2.12 Sum and Product of Roots (HL)

1. Fundamental Concepts

The Fundamental Theorem of Algebra establishes that any polynomial of degree $n$ inherently possesses exactly $n$ complex roots. The following notation maps the summation and product of these generalized roots:

$S = r_1 + r_2 + \dots + r_n$ (The sum of the roots)

$P = r_1 r_2 \dots r_n$ (The product of the roots)

2. Quadratic Functions (Vieta Formulas)

For a standard quadratic function $f(x) = ax^2 + bx + c$, there are strictly two complex roots $r_1$ and $r_2$. The factorization over $\mathbb{C}$ maps to $f(x) = a(x - r_1)(x - r_2)$.

Expanding the factored form yields:

$f(x) = a(x^2 - r_1 x - r_2 x + r_1 r_2)$

$= ax^2 - a(r_1 + r_2)x + ar_1 r_2$

Equating structural coefficients from the original polynomial $ax^2 + bx + c$ dictates:

$b = -a(r_1 + r_2) \Rightarrow S = r_1 + r_2 = -\frac{b}{a}$

$c = a(r_1 r_2) \Rightarrow P = r_1 r_2 = \frac{c}{a}$

3. Cubic Functions

Consider the cubic function $f(x) = ax^3 + bx^2 + cx + d$. The factorization over $\mathbb{C}$ yields $f(x) = a(x - r_1)(x - r_2)(x - r_3)$.

Evaluating the constant term:
The expanded constant is $-a r_1 r_2 r_3$. Equating to $d$ produces $d = -a r_1 r_2 r_3 \Rightarrow r_1 r_2 r_3 = -\frac{d}{a}$.

Evaluating the coefficient of $x^2$:
The expanded quadratic coefficient is $-a(r_1 + r_2 + r_3)$. Equating to $b$ produces $b = -a(r_1 + r_2 + r_3) \Rightarrow r_1 + r_2 + r_3 = -\frac{b}{a}$.

Standard Index Notation: Cubic functions are frequently expressed as $f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$. Under this framework, the formulas transition to:

$S = -\frac{a_2}{a_3} \quad \text{and} \quad P = -\frac{a_0}{a_3}$

4. The General Case for Degree $n$

For any general polynomial function $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, factorization scales structurally to $a_n(x - r_1)(x - r_2) \dots (x - r_n)$.

Matching the expanded terms dictates the following universal principles:

Sum of all roots: Derived from the $x^{n-1}$ coefficient.
$a_{n-1} = -a_n(r_1 + r_2 + \dots + r_n) \Rightarrow \mathbf{S = -\frac{a_{n-1}}{a_n}}$.

Product of all roots: Derived strictly from the constant component $a_0$.
$a_0 = (-1)^n a_n(r_1 r_2 \dots r_n) \Rightarrow \mathbf{P = (-1)^n \frac{a_0}{a_n}}$.

Note: Additional intermediate relations also govern the sum of all possible pairs ($S_2 = \frac{a_{n-2}}{a_n}$) and the sum of all triples ($S_3 = -\frac{a_{n-3}}{a_n}$).

5. Analytical Applications

EXAMPLE 1 (Cubic Application)

Let $f(x) = 2x^3 + ax^2 + bx + c$. It is given that the sum of the roots evaluates to 3.5, the product of the roots is -5, and the polynomial is perfectly divisible by $(x+2)$. Find $a, b, c$.

Solution:

Utilizing the sum formula: $S = -\frac{a_2}{a_3} \Rightarrow -\frac{a}{2} = 3.5 \Rightarrow \mathbf{a = -7}$.

Utilizing the product formula: $P = (-1)^3 \frac{a_0}{a_3} \Rightarrow -\frac{c}{2} = -5 \Rightarrow \mathbf{c = 10}$.

Applying the factor theorem for divisibility by $(x+2)$ dictates $f(-2) = 0$:
$2(-8) + a(4) + b(-2) + c = 0$
$-16 - 28 - 2b + 10 = 0$
$-34 - 2b = 0 \Rightarrow \mathbf{b = -17}$.

EXAMPLE 2 (Quartic Application)

Let $f(x) = ax^4 - 10x^3 + bx + c$. The sum of the roots evaluates to 2, the product of the roots is 5, and the polynomial is perfectly divisible by $(x-1)$. Find $a, b, c$.

Solution:

Utilizing the sum formula for a quartic ($n=4$): $S = -\frac{a_3}{a_4} \Rightarrow -\frac{-10}{a} = 2 \Rightarrow \frac{10}{a} = 2 \Rightarrow \mathbf{a = 5}$.

Utilizing the product formula: $P = (-1)^4 \frac{a_0}{a_4} \Rightarrow \frac{c}{a} = 5 \Rightarrow \frac{c}{5} = 5 \Rightarrow \mathbf{c = 25}$.

Applying the factor theorem mandates $f(1) = 0$:
$a(1) - 10(1) + b(1) + c = 0$
$5 - 10 + b + 25 = 0$
$20 + b = 0 \Rightarrow \mathbf{b = -20}$. (Note: Calculating from the raw document source yielded $+25$, maintaining strict algebraic logic here).