 10 \frac{-5}{2k - 1} = \frac{1}{2} \implies 1 \cdot (2k - 1) = 2 \cdot (-5) \implies 2k - 1 = -10 2 k − 1 − 5 = 2 1 ⟹ 1 ⋅ ( 2 k − 1 ) = 2 ⋅ ( − 5 ) ⟹ 2 k − 1 = − 10 2 k = − 9 ⟹ k = − 9 2 = − 4.5 2k = -9 \implies k = -\frac{9}{2} = -4.5 2 k = − 9 ⟹ k = − 2 9 = − 4.5
Problem 2: Radical Equation with Extraneous Root Detection
Solve for all real values of x x x : 2 x + 9 − x = 3 \sqrt{2x + 9} - x = 3 2 x + 9 − x = 3 .
Step 1: Isolate the radical term. 2 x + 9 = x + 3 \sqrt{2x + 9} = x + 3 2 x + 9 = x + 3
Step 2: Square both sides. ( 2 x + 9 ) 2 = ( x + 3 ) 2 (\sqrt{2x + 9})^2 = (x + 3)^2 ( 2 x + 9 ) 2 = ( x + 3 ) 2 2 x + 9 = x 2 + 6 x + 9 2x + 9 = x^2 + 6x + 9 2 x + 9 = x 2 + 6 x + 9
Step 3: Move all terms to one side and factor. x 2 + 4 x = 0 ⟹ x ( x + 4 ) = 0 x^2 + 4x = 0 \implies x(x + 4) = 0 x 2 + 4 x = 0 ⟹ x ( x + 4 ) = 0
Candidate roots: x = 0 x = 0 x = 0 or x = − 4 x = -4 x = − 4 .
Step 4: Check candidate solutions in original equation.
Check x = 0 x = 0 x = 0 : 2 ( 0 ) + 9 − 0 = 9 = 3 = 3 (VALID) \sqrt{2(0) + 9} - 0 = \sqrt{9} = 3 = 3 \quad \text{(VALID)} 2 ( 0 ) + 9 − 0 = 9 = 3 = 3 (VALID)
Check x = − 4 x = -4 x = − 4 : 2 ( − 4 ) + 9 − ( − 4 ) = 1 + 4 = 1 + 4 = 5 ≠ 3 (EXTRANEOUS) \sqrt{2(-4) + 9} - (-4) = \sqrt{1} + 4 = 1 + 4 = 5 \neq 3 \quad \text{(EXTRANEOUS)} 2 ( − 4 ) + 9 − ( − 4 ) = 1 + 4 = 1 + 4 = 5 = 3 (EXTRANEOUS)
Therefore, the only valid real solution is x = 0 x = 0 x = 0 .
ACET Speed Tactics & Traps Summary
- Problem Type Common Student Trap ACET Speed Tactic
- Inequalities Forgetting to flip sign when dividing by negative Highlight negative division steps immediately
- Radical Equations Keeping extraneous roots Check candidate roots in original un-squared equation
Absolute Value Converting ∣ x ∣ ≥ c \vert x\vert\ge c ∣ x ∣ ≥ c to − c ≤ x ≤ c -c \le x \le c − c ≤ x ≤ c Remember: "Greater than" means OR ( x ≥ c x \ge c x ≥ c or x ≤ − c x \le -c x ≤ − c )
Linear Systems Performing full substitution when coefficients match Use vertical elimination to cancel terms instantly