Comprehensive Study Notes on Polynomials: Definitions, Operations, and Theorems

Bi-quadratic Polynomial: In general, a bi-quadratic polynomial is defined by the function $f(x) = ax^4 + bx^3 + cx^2 + dx + e$ , where $a \neq 0$ .
Polynomials in One vs. Multiple Variables:
- While much of introductory mathematics focuses on polynomials in a single variable (e.g., $x$ ), polynomials can exist in more than one variable.
- Example in Three Variables: $x^2 + y^2 + 2xyz$ is a polynomial containing the variables $x$ , $y$ , and $z$ .
- Example in Three Variables (Alternative): If $u$ , $v$ , and $w$ are variables, then $u^2 + v^8 + 3w$ constitutes a polynomial in three variables.
Degree and Coefficients:
- Degree: The highest power of the variable in the polynomial expression.
- Coefficients: The numerical or constant multipliers of the variable terms. For instance, in the expression $17 - 2x + 7x^2$ , the coefficient of $x^2$ is $7$ .

Classification by Number of Terms:
- Monomial: A polynomial consisting of exactly one term (e.g., $12x$ , $2x^{100}$ , or a constant).
- Binomial: A polynomial consisting of exactly two terms (e.g., $7x^{35} - 5$ , $3x^{20} + 5$ ).
- Trinomial: A polynomial consisting of exactly three terms (e.g., $3x^2 + 2x + 5$ ).
Classification by Degree:
- Constant Polynomial: Degree is $0$ (e.g., $f(x) = 7$ or $f(x) = 0$ ).
- Linear Polynomial: Degree is $1$ (e.g., $3x - 2$ , $4x + 3$ , $2z + 1$ ).
- Quadratic Polynomial: Degree is $2$ (e.g., $x^2 + x^2 + 4$ , $2x^2 - x + 4$ , $3x^2 - 5$ ).
- Cubic Polynomial: Degree is $3$ (e.g., $12 - x + 2x^3$ , $2x^3 - 7x + 4$ , $3x + 4x^2 + 5x^3 - 7$ ).
- Bi-quadratic Polynomial: Degree is $4$ (e.g., $7x^4 + 4x^3 + 3x^2 - 2$ ).

Value of a Polynomial: The value of a polynomial $f(x)$ at a specific point $x = \alpha$ is obtained by substituting $\alpha$ for $x$ throughout the expression and is denoted by $f(\alpha)$ .
- Example Calculation: For $f(x) = 2x^3 - 3x^2 + 4x - 2$ , to find the value at $x = 2$ : $f(2) = 2 \times (2)^3 - 3 \times (2)^2 + 4 \times (2) - 2$ $f(2) = 16 - 12 + 8 - 2 = 10$ Thus, the value of $f(x)$ at $x = 2$ is $10$ .
Zeros (Roots) of a Polynomial: A real number $\alpha$ is a zero or root of the polynomial $f(x)$ if $f(\alpha) = 0$ .
- Verification Example: To check if $x = 1, 2, 3$ are zeros of $p(x) = x^3 - 6x^2 + 11x - 6$ :
  - $p(1) = 1 - 6 + 11 - 6 = 0$
  - $p(2) = 8 - 24 + 22 - 6 = 0$
  - $p(3) = 27 - 54 + 33 - 6 = 0$ Since all results are zero, $1, 2, 3$ are indeed roots.

Conceptual Basis: In integer division, the remainder is either zero or less than the divisor. Similarly, in polynomial division, when a polynomial $f(x)$ is divided by a polynomial $g(x)$ , the degree of the remainder is either zero or strictly less than the degree of the divisor.
Specific Application to Linear Divisors: When a polynomial $f(x)$ is divided by a linear polynomial (degree 1), the remainder must be of degree zero (a constant).
Formulaic Definition of the Remainder Theorem: Let $p(x)$ be any polynomial of degree greater than or equal to one and let $a$ be any real number. If $p(x)$ is divided by the linear polynomial $(x - a)$ , then the remainder is $p(a)$ .
Proof Outline:
- Let $q(x)$ be the quotient and $r(x)$ be the remainder when $p(x)$ is divided by $(x - a)$ .
- The division algorithm states: $p(x) = (x - a) \times q(x) + r(x)$ .
- Since the degree of $(x - a)$ is $1$ , the degree of $r(x)$ must be less than $1$ , meaning $r(x)$ is a constant $r$ .
- By substituting $x = a$ into the equation: $p(a) = (a - a) \times q(a) + r$ , which simplifies to $p(a) = 0 \times q(a) + r$ , therefore $p(a) = r$ .

Integral Roots: To find the integral roots of a polynomial like $f(x) = x^3 + 6x^2 + 11x + 6$ , one tests integer factors of the constant term ( $6$ ). In this case, the roots are $-1, -2, -3$ .
Rational Roots: For polynomials like $f(x) = 2x^3 + x^2 - 7x - 6$ , the rational roots are found by testing values in the form of $\frac{b}{c}$ , where $b$ is a factor of the constant term and $c$ is a factor of the leading coefficient. The rational roots for this specific equation are $2, -1, \text{ and } -\frac{3}{2}$ .
Theorem on Factors: If $\frac{b}{c}$ is a rational root of the polynomial (in simplest form), then $(cx - b)$ is a factor of the polynomial.

Identification Tasks:
- $3x^2 - 4x + 15$ : Polynomial in one variable.
- $y^2 + 2\sqrt{3}$ : Polynomial in one variable.
- $3\sqrt{x} + \sqrt{2}x$ : Not a polynomial (variable under a radical/fractional power).
- $x - \frac{4}{x}$ : Not a polynomial (variable in denominator/negative exponent).
- $t^2 + y^3 + 150$ : Polynomial in two variables (originally lists three variables in some contexts depending on notation).
- $1 - \sqrt{5x}$ : Not a polynomial.
- $\frac{1}{x+1}$ : Not a polynomial.
- $\frac{1}{x}$ : Not a polynomial.
- $2 + 5x + 7$ : Polynomial in one variable.
Determining Coefficients for $x^2$ :
- $17 - 2x + 7x^2$ : Coefficient is $7$ .
- $9 - 12x + x^3$ : Coefficient is $0$ (term is absent).
- $x^2 - 3x + 4$ : Coefficient is $1$ .
- $\sqrt{3}x - 7$ : Coefficient is $0$ .
- $(x-1)(3x-4)$ : Expanding gives $3x^2 - 7x + 4$ , so the coefficient is $3$ .
- $(2x-5)(2x^2 - 3x + 1)$ : Expanding yields $-6x^2 - 10x^2$ as part of the terms. Specifically, the coefficient of $x^2$ in this expansion is $-16$ .

Evaluating $f(x) = 2x^3 - 13x^2 + 17x + 12$ :
- (i) $f(2)$ : $2(8) - 13(4) + 17(2) + 12 = 16 - 52 + 34 + 12 = 10$ .
- (ii) $f(-3)$ : $2(-27) - 13(9) + 17(-3) + 12 = -54 - 117 - 51 + 12 = -210$ .
- (iii) $f(0)$ : $2(0) - 13(0) + 17(0) + 12 = 12$ .
Solving for Unknown Constants ( $a$ and $b$ ):
- Problem: If $x = 2$ is a root of $f(x) = 2x^2 - 3x + 7a$ , find $a$ .
  - Substitute $x = 2$ : $2(4) - 3(2) + 7a = 0 \rightarrow 8 - 6 + 7a = 0 \rightarrow 2 + 7a = 0 \rightarrow a = -\frac{2}{7}$ .
- Problem: If $x = \frac{1}{2}$ is a zero of $p(x) = 8x^3 - ax^2 - x + 2$ , find $a$ .
  - $8(\frac{1}{8}) - a(\frac{1}{4}) - \frac{1}{2} + 2 = 0 \rightarrow 1 - \frac{a}{4} - 0.5 + 2 = 0 \rightarrow 2.5 = \frac{a}{4} \rightarrow a = 10$ .
- Problem: If $x = 0$ and $x = -1$ are roots of $f(x) = 2x^3 - 3x^2 + ax + b$ , find $a$ and $b$ .
  - For $x = 0$ : $f(0) = b = 0$ .
  - For $x = -1$ : $f(-1) = 2(-1) - 3(1) + a(-1) + 0 = 0 \rightarrow -2 - 3 - a = 0 \rightarrow a = -5$ .
  - Final values: $a = -5, b = 0$ .