5.2 Quadratic Equations, Polynomials & Algebraic Factoring

5.2 Quadratic Equations, Polynomials & Algebraic Factoring

While linear equations model steady-state rates and direct proportionalities, physical and tactical environments frequently exhibit non-linear dynamics. Projectile ballistics, mortar trajectories, radar echo dissipation, and defensive perimeter geometries follow quadratic and higher-degree polynomial laws. In the GAF Officer Cadet Written Examination, questions covering algebraic factoring, quadratic equations, and polynomial theorems appear with high frequency. Officers must be capable of deconstructing complex algebraic expressions into irreducible components, diagnosing the nature of solutions using discriminant analysis, and applying parabolic modeling to field challenges.

---

Quadratic Equations: Canonical Form & Factoring Techniques

A quadratic equation is a second-degree polynomial equation in a single variable, expressed in standard form as:

$ax^2 + bx + c = 0$

where $a, b, c \in \mathbb{R}$ and $a \neq 0$. The coefficient $a$ is the quadratic coefficient, $b$ is the linear coefficient, and $c$ is the constant term. If $c = 0$, the equation is incomplete ($ax^2 + bx = 0$); if $b = 0$, it represents a pure quadratic ($ax^2 + c = 0$).

The Four Core Factoring Archetypes

Factoring transforms a polynomial sum into a product of simpler linear binomials, exploiting the Zero-Product Property: if $A \times B = 0$, then either $A = 0$, $B = 0$, or both are zero.

Worked Example 1: Factoring Non-Monic Quadratic Trinomial

Solve the quadratic equation using the split-the-middle-term factoring method:

$6x^2 - 11x - 10 = 0$

Step 1: Calculate the key product $ac$. Here, $a = 6$, $b = -11$, and $c = -10$. The product is:

$ac = 6 \times (-10) = -60$

Step 2: Identify two factors $p$ and $q$ whose product is $-60$ and whose sum is $-11$. Examine factor pairs of $-60$:

• $1 \times (-60) \implies \text{sum} = -59$
• $2 \times (-30) \implies \text{sum} = -28$
• $3 \times (-20) \implies \text{sum} = -17$
• $4 \times (-15) \implies \text{sum} = 4 + (-15) = -11$

The correct factor pair is $+4$ and $-15$.

Step 3: Split the linear term $-11x$ into $+4x - 15x$.

$6x^2 + 4x - 15x - 10 = 0$

Step 4: Factor by grouping in pairs.

$2x(3x + 2) - 5(3x + 2) = 0$

Factor out the shared binomial factor $(3x + 2)$:

$(3x + 2)(2x - 5) = 0$

Step 5: Apply the Zero-Product Property.

$3x + 2 = 0 \implies x = -\frac{2}{3}$ $2x - 5 = 0 \implies x = \frac{5}{2}$

The roots are $x = -\frac{2}{3}$ and $x = \frac{5}{2}$.

---

Completing the Square & Derivation of the Quadratic Formula

When a quadratic trinomial cannot be easily factored using rational integers, we use completing the square—the mathematical mechanism that generates the universal quadratic formula.

Step-by-Step Derivation

Start with the general equation:

$ax^2 + bx + c = 0$

1. Divide all terms by $a$ ($a \neq 0$) and transpose the constant:

$x^2 + \frac{b}{a}x = -\frac{c}{a}$

1. Add the square of half the linear coefficient, $\left( \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2}$, to both sides:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}$

1. The LHS is now a perfect square trinomial. Combine the RHS over a common denominator:

$\left( x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}$

1. Extract the square root of both sides:

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

1. Isolate $x$ to yield the Quadratic Formula:

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

---

Discriminant Analysis: Diagnosing Root Nature

The expression residing under the radical symbol in the quadratic formula is termed the discriminant, denoted by the Greek letter Delta ($\Delta$):

$\Delta = b^2 - 4ac$

The numerical value and sign of $\Delta$ completely determine the nature and geometric behavior of the quadratic function's roots without requiring full evaluation of the roots.

Worked Example 2: Determining an Unknown Parameter for Equal Roots

For what values of the constant $k$ will the quadratic equation possess real and equal (coincident) roots?

$2x^2 + kx + 18 = 0$

Step 1: Identify coefficients. Here, $a = 2$, $b = k$, and $c = 18$.

Step 2: Apply the condition for real and equal roots. Equal roots mandate that the discriminant equals zero: $\Delta = b^2 - 4ac = 0$.

$k^2 - 4(2)(18) = 0$

$k^2 - 144 = 0$

$k^2 = 144$

$k = \pm \sqrt{144} = \pm 12$

The equation possesses equal roots when $k = 12$ or $k = -12$.

---

Vieta's Formulas & Symmetric Root Expressions

Named after the French mathematician François Viète, Vieta's formulas establish fundamental relationships between the roots of a polynomial and its coefficients.

If $\alpha$ and $\beta$ denote the two roots of $ax^2 + bx + c = 0$, then factoring yields $a(x - \alpha)(x - \beta) = a(x^2 - (\alpha + \beta)x + \alpha \beta) = 0$. Equating coefficients with $ax^2 + bx + c = 0$ gives:

$\text{Sum of Roots: } \alpha + \beta = -\frac{b}{a}$ $\text{Product of Roots: } \alpha \beta = \frac{c}{a}$

Constructing Equations from Roots

A quadratic equation whose roots are $\alpha$ and $\beta$ can be constructed directly:

$x^2 - (\text{Sum of Roots})x + (\text{Product of Roots}) = 0$

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

Symmetric Functions of Roots

Exam questions routinely require evaluating symmetric expressions without calculating the roots themselves:

• Sum of Squares: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$
• Sum of Reciprocals: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$
• Difference Squared: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$

Worked Example 3: Evaluating Symmetric Functions and Constructing New Systems

The roots of the quadratic equation $3x^2 - 7x + 2 = 0$ are $\alpha$ and $\beta$. Without solving for $\alpha$ and $\beta$, calculate:

1. The value of $\alpha^2 + \beta^2$
2. The quadratic equation with integer coefficients whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$

Step 1: Extract basic Vieta parameters. From $3x^2 - 7x + 2 = 0$, $a = 3$, $b = -7$, $c = 2$:

$\alpha + \beta = -\frac{-7}{3} = \frac{7}{3}$ $\alpha \beta = \frac{2}{3}$

Step 2: Compute $\alpha^2 + \beta^2$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left( \frac{7}{3} \right)^2 - 2\left( \frac{2}{3} \right)$

$\alpha^2 + \beta^2 = \frac{49}{9} - \frac{4}{3} = \frac{49}{9} - \frac{12}{9} = \frac{37}{9}$

Step 3: Construct the equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. Find the new sum of roots ($S_{\text{new}}$) and new product of roots ($P_{\text{new}}$):

$S_{\text{new}} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{7/3}{2/3} = \frac{7}{2}$

$P_{\text{new}} = \left( \frac{1}{\alpha} \right) \left( \frac{1}{\beta} \right) = \frac{1}{\alpha \beta} = \frac{1}{2/3} = \frac{3}{2}$

Apply the construction formula:

$x^2 - S_{\text{new}}x + P_{\text{new}} = 0 \implies x^2 - \frac{7}{2}x + \frac{3}{2} = 0$

Multiply by 2 to clear fractions and ensure integer coefficients:

$2x^2 - 7x + 3 = 0$

---

Polynomial Degrees, Division & Theorems

A polynomial in $x$ is an algebraic expression of the form $P(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$, where $n$ is a non-negative integer and $a_n \neq 0$. The exponent $n$ designates the degree of the polynomial:

• Degree 1: Linear ($ax + b$)
• Degree 2: Quadratic ($ax^2 + bx + c$)
• Degree 3: Cubic ($ax^3 + bx^2 + cx + d$)
• Degree 4: Quartic ($ax^4 + bx^3 + cx^2 + dx + e$)

The Division Algorithm for Polynomials

For any polynomial dividend $P(x)$ and non-zero divisor $D(x)$, there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that:

$P(x) = D(x) \cdot Q(x) + R(x)$

where either $R(x) = 0$ or the degree of $R(x)$ is strictly less than the degree of $D(x)$. If the divisor is linear, $D(x) = x - c$, then the remainder must be a constant $R$.

The Remainder Theorem

When a polynomial $P(x)$ is divided by a linear divisor $(x - c)$, the constant remainder $R$ is equal to $P(c)$:

$R = P(c)$

Proof: From $P(x) = (x - c)Q(x) + R$, substitute $x = c$: $P(c) = (c - c)Q(c) + R = 0 \cdot Q(c) + R = R$.

The Factor Theorem

A linear expression $(x - c)$ is an exact factor of a polynomial $P(x)$ if and only if $P(c) = 0$. Conversely, if $P(c) = 0$, then $x = c$ is a root of the equation $P(x) = 0$, and $(x - c)$ is an exact factor.

Worked Example 4: Cubic Polynomial Factoring via Factor Theorem

Find all real roots of the cubic polynomial equation:

$P(x) = 2x^3 - 3x^2 - 11x + 6 = 0$

Step 1: Test integer factors of the constant term $+6$. Possible rational roots $\pm 1, \pm 2, \pm 3, \pm 6$:

• $P(1) = 2(1)^3 - 3(1)^2 - 11(1) + 6 = 2 - 3 - 11 + 6 = -6 \neq 0$
• $P(-1) = 2(-1)^3 - 3(-1)^2 - 11(-1) + 6 = -2 - 3 + 11 + 6 = 12 \neq 0$
• $P(2) = 2(8) - 3(4) - 11(2) + 6 = 16 - 12 - 22 + 6 = -12 \neq 0$
• $P(-2) = 2(-8) - 3(4) - 11(-2) + 6 = -16 - 12 + 22 + 6 = 0$

Since $P(-2) = 0$, by the Factor Theorem, $(x + 2)$ is an exact factor of $P(x)$.

Step 2: Divide $P(x)$ by $(x + 2)$ using polynomial long division or synthetic division. Dividing $2x^3 - 3x^2 - 11x + 6$ by $(x + 2)$ yields the quadratic quotient:

$Q(x) = 2x^2 - 7x + 3$

Step 3: Factor the quadratic quotient. Split the middle term of $2x^2 - 7x + 3$ (product $ac = 6$, sum $-7 \implies -6$ and $-1$):

$2x^2 - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3)$

Step 4: State the complete factorization and roots.

$P(x) = (x + 2)(2x - 1)(x - 3) = 0$

The three real roots are $x = -2$, $x = \frac{1}{2}$, and $x = 3$.

---

Applied Tactical Modeling: Mortar Trajectory & Ballistics

Under classical Newtonian mechanics, neglecting air resistance, the vertical altitude $h(t)$ in meters of a ballistic projectile fired upward from initial height $h_0$ with vertical velocity $v_0$ under gravitational acceleration $g \approx 10\text{ m/s}^2$ is modeled by the quadratic kinematics equation:

$h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0 = -5t^2 + v_0 t + h_0$

Worked Example 5: Mortar Illumination Flare Trajectory

During a night reconnaissance patrol in the Northern Command area, an officer orders the deployment of an illumination flare from a ridge 40 meters above a valley. The flare is launched with an initial upward velocity of 60 m/s. Its altitude $h(t)$ in meters after $t$ seconds is modeled by:

$h(t) = -5t^2 + 60t + 40$

1. At what time $t$ does the flare reach its maximum apogee (peak altitude), and what is this peak altitude?
2. How long does the flare remain airborne before impacting the valley floor ($h = 0$)?

Part 1: Determining Peak Apogee The maximum value of a downward-opening parabola $h(t) = at^2 + bt + c$ occurs at the vertex time:

$t_{\text{peak}} = -\frac{b}{2a} = -\frac{60}{2(-5)} = \frac{60}{10} = 6\text{ seconds}$

Substitute $t = 6$ seconds into the altitude equation to find peak height:

$h(6) = -5(6)^2 + 60(6) + 40 = -5(36) + 360 + 40 = -180 + 360 + 40 = 220\text{ meters}$

The flare achieves its maximum apogee of 220 meters at $t = 6$ seconds.

Part 2: Time to Impact ($h = 0$) Set $h(t) = 0$:

$-5t^2 + 60t + 40 = 0$

Divide the entire equation by $-5$ to simplify coefficients:

$t^2 - 12t - 8 = 0$

Apply the quadratic formula with $a = 1$, $b = -12$, $c = -8$:

$t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(-8)}}{2(1)} = \frac{12 \pm \sqrt{144 + 32}}{2} = \frac{12 \pm \sqrt{176}}{2}$

Simplify the radical: $\sqrt{176} = \sqrt{16 \times 11} = 4\sqrt{11}$:

$t = \frac{12 \pm 4\sqrt{11}}{2} = 6 \pm 2\sqrt{11}$

Because physical time cannot be negative ($6 - 2\sqrt{11} \approx 6 - 6.63 = -0.63\text{ s}$, which represents the backward trajectory prior to launch), we take the positive real root:

$t = 6 + 2\sqrt{11} \approx 6 + 6.63 = 12.63\text{ seconds}$

The flare remains airborne for approximately 12.63 seconds before impacting the valley floor.