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ACET (Ateneo College Entrance Test)

Practice QuestionsStudy Guide

1.2 Algebraic Expressions, Polynomials & Linear Equations

Key Takeaways

  • 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:

  1. Difference of Two Squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)
  2. Perfect Square Trinomials: a2±2ab+b2=(a±b)2a^2 \pm 2ab + b^2 = (a \pm b)^2
  3. Sum and Difference of Two Cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2) a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)
  4. Square of a Trinomial: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
  5. Cube of a Binomial: (a±b)3=a3±3a2b+3ab2±b3(a \pm b)^3 = a^3 \pm 3a^2 b + 3ab^2 \pm b^3

ACET Speed Tactic (Expression Simplification): Simplify $\frac{x^3 - 8}{x^2 - 4}$. Recognize $x^3 - 8$ as a difference of cubes $(x - 2)(x^2 + 2x + 4)$ and $x^2 - 4$ as a difference of squares $(x - 2)(x + 2)$. Canceling $(x - 2)$ yields $\frac{x^2 + 2x + 4}{x + 2}$ immediately.

2. Polynomial Arithmetic & Division

Polynomial Long Division vs. Synthetic Division

Synthetic division provides a streamlined algorithm when dividing a polynomial $P(x)$ by a linear factor of the form $(x - c)$:

To divide $P(x) = 2x^3 - 5x^2 + 4x - 7$ by $(x - 3)$:

  1. Write the root $c = 3$ on the left.
  2. List the coefficients: 2, -5, 4, -7.
  3. Bring down the leading coefficient 2.
  4. Multiply by $c=3$ and add down iteratively:
3 & 2 & -5 & 4 & -7 \\ & & 6 & 3 & 21 \\ \hline & 2 & 1 & 7 & 14 \end{array}$$ Quotient: $2x^2 + x + 7$, Remainder: $14$. --- ## 3. Systems of Linear Equations in Two and Three Variables For a $2 \times 2$ system: $$\begin{aligned} a_1 x + b_1 y &= c_1 \\ a_2 x + b_2 y &= c_2 \end{aligned}$$ ### System Classifications & Geometric Interpretations - **Consistent & Independent:** Unique solution exists. Lines intersect at one point. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. - **Inconsistent:** No solution exists. Lines are parallel and distinct. $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. - **Consistent & Dependent:** Infinitely many solutions. Lines are coincident. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. ### Determinants & Cramer's Rule ($2 \times 2$ Matrix) $$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = 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}$$ --- ## 4. Absolute Value Equations & Inequalities Absolute value represents distance on a real number line: $|x| = \sqrt{x^2} \ge 0$. ### Key Analytical Rules 1. **Equality:** $|ax + b| = c \iff ax + b = c \quad \text{or} \quad ax + b = -c \quad (c \ge 0)$. 2. **Bounded Inequality ("Less Than"):** $|ax + b| \le c \iff -c \le ax + b \le c$. 3. **Unbounded Inequality ("Greater Than"):** $|ax + b| \ge c \iff ax + b \ge c \quad \text{or} \quad ax + b \le -c$. 4. **Triangle Inequality:** $|a + b| \le |a| + |b|$. > **CRITICAL ACET TRAP:** When solving $-4 |2x - 3| \ge -12$, you MUST divide by $-4$ first and **flip the inequality sign** before removing absolute value bars: $|2x - 3| \le 3$. --- ## 5. Linear & Compound Inequalities When multiplying or dividing both sides of an inequality by a negative quantity, the inequality symbol **must be reversed**. ### Interval Notation - Open interval: $(a, b) \iff a < x < b$ - Closed interval: $[a, b] \iff a \le x \le b$ - Half-open interval: $[a, b) \iff a \le x < b$ --- ## 6. Rational & Radical Equations: Spotting Extraneous Roots Extraneous roots occur when mathematical transformations (such as squaring both sides or multiplying by a variable denominator) introduce invalid solutions that violate domain constraints. ### Radical Equation Rule Squaring both sides of $\sqrt{f(x)} = g(x)$ requires $g(x) \ge 0$. Always substitute candidate roots back into the **original un-squared equation**. --- ## Worked Step-by-Step ACET Exam Problems ### Problem 1: System of Equations with Dependent Parameters *Find the value of the constant $k$ such that the following system has infinitely many solutions:* $$\begin{aligned} 3x - 5y &= 12 \\ 6x + (2k - 1)y &= 24 \end{aligned}$$ **Step 1: Identify dependent system conditions.** For infinitely many solutions, the ratio of coefficients must be equal: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ **Step 2: Set up the ratio equality.** $$\frac{3}{6} = \frac{-5}{2k - 1} = \frac{12}{24}$$ Notice $\frac{3}{6} = \frac{1}{2}$ and $\frac{12}{24} = \frac{1}{2}$. **Step 3: Solve for $k$.** $$\frac{-5}{2k - 1} = \frac{1}{2} \implies 1 \cdot (2k - 1) = 2 \cdot (-5) \implies 2k - 1 = -10$$ $$2k = -9 \implies k = -\frac{9}{2} = -4.5$$ --- ### Problem 2: Radical Equation with Extraneous Root Detection *Solve for all real values of $x$: $\sqrt{2x + 9} - x = 3$.* **Step 1: Isolate the radical term.** $$\sqrt{2x + 9} = x + 3$$ **Step 2: Square both sides.** $$(\sqrt{2x + 9})^2 = (x + 3)^2$$ $$2x + 9 = x^2 + 6x + 9$$ **Step 3: Move all terms to one side and factor.** $$x^2 + 4x = 0 \implies x(x + 4) = 0$$ Candidate roots: $x = 0$ or $x = -4$. **Step 4: Check candidate solutions in original equation.** - Check $x = 0$: $$\sqrt{2(0) + 9} - 0 = \sqrt{9} = 3 = 3 \quad \text{(VALID)}$$ - Check $x = -4$: $$\sqrt{2(-4) + 9} - (-4) = \sqrt{1} + 4 = 1 + 4 = 5 \neq 3 \quad \text{(EXTRANEOUS)}$$ Therefore, the only valid real solution is $x = 0$. --- ## 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 $|x| \ge c$ to $-c \le x \le c$ | Remember: "Greater than" means OR ($x \ge c$ or $x \le -c$) | | **Linear Systems** | Performing full substitution when coefficients match | Use vertical elimination to cancel terms instantly |
Test Your Knowledge

What is the complete factorization of the expression 3x^3 - 12x?

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Test Your Knowledge

Solve the compound inequality for x: -7 < 3 - 2x <= 11.

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Test Your Knowledge

If x + y = 10 and x^2 - y^2 = 40, what is the value of x - y?

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1.3 Functions, Domain/Range, Graphs & Transformations

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