Polynomial Division and Roots: Chapter Review

Vocabulary and notation

Polynomial

A finite sum of terms $c\,x^k$ with real $c$ and whole-number $k$; domain all reals. Ruled out: $\tfrac{3}{x}$, $\sqrt{x}$, $2^x$. The restriction falls on $x$ only, so $\tfrac{x^2}{3}+\sqrt{2}$ qualifies.

Degree names and term counts

Degrees $0$ to $5$: constant, linear, quadratic, cubic, quartic, quintic. Terms counted: monomial, binomial, trinomial. So $x^4-5x^2+4$ is a quartic trinomial.

Standard form, degree, leading coefficient, constant term

Terms in descending powers. Degree is the highest power with a nonzero coefficient, the leading coefficient sits on it, the constant term is $a_0$ with its sign. Combine like terms and order them BEFORE reading any of these; the zero polynomial has no degree.

Turning point

Where a graph switches between rising and falling: a local hilltop or valley floor.

Root (zero) versus factor

$c$ is a root when the single evaluation $P(c)$ gives $0$; $D$ is a factor when $P = DQ$ exactly, an identity at every $x$. Different claims, and the Factor Theorem is the exchange rate.

Corner number

The $c$ in a synthetic tableau's corner: the value making the divisor zero, so $x + 3$ takes $c = -3$.

Placeholder

An explicit $0x^k$ written for every power missing from a dividend, so like powers stay aligned.

Formulas and theorems

• Degree of a product and of a sum

  $$\begin{gathered} \deg(fg) = \deg f + \deg g \\ \deg(f+g) \le \max(\deg f, \deg g) \end{gathered}$$

  Use when $f, g$ nonzero. Leading coefficients MULTIPLY, so a factored form gives its degree and leading term unexpanded. A sum's degree DROPS only when leading terms cancel, possibly to the zero polynomial, which has no degree.

  e.g. $(3x^2-2)(x^3+4x)$ has degree $5$, leading term $3x^5$; $(x^3+2x)-(x^3-x) = 3x$ has degree $1$.

• Two free evaluations

  $$\begin{gathered} f(0) = a_0 \\ f(1) = a_n + a_{n-1} + \cdots + a_0 \end{gathered}$$

  Use when Any polynomial. $f(0)$ is the constant term with its sign, the $y$-intercept; $f(1)$ adds every coefficient, a missing power contributing $0$.

  e.g. $2x^4-x^3+3x-9$ gives $f(0) = -9$ and $f(1) = -5$.

• End behavior belongs to the leading term

  $$\begin{gathered} \text{far right follows the sign of } a_n \\ \text{far left agrees if } n \text{ is even} \\ \text{far left opposes if } n \text{ is odd} \end{gathered}$$

  Text description

  Four miniature graphs: an even degree sends both ends the same way, up when the leading coefficient is positive and down when it is negative, while an odd degree sends the two ends opposite ways.

  Use when $n \ge 1$, $a_n \neq 0$. Only the highest exponent speaks far out, whatever the lower coefficients' size. So every odd-degree polynomial has a real zero, and an even-degree graph has an ODD number of turning points, hence at least one.

  e.g. $-3x^4+5x^3-1$ falls on both ends; $-x^5+100x^4$ obeys $-x^5$.

• Counting ceilings

  $$\begin{gathered} \#\{\text{distinct roots}\} \le n \\ \#\{\text{turning points}\} \le n-1 \end{gathered}$$

  Use when Nonzero polynomial, degree $n \ge 1$; roots counted distinct, real or nonreal. Ceilings, not quotas: $x^2+1$ has no real zero, $x^3$ no turning point. There is also a floor: $k$ distinct real zeros force at least $k-1$ turns.

• A sign change traps a zero

  $$\begin{gathered} f(a) < 0 < f(b) \\ \Longrightarrow \ f \ \text{ has a zero between } a \text{ and } b \end{gathered}$$

  Text description

  An unbroken curve running from a point below the axis at a to a point above it at b, with the crossing it is forced to make highlighted between them.

  Use when Polynomial graphs are smooth and unbroken, which is what powers this; a corner or a break rules a curve out. One-way only: $x^2$ has a zero with no sign change.

• Division algorithm, with the quotient's size

  $$\begin{gathered} P = DQ + R \\ R = 0 \ \text{ or } \ \deg R < \deg D \\ \deg Q = \deg P - \deg D \end{gathered}$$

  Use when $D \neq 0$; the last equation needs $\deg P \ge \deg D$. The degree bound makes the answer finished and $(Q, R)$ unique; $R = 0$ is separate since the zero polynomial has no degree. $Q$'s leading coefficient is $P$'s divided by $D$'s, and fractions there are legal when $D$ is not monic. If $\deg P < \deg D$ then $Q = 0$, $R = P$. Dividing through gives $\tfrac{P}{D} = Q + \tfrac{R}{D}$ where $D(x) \neq 0$, remainder still OVER the divisor.

  e.g. $x^3 - 4x + 1$ over $x - 2$ gives the quadratic $x^2 + 2x$ with remainder $1$.

• Synthetic division recurrence

  $$\begin{gathered} b_{n-1} = a_n \\ b_{k-1} = a_k + c\,b_k \\ r = a_0 + c\,b_0 \end{gathered}$$

  Use when The divisor must be exactly $x - c$: degree $1$ AND monic (leading coefficient $1$). The top row lists every coefficient with $0$ placeholders, and the tableau only ever ADDS, the subtraction baked into the corner's flipped sign. The bottom row is the quotient, one degree lower, then the remainder, and it also computes $P(c)$ in $n$ multiplications (Horner's method).

• Remainder Theorem

  $$P(x) \div (x-c) \ \text{ leaves remainder } \ P(c)$$

  Use when Any polynomial and any $c$, real or complex; true whether or not that remainder is zero. For $ax - b$ with $a \neq 0$ the remainder is $P\!\left(\tfrac{b}{a}\right)$.

  e.g. $2x^3-3x^2+4$ over $x-2$ leaves $P(2) = 8$.

• Factor Theorem

  $$\begin{gathered} x - c \ \text{ is a factor of } \ P(x) \\ \iff P(c) = 0 \end{gathered}$$

  Use when Any polynomial, any $c$. Both directions hold, so one evaluation settles it: $P(c) \neq 0$ PROVES $x - c$ is not a factor. For $a \neq 0$, $ax - b$ is a factor exactly when $P\!\left(\tfrac{b}{a}\right) = 0$, so test the value killing the candidate factor. It converts roots and factors, never asserting a root exists.

• Build a polynomial from prescribed roots

  $$\begin{gathered} P(x) = a\,(x-r_1)(x-r_2)\cdots(x-r_k) \\ a \neq 0 \end{gathered}$$

  Use when Degree $k$, with exactly those roots; a repeated root repeats its factor. Roots fix $P$ only up to $a$: $(x-1)(x-3)$ and $7(x-1)(x-3)$ tie until a point or the leading coefficient chooses.

• Nonreal roots come in conjugate pairs

  $$P \ \text{real}, \quad P(z) = 0 \ \Longrightarrow \ P(\bar{z}) = 0$$

  Use when EVERY coefficient of $P$ real; otherwise it fails, and $x^2-2ix-1 = (x-i)^2$ has $i$ twice and $-i$ never. The pair merges into the real quadratic $x^2-2px+p^2+q^2$, discriminant negative.

  e.g. The monic cubic with roots $2$ and $1-i$ is $(x-2)(x^2-2x+2) = x^3-4x^2+6x-4$.

• Rational Root Theorem

  $$\tfrac{p}{q} \ \text{ a root } \ \Longrightarrow \ p \mid a_0 \ \text{ and } \ q \mid a_n$$

  Use when EVERY coefficient an integer, $a_n \neq 0$, $\tfrac{p}{q}$ in lowest terms; clear fractional coefficients first, which changes no root. Numerators come from the CONSTANT term, denominators from the LEADING coefficient. One-way only, and blind to irrational and nonreal roots. If $a_0 = 0$ it filters nothing: factor out the largest power of $x$ first.

• Monic case (integer root theorem)

  $$\begin{gathered} a_n = 1 \\ \Longrightarrow \ \text{every rational root} \\ \text{is an integer dividing } a_0 \end{gathered}$$

  Use when Integer coefficients, leading coefficient $1$; no fraction can occur. A fully failed list PROVES no rational root, the standard proof that $\sqrt{n}$ is irrational for non-square $n$.

  e.g. $x^3-4x+2$: candidates $\pm 1, \pm 2$ all fail, so there is no rational root.

Problem types, step by step

Exam traps

• Trap Putting the divisor's own constant in the corner: dividing $x^3+5x-4$ by $x+1$ with corner $1$, reporting remainder $2$.

  Fix The corner holds the value making the divisor zero, so $x+1$ takes $-1$ and the remainder is $P(-1) = -10$.

• Trap Feeding a divisor that is not $x - c$ into the corner, or halving the remainder along with the quotient.

  Fix For $ax - b$ use corner $\tfrac{b}{a}$, divide the QUOTIENT by $a$, leave the remainder alone: $3x^3+2x^2-7x+4$ over $3x-1$ has remainder $2$, not $\tfrac{2}{3}$. Degree $2$ or more has no tableau; use long division.

• Trap Omitting a placeholder: running $x^3-7x+6$ over $x-2$ from the row $1, -7, 6$.

  Fix The row is $1, 0, -7, 6$. The short row returns $x-5$ remainder $-4$ instead of $x^2+2x-3$ remainder $0$.

• Trap Subtracting only the leading term of $q\,D$, or letting a subtracted negative keep its sign.

  Fix Subtract the entire product, so subtracting $-3x$ ADDS $3x$. A row that fails to shift right means the leading term never cancelled.

• Trap Promoting the candidate list to an answer: reporting $\pm 1, \pm 2, \pm 4, \pm\tfrac{1}{2}$ as the roots of $2x^3-x^2-8x+4$.

  Fix Those eight are candidates; substituting leaves only $\tfrac{1}{2}$, $2$, and $-2$. Five fail, and a polynomial can fail all eight.

• Trap Reading a fully failed candidate list as "no roots".

  Fix It proves no RATIONAL roots, nothing more. Every candidate of $x^2-2$ fails, yet it has the real roots $\pm\sqrt{2}$.

• Trap Hunting the next root back in $P$, or assuming a found root cannot repeat.

  Fix Search inside the quotient and retest $c$ there: $4x^3-8x^2+5x-1 = (x-1)(2x-1)^2$, where $\tfrac{1}{2}$ is a root twice.