1.2 Algebraic Expressions, Polynomials & Linear Equations

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.

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1. Core Algebraic Identities & Factoring Patterns

Instant recognition of factoring structures enables you to reduce rational expressions within seconds:

1. Difference of Two Squares: $$a^2 - b^2 = (a - b)(a + b)$$
2. Perfect Square Trinomials: $$a^2 \pm 2ab + b^2 = (a \pm b)^2$$
3. Sum and Difference of Two Cubes: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
4. Square of a Trinomial: $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
5. Cube of a Binomial: $$(a \pm b)^3 = a^3 \pm 3a^2 b + 3ab^2 \pm b^3$$

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.

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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:

$$\begin{array}{r|rrrr} 3 & 2 & -5 & 4 & -7 \\ & & 6 & 3 & 21 \\ \hline & 2 & 1 & 7 & 14 \end{array}$$

Quotient: $2x^2 + x + 7$, Remainder: $14$.

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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)

$$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$$ $$x = \frac{D_x}{D} = \frac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{D_y}{D} = \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}$$

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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: $|2x - 3| \le 3$.

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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: $[a, b) \iff a \le x < b$

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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.

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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 $k$.

$$\frac{-5}{2k - 1} = \frac{1}{2} \implies 1 \cdot (2k - 1) = 2 \cdot (-5) \implies 2k - 1 = -10$$ $$2k = -9 \implies k = -\frac{9}{2} = -4.5$$

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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 $x = 0$.

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ACET Speed Tactics & Traps Summary