7.1 Elementary Algebra, Polynomials, Roots, and Factorization

7.1 Polynomial Algebra, Roots, and the Fundamental Theorem of Algebra

Polynomial algebra on the GRE Mathematics Subject Test emphasizes root structure, symmetric polynomials, and field-dependent irreducibility criteria. Mastery of Vieta's formulas, Descartes' Rule of Signs, and Eisenstein's criterion enables solving advanced algebra items efficiently.

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Division Algorithm, Remainder Theorem, and Factor Theorem

Polynomial rings over a field $\mathbb{F}$ (such as $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$) possess a Euclidean division structure analogous to $\mathbb{Z}$.

Division Algorithm

For $P(x), D(x) \in \mathbb{F}[x]$ with $D(x) \neq 0$, there exist unique polynomials $Q(x), R(x) \in \mathbb{F}[x]$ satisfying: $P(x) = D(x)Q(x) + R(x) \quad \text{where } R(x) = 0 \text{ or } \deg(R) < \deg(D)$

Remainder and Factor Theorems

When the divisor is linear, $D(x) = x - c$:

• Remainder Theorem: Dividing $P(x)$ by $x - c$ yields a constant remainder equal to $P(c)$: $P(x) = (x - c)Q(x) + P(c)$.
• Factor Theorem: $x - c$ divides $P(x)$ in $\mathbb{F}[x]$ if and only if $P(c) = 0$.
• Multiplicity: Root $c$ has multiplicity $m \ge 1$ if $(x - c)^m \mid P(x)$ and $(x - c)^{m+1} \nmid P(x)$, equivalent to $P(c) = \dots = P^{(m-1)}(c) = 0$ with $P^{(m)}(c) \neq 0$.

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Rational Root Theorem and Descartes' Rule of Signs

Rational Root Theorem

Let $P(x) = a_n x^n + \dots + a_1 x + a_0 \in \mathbb{Z}[x]$ with $a_n \neq 0, a_0 \neq 0$. If $r = p/q \in \mathbb{Q}$ is a rational root in lowest terms ($\gcd(p, q) = 1$), then: $p \mid a_0 \quad \text{and} \quad q \mid a_n$ For monic polynomials ($a_n = 1$), rational roots must be integer divisors of $a_0$.

Descartes' Rule of Signs

Let $P(x) \in \mathbb{R}[x]$ have non-zero real coefficients:

• Positive Real Roots: The number of positive real roots (counting multiplicity) equals sign changes $v$ between consecutive non-zero coefficients of $P(x)$, or differs from $v$ by an even integer ($v - 2k$, $k \ge 0$).
• Negative Real Roots: The number of negative real roots equals sign variations in coefficients of $P(-x)$, or differs by an even integer.

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Vieta's Formulas and Symmetric Polynomials

Vieta's formulas connect the roots $r_1, \dots, r_n \in \mathbb{C}$ of $P(x) = a_n x^n + \dots + a_0 = a_n \prod_{i=1}^n (x - r_i)$ to its coefficients:

• Sum of roots: $e_1 = \sum_{i=1}^n r_i = -\frac{a_{n-1}}{a_n}$
• Pairwise products: $e_2 = \sum_{1 \le i < j \le n} r_i r_j = \frac{a_{n-2}}{a_n}$
• Degree-$k$ sum: $e_k = \sum_{1 \le i_1 < \dots < i_k \le n} r_{i_1} \dots r_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}$
• Product of all roots: $e_n = \prod_{i=1}^n r_i = (-1)^n \frac{a_0}{a_n}$

Power Sums and Newton's Identities

Power sums $p_k = \sum_{i=1}^n r_i^k$ satisfy $p_1 = e_1$, $p_2 = e_1^2 - 2e_2$, and $\sum_{i=1}^n \frac{1}{r_i} = \frac{e_{n-1}}{e_n} = -\frac{a_1}{a_0}$.

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Fundamental Theorem of Algebra and Irreducibility Over Fields

A polynomial is irreducible in $\mathbb{F}[x]$ if it cannot factor into non-constant polynomials of lower degree in $\mathbb{F}[x]$.

Gauss's Lemma and Eisenstein's Criterion

• Gauss's Lemma: A primitive polynomial $P(x) \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Z}[x]$ if and only if it is irreducible in $\mathbb{Q}[x]$.
• Eisenstein's Criterion: Let $P(x) = a_n x^n + \dots + a_0 \in \mathbb{Z}[x]$. If a prime $p$ satisfies:
  1. $p \nmid a_n$
  2. $p \mid a_i$ for all $i \in {0, 1, \dots, n-1}$
  3. $p^2 \nmid a_0$ then $P(x)$ is irreducible over $\mathbb{Q}$.
• Shifted Eisenstein: Substituting $x = y + c$ can reveal irreducibility. The cyclotomic polynomial $\Phi_p(x) = \frac{x^p - 1}{x - 1}$ with $x = y + 1$ satisfies Eisenstein with prime $p$.
• Reduction Modulo $p$: If monic $P(x) \in \mathbb{Z}[x]$ reduces modulo prime $p$ to an irreducible $\bar{P}(x) \in \mathbb{F}_p[x]$ of degree $n$, then $P(x)$ is irreducible over $\mathbb{Q}$.

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Step-by-Step Worked Problem

Problem: Let $P(x) = x^3 - 3x^2 + 4x - 5$ have roots $r_1, r_2, r_3 \in \mathbb{C}$. Compute $\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2}$.

Solution:

1. Identify coefficients: $a_3 = 1, a_2 = -3, a_1 = 4, a_0 = -5$.
2. Compute Vieta values: $e_1 = -(-3)/1 = 3$, $e_2 = 4/1 = 4$, $e_3 = -(-5)/1 = 5$.
3. Expand using common denominator $e_3^2$: $\sum_{i=1}^3 \frac{1}{r_i^2} = \frac{e_2^2 - 2e_1 e_3}{e_3^2}$
4. Evaluate: $\sum_{i=1}^3 \frac{1}{r_i^2} = \frac{4^2 - 2(3)(5)}{5^2} = \frac{16 - 30}{25} = -\frac{14}{25}$

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GRE Exam Traps & Pitfalls

Trap 1: Assuming Vieta Sums of Squares Must Be Positive The sum of squared reciprocals evaluates to $-14/25$. Non-real complex roots have squares with negative real components. Do not discard negative power sums.

Trap 2: Treating Descartes' Rule as an Exact Count Three sign variations in $P(x)$ guarantees 3 or 1 positive roots, never an unconditional count of 3.

Trap 3: Overlooking $p^2 \nmid a_0$ in Eisenstein If $p^2 \mid a_0$, Eisenstein fails and provides no information regarding irreducibility.

Trap 4: Conflating Real and Rational Irreducibility While $x^4 + 1$ is irreducible over $\mathbb{Q}$, it factors over $\mathbb{R}$ as $(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$. Over $\mathbb{R}$, no polynomial of degree $\ge 3$ is irreducible.

Elementary Algebra Toolkit

Before using polynomial theorems, simplify without changing the solution set. Every manipulation carries conditions: a denominator cannot be zero, an even root requires a nonnegative radicand over the reals, and a logarithm requires a positive argument.

Exponents, radicals, and rational expressions

For positive bases where needed, $a^m a^n=a^{m+n},\qquad (a^m)^n=a^{mn},\qquad a^{-n}=\frac1{a^n},\qquad a^{p/q}=\sqrt[q]{a^p}.$ An even root represents the nonnegative principal root, so $\sqrt{x^2}=|x|$, not always $x$. When rationalizing $1/(\sqrt a+\sqrt b)$, multiply by the conjugate $\sqrt a-\sqrt b$.

To combine rational expressions, factor first and use a common denominator. Cancellation removes factors, not terms: $\frac{x^2-9}{x^2-x-6}=\frac{(x-3)(x+3)}{(x-3)(x+2)}=\frac{x+3}{x+2},$ but the original domain still excludes both $x=3$ and $x=-2$. A rational equation should be multiplied by its least common denominator only after excluded values are recorded; any resulting candidate must be checked in the original equation.

Equations and inequalities

Completing the square gives $ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a},$ and yields the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ The discriminant determines two distinct real roots, one repeated real root, or a nonreal conjugate pair according as it is positive, zero, or negative.

Squaring both sides or multiplying by an expression that might vanish can introduce extraneous roots. For example, $\sqrt{x+1}=x-1$ first requires $x\ge1$; squaring gives $x+1=(x-1)^2$, but every candidate must meet the domain restriction and original equation.

For inequalities, multiplying or dividing by a negative quantity reverses the inequality. Polynomial and rational inequalities are reliably solved with a sign chart: move all terms to one side, factor, mark zeros and undefined points, test each interval, and decide whether endpoints are included. For $\frac{(x-1)(x+2)}{x-3}\le0,$ critical values are $-2,1,3$; $3$ is never included because the expression is undefined.

Absolute value is distance. For $a>0$, $|u|<a\iff -a<u<a,\qquad |u|>a\iff u<-a\text{ or }u>a.$ More generally, $|x-c|\le r$ describes the closed interval $[c-r,c+r]$.

Exponential and logarithmic equations

For $b>0$, $b\ne1$, the exponential $b^x$ is one-to-one and has inverse $\log_b x$. Core identities are

\log_b(x/y)=\log_bx-\log_by,\quad \log_b(x^r)=r\log_bx,$$ with every logarithm argument positive. Thus $$\log_2(x-1)+\log_2(x-3)=3$$ requires $x>3$ and becomes $(x-1)(x-3)=8$. The algebraic candidates are $x=5$ and $x=-1$, but only $x=5$ satisfies the logarithmic domain. When unlike exponential bases cannot be rewritten compatibly, logarithms isolate the variable: $3^{2x-1}=7$ gives $x=(1+\log_3 7)/2$. ### Common traps Do not distribute a root across addition: $\sqrt{a+b}$ is generally not $\sqrt a+\sqrt b$. Do not cancel across sums. Retain excluded values after factor cancellation. Reverse an inequality only when multiplying or dividing by a negative number, and check all candidates produced by squaring, clearing denominators, or applying logarithms.