2.2 Algebra, Polynomials, and Linear/Quadratic Equations

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

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

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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: $6x^3 - 12x^2 + 18x = 6x(x^2 - 2x + 3)$.
• Difference of Squares: Any expression of the form $a^2 - b^2$ factors directly into $(a - b)(a + b)$.
  Example: $16x^2 - 49 = (4x - 7)(4x + 7)$. Note that a sum of squares ($a^2 + b^2$) does not factor over real numbers.
• Factoring Quadratic Trinomials ($ax^2 + bx + c$):
  • When $a = 1$: Find two numbers $r$ and $s$ such that $r \cdot s = c$ and $r + s = b$. Then $x^2 + bx + c = (x + r)(x + s)$.
  • When $a \neq 1$: Use the grouping method. Find two numbers whose product is $ac$ and whose sum is $b$, split the middle term, and factor by grouping.

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3. Solving Linear Equations and Systems

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

Systems of Linear Equations

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

1. Substitution Method: Isolate one variable in one equation and substitute it into the second equation.
2. 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:

$$\begin{aligned} 2x + 3y &= 18 \quad \text{--- (Equation 1)} \\ 5x - 2y &= 7 \quad \text{--- (Equation 2)} \end{aligned}$$

• Step 1: Multiply Equation 1 by $2$ and Equation 2 by $3$ to align the coefficients of $y$: $$4x + 6y = 36$$ $$15x - 6y = 21$$
• Step 2: Add the two resulting equations to eliminate $y$: $$(4x + 15x) + (6y - 6y) = 36 + 21 \implies 19x = 57 \implies x = 3$$
• Step 3: Substitute $x = 3$ back into Equation 1: $$2(3) + 3y = 18 \implies 6 + 3y = 18 \implies 3y = 12 \implies y = 4$$
• Solution: $(x, y) = (3, 4)$.

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4. Quadratic Equations and the Discriminant

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

The Quadratic Formula

When factoring is difficult or impossible over integers, use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The Discriminant ($\Delta = b^2 - 4ac$)

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

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5. Fundamentals of Logarithms

A logarithm is the inverse operation of exponentiation. The fundamental equivalence is:

$$\log_b(x) = y \iff b^y = x \quad (b > 0, b \neq 1, x > 0)$$

Essential Logarithmic Properties

• Product Rule: $\log_b(M \cdot N) = \log_b(M) + \log_b(N)$
• Quotient Rule: $\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$
• Power Rule: $\log_b(M^k) = k \cdot \log_b(M)$
• Identity & Zero Rules: $\log_b(b) = 1$ and $\log_b(1) = 0$
• Change of Base Formula: $\log_b(x) = \frac{\log_c(x)}{\log_c(b)}$

Worked Example: Logarithmic Equation

Solve for $x$: $\log_2(x) + \log_2(x - 3) = 2$.

• Apply Product Rule: $\log_2(x(x - 3)) = 2$.
• Convert to exponential form: $x(x - 3) = 2^2 \implies x^2 - 3x = 4 \implies x^2 - 3x - 4 = 0$.
• Factor: $(x - 4)(x + 1) = 0 \implies x = 4$ or $x = -1$.
• Check domain ($x > 0$ and $x - 3 > 0$): $x = -1$ is extraneous. Thus, the only valid solution is $x = 4$.

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6. Applied Multi-Step Word Problems

AdUCET word problems translate English scenarios into algebraic expressions.

Problem Types & Models

1. Cost, Revenue, and Profit:
   Profit $P(x) = R(x) - C(x)$, where Revenue $R(x) = p \cdot x$ and Total Cost $C(x) = \text{Fixed Cost} + \text{Variable Cost} \cdot x$. Break-even occurs when $P(x) = 0$.
2. Work Rate Model:
   If Worker A finishes a job in $t_1$ hours and Worker B finishes in $t_2$ hours, their combined time $T$ satisfies: $$\frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} \implies T = \frac{t_1 t_2}{t_1 + t_2}$$
3. Motion / Distance-Rate-Time:
   $d = r \cdot t$. 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 $\frac{1}{6}$ reservoir/hour; Pipe B's rate is $\frac{1}{12}$ reservoir/hour.
• Step 2: Combine rates: $\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$ reservoir/hour.
• Step 3: Calculate total time $T = \frac{1}{\text{Combined Rate}} = \frac{1}{1/4} = 4$ 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 $d$. Father's current age is $3d$.
• Step 2: Express future ages in 12 years. Daughter will be $d + 12$; Father will be $3d + 12$.
• Step 3: Formulate and solve the sum equation: $$(d + 12) + (3d + 12) = 80 \implies 4d + 24 = 80 \implies 4d = 56 \implies d = 14$$ The daughter is currently 14 years old.