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Complete Course of Mathematics
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Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
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Topic 16: Statistics & Probability


Content On This Page
Algebraic Expressions and Polynomials: Introduction and Definition Terms Related to Polynomials (Degree, leading coefficient, Monomial, Binomial, Trinomial) Value of a Polynomial
Zero of a Polynomial Geometrical Representation of a Polynomial (Graphing Zeroes) Relationship Between Zeroes and Coefficients of a Polynomial


Polynomials: Definition, Types, and Properties



Algebraic Expressions and Polynomials: Introduction and Definition

Building from Algebraic Expressions

Before defining polynomials, it's helpful to recall what an algebraic expression is. An algebraic expression is a mathematical phrase that combines numbers (constants), variables (symbols representing quantities that can change or are unknown), and the basic arithmetic operations: addition ($+$), subtraction ($-$), multiplication ($\times$), and division ($\div$).

Examples of algebraic expressions include:

Polynomials represent a special and very important category of algebraic expressions. They have a specific structure and follow certain rules regarding the exponents of the variables.


Formal Definition of a Polynomial

A polynomial is an algebraic expression that consists of one or more terms added together, where each term is a product of a constant (called the coefficient) and one or more variables raised to non-negative integer powers.

Let's break this down and look at the formal definition, starting with polynomials in a single variable.

A polynomial in one variable, say $x$, is an expression of the form:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_2 x^2 + a_1 x^1 + a_0 x^0$

[General form of a polynomial in one variable $x$]

In this general form, the following conditions must be met:

The definition of a polynomial can be extended to include multiple variables. A term in a polynomial with multiple variables is a product of a constant coefficient and one or more variables, where each variable is raised to a non-negative integer power. A polynomial in multiple variables is a sum of such terms.

Example of a polynomial in multiple variables: $5x^2 y^3 - 2xy + 7z - 4$.

Since all terms fit the description (constant $\times$ variables raised to non-negative integer powers), the entire expression is a polynomial.


Examples of Polynomials and Non-Polynomials

To determine if an algebraic expression is a polynomial, you need to examine each term. The exponents of the variables must be non-negative integers ($0, 1, 2, 3, \ldots$). Expressions involving division by variables, variables under radical signs (unless they simplify to integer exponents), or variables in the exponents are generally NOT polynomials.

Examples of expressions that ARE polynomials:

Examples of expressions that are NOT polynomials (with reasons):

Understanding the specific requirements for an expression to be a polynomial (non-negative integer exponents on variables, no division by variables, no variables under radicals etc.) is crucial for classifying algebraic expressions and for determining which properties and theorems of polynomials can be applied to them.



Terms Related to Polynomials (Degree, leading coefficient, Monomial, Binomial, Trinomial)

Terminology for Describing Polynomials

To effectively work with and discuss polynomials, it's essential to use precise terminology that describes their structure and characteristics. These terms allow mathematicians to classify polynomials, predict their behavior, and apply appropriate methods for solving related problems (like finding roots or factorising).


1. Term of a Polynomial

As established in the discussion on algebraic expressions, a term is a single number, a single variable, or the product of numbers and variables raised to non-negative integer powers. In a polynomial, the terms are the individual components that are added or subtracted.

Example: Consider the polynomial $7x^4 - 5x^2 y + 3y - 12$. The terms of this polynomial are $7x^4$, $-5x^2 y$, $3y$, and $-12$. Each of these is a valid monomial that fits the definition of a term in a polynomial.

2. Coefficient of a Term

The coefficient of a term is the numerical factor that multiplies the variable part of that term. It's the constant number associated with the term.

Example: In the polynomial $7x^4 - 5x^2 y + 3y - 12$:

3. Degree of a Term

The degree of a term is the sum of the exponents of all the variables present in that specific term. For a constant term (a term without any variables), the degree is considered to be $0$, as it can be thought of as being multiplied by a variable raised to the power of $0$ (e.g., $5 = 5x^0$).

Example: In the polynomial $10x^5 - 4x^2 y^3 + 6x - 1$:

4. Degree of a Polynomial

The degree of a polynomial is the highest degree among all of its terms that have non-zero coefficients. This single number is a fundamental characteristic of the polynomial and significantly influences its properties and the methods used to work with it.

To find the degree of a polynomial:

  1. Find the degree of each individual term (sum of variable exponents).
  2. The highest of these term degrees is the degree of the polynomial. (Ignore terms with a coefficient of zero, as they don't affect the degree).

Example: Consider the polynomial $10x^5 - 4x^2 y^3 + 6x - 1$. We found the degrees of the terms to be $5, 5, 1, 0$. The highest degree among these is $5$. Therefore, the degree of the polynomial is $\textbf{5}$.

Example: Consider the polynomial $P(x) = 7x^2 - 3x^5 + 2x^3 + 9$. The terms are $7x^2, -3x^5, 2x^3, 9$. Their degrees are $2, 5, 3, 0$, respectively. The highest degree is $5$. The degree of the polynomial $P(x)$ is $\textbf{5}$. When written in standard form (decreasing powers of $x$), $P(x) = -3x^5 + 2x^3 + 7x^2 + 9$, the highest degree term is clearly visible.

Example: A constant polynomial $c$, where $c \neq 0$ (e.g., $25$). This can be written as $25x^0$. The only term has degree $0$. The degree of a non-zero constant polynomial is $\textbf{0}$.

Example: The zero polynomial, $P(x) = 0$. This polynomial has no non-zero terms. Its degree is a special case and is usually considered to be undefined. Sometimes, for certain applications, it might be defined as $-\infty$, but "undefined" is the most common convention.

5. Classification by Degree

Polynomials are frequently categorized based on their degree, as the degree often indicates the number of roots (solutions) a polynomial equation might have and influences the shape of its graph. Common classifications include:

6. Leading Term and Leading Coefficient

For a polynomial in a single variable, when the terms are written in order of decreasing powers of the variable (this is called the standard form of the polynomial), the term with the highest degree is called the leading term. The coefficient of this leading term is called the leading coefficient.

The leading term and leading coefficient are particularly important as they determine the polynomial's behavior for very large positive or negative values of the variable (the end behavior of the graph).

Example: Consider the polynomial $P(x) = 8x^3 - 5x^7 + 2x - 1$. First, write it in standard form (decreasing powers of $x$):

$$ P(x) = -5x^7 + 8x^3 + 2x - 1 $$

The highest degree term is $-5x^7$. The leading term is $\textbf{-5x\textsuperscript{7}}$. The coefficient of this term is $-5$. So, the leading coefficient is $\textbf{-5}$. The degree of this polynomial is $7$.

Example: Consider the polynomial $Q(y) = y^2 - 5y + 1$. It is already in standard form. The highest degree term is $y^2$. The leading term is $\textbf{y\textsuperscript{2}}$. The coefficient of $y^2$ is $1$. So, the leading coefficient is $\textbf{1}$. The degree of this polynomial is $2$.

7. Classification by Number of Terms

Polynomials are also classified based on how many terms they contain after all like terms have been combined and the expression is simplified. This classification is simpler than by degree but also useful for identification.

Mastering these terms – term, coefficient, degree of a term, degree of a polynomial, leading term, leading coefficient, and the classification by the number of terms – provides the essential vocabulary for understanding and working with polynomials in algebra.



Value of a Polynomial

Evaluating a Polynomial at a Specific Point

Just like a general algebraic expression, the numerical value of a polynomial is not fixed. It changes depending on the specific numbers assigned to the variables within it. The process of finding this numerical value is called evaluating the polynomial. The resulting number is known as the value of the polynomial for those particular variable assignments.

If we have a polynomial denoted by $P(x)$, where $x$ is the variable, and we want to find its value when $x$ takes on a specific number, say $c$, we write this as $P(c)$. To calculate $P(c)$, we substitute the number $c$ for every occurrence of the variable $x$ in the polynomial expression and then carry out all the indicated arithmetic operations according to the standard order.

For a polynomial in multiple variables, say $Q(x, y)$, evaluating it at specific values, for example, $x=a$ and $y=b$, is denoted as $Q(a, b)$. We substitute $a$ for every $x$ and $b$ for every $y$ and calculate the result.


Procedure for Evaluating a Polynomial

To find the value of a polynomial for given numerical values of its variables, follow these steps:

  1. Substitution:

    Replace each variable in the polynomial with its corresponding given numerical value. It is a good practice to enclose the substituted number within parentheses, especially when dealing with negative numbers, fractions, or when the variable is raised to a power or multiplied by a coefficient. This helps maintain the correct signs and ensures operations are performed correctly.
  2. Calculation:

    Compute the value of the resulting numerical expression using the correct order of operations (BODMAS / PEMDAS).
    • Brackets / Parentheses: Evaluate expressions within grouping symbols first.
    • Orders / Exponents: Calculate powers and roots.
    • Division and Multiplication: Perform these from left to right.
    • Addition and Subtraction: Perform these from left to right.

Polynomials with real coefficients are defined for all real numbers. Therefore, you can substitute any real number for the variables, and the resulting arithmetic expression will always be defined.

Example 1. Find the value of the polynomial $P(x) = 3x^2 - 5x + 7$ when $x = 2$ and when $x = -1$.

Answer:

Case 1: Evaluate $P(x)$ when $x = 2$.

Substitute $x=2$ into the polynomial $P(x) = 3x^2 - 5x + 7$. We write this as $P(2)$.

$$ P(2) = 3(2)^2 - 5(2) + 7 $$

Now, evaluate the arithmetic expression following BODMAS:

First, calculate the exponent:

$$ (2)^2 = 4 $$

Substitute this back:

$$ P(2) = 3(4) - 5(2) + 7 $$

Next, perform the multiplications from left to right:

$$ 3(4) = 12 $$ $$ 5(2) = 10 $$

Substitute these back:

$$ P(2) = 12 - 10 + 7 $$

Finally, perform addition and subtraction from left to right:

$$ 12 - 10 = 2 $$ $$ 2 + 7 = 9 $$ $$ P(2) = 9 $$

The value of the polynomial $P(x)$ is $\textbf{9}$ when $x=2$.


Case 2: Evaluate $P(x)$ when $x = -1$.

Substitute $x=-1$ into the polynomial $P(x) = 3x^2 - 5x + 7$. We write this as $P(-1)$.

$$ P(-1) = 3(-1)^2 - 5(-1) + 7 $$

Now, evaluate the arithmetic expression following BODMAS:

First, calculate the exponent:

$$ (-1)^2 = (-1) \times (-1) = 1 $$

Substitute this back:

$$ P(-1) = 3(1) - 5(-1) + 7 $$

Next, perform the multiplications from left to right:

$$ 3(1) = 3 $$ $$ -5(-1) = (-5) \times (-1) = +5 $$

Substitute these back:

$$ P(-1) = 3 - (+5) + 7 $$

This simplifies to:

$$ P(-1) = 3 + 5 + 7 $$

Finally, perform addition and subtraction from left to right:

$$ 3 + 5 = 8 $$ $$ 8 + 7 = 15 $$

Correction Note:

Let's re-check the substitution and signs carefully:

$$ P(-1) = 3(-1)^2 - 5(-1) + 7 $$ $$ P(-1) = 3(1) - (-5) + 7 $$ $$ P(-1) = 3 + 5 + 7 $$ $$ P(-1) = 8 + 7 = 15 $$

My prior calculation $3 - (5) + 7$ was incorrect; $-5(-1)$ results in $+5$, so the expression should be $3 + 5 + 7$. Apologies for the error.

$$ P(-1) = 3(1) - 5(-1) + 7 = 3 - (-5) + 7 = 3 + 5 + 7 = 15 $$

The value of the polynomial $P(x)$ is $\textbf{15}$ when $x=-1$.


Example 2. Evaluate the polynomial $Q(a, b) = a^2 + 2ab + b^2$ when $a = 5$ and $b = -3$.

Answer:

We need to find the value of $Q(a, b)$ when $a=5$ and $b=-3$. Substitute these values into the expression $Q(a, b) = a^2 + 2ab + b^2$. We denote this as $Q(5, -3)$.

$$ Q(5, -3) = (5)^2 + 2(5)(-3) + (-3)^2 $$

Evaluate the arithmetic expression following BODMAS:

First, calculate the exponents:

$$ (5)^2 = 25 $$ $$ (-3)^2 = (-3) \times (-3) = 9 $$

Substitute these back:

$$ Q(5, -3) = 25 + 2(5)(-3) + 9 $$

Next, perform the multiplication:

$$ 2(5)(-3) = (2 \times 5) \times (-3) = 10 \times (-3) = -30 $$

Substitute this back:

$$ Q(5, -3) = 25 + (-30) + 9 $$

This simplifies to:

$$ Q(5, -3) = 25 - 30 + 9 $$

Finally, perform addition and subtraction from left to right:

$$ 25 - 30 = -5 $$ $$ -5 + 9 = 4 $$ $$ Q(5, -3) = 4 $$

The value of the polynomial $Q(a, b) = a^2 + 2ab + b^2$ is $\textbf{4}$ when $a=5$ and $b=-3$. (As a check, note that $Q(a,b)$ is the expansion of $(a+b)^2$. Substituting $a=5, b=-3$ into $(a+b)^2$ gives $(5+(-3))^2 = (5-3)^2 = 2^2 = 4$, which matches our calculated value).

Evaluating polynomials is a fundamental operation used in various contexts, such as finding points on the graph of a polynomial function, checking solutions to polynomial equations, or calculating specific outcomes in applied problems modelled by polynomials.



Zero of a Polynomial

Defining the Zeros (Roots)

A key concept in the study of polynomials is the idea of a "zero". A number is called a zero of a polynomial if, when that number is substituted for the variable in the polynomial expression, the value of the polynomial becomes $0$.

More formally, for a polynomial $P(x)$ in one variable $x$, a number $c$ is a zero of $P(x)$ if and only if evaluating the polynomial at $x=c$ results in $0$. That is:

A number $c$ is a zero of $P(x)$ $\iff P(c) = 0$.

The terms zero of a polynomial and root of a polynomial are often used interchangeably. The term "root" is more commonly used when the polynomial is set equal to zero to form an equation. Thus, the zeros of the polynomial $P(x)$ are the solutions or roots of the polynomial equation $P(x) = 0$. Finding these zeros or roots is a central problem in algebra with wide applications.


Methods for Finding Zeros (Simple Cases)

Finding the zeros of a polynomial means determining the specific value(s) of the variable(s) that make the polynomial expression evaluate to zero. The general strategy is to set the polynomial expression equal to zero and solve the resulting equation. The complexity of finding the zeros depends heavily on the degree of the polynomial.

To find the zeros of the polynomial $P(x)$, we solve the equation $P(x) = 0$.

[Equation to find Zeros]

1. Zero of a Constant Polynomial

A non-zero constant polynomial has the form $P(x) = c$, where $c$ is a specific constant number, and $c \neq 0$. To find its zeros, we set it equal to zero:

$$ P(x) = 0 \implies c = 0 $$

Since we assumed $c \neq 0$, this equation ($c=0$) is a contradiction. It has no solution for $x$. Therefore, a non-zero constant polynomial (like $P(x)=5$ or $P(x)=-2$) has no zeros.

The only constant polynomial that has zeros is the zero polynomial, $P(x) = 0$. When we set $P(x) = 0$, we get $0 = 0$, which is true for any value of $x$. Therefore, every real number is a zero of the zero polynomial.

2. Zero of a Linear Polynomial

A linear polynomial in one variable has the general form $P(x) = ax + b$, where $a$ and $b$ are constants, and crucially, $a \neq 0$. To find its zero, we set the polynomial equal to zero and solve the resulting linear equation:

$$ P(x) = 0 $$ $$ ax + b = 0 $$

Subtract $b$ from both sides:

$$ ax = -b $$

Divide by $a$ (which is non-zero):

$x = -\frac{b}{a} $

[Zero of a linear polynomial]

A linear polynomial with a non-zero leading coefficient ($a \neq 0$) always has exactly one unique real zero, given by the formula $x = -\frac{b}{a}$.

Example 1. Find the zero of the polynomial $P(x) = 5x + 10$.

Answer:

To find the zero, set $P(x)$ equal to zero:

$$ 5x + 10 = 0 $$

This is a linear equation. We need to isolate $x$. Subtract $10$ from both sides:

$5x + 10 - 10 = 0 - 10$

$5x = -10$

Divide both sides by $5$:

$\frac{5x}{5} = \frac{-10}{5}$

$x = -2$

The zero of the polynomial $P(x) = 5x + 10$ is $\textbf{x = -2}$.

Check: Evaluate $P(x)$ at $x=-2$. $P(-2) = 5(-2) + 10 = -10 + 10 = 0$. Since $P(-2) = 0$, our answer is correct.

3. Zeros of a Quadratic Polynomial

A quadratic polynomial in one variable has the general form $P(x) = ax^2 + bx + c$, where $a, b, c$ are constants and $a \neq 0$. Finding its zeros involves solving the quadratic equation $ax^2 + bx + c = 0$. Quadratic equations can have different types and numbers of solutions (zeros):

There are several methods to find the zeros of a quadratic polynomial:

Example 2. Find the zeros of the polynomial $P(x) = x^2 - 5x + 6$ using factoring and the quadratic formula.

Answer:

We need to find the values of $x$ for which $P(x) = 0$. Set the polynomial equal to zero:

$$ x^2 - 5x + 6 = 0 $$

This is a quadratic equation. It is in the form $ax^2 + bx + c = 0$, with $a=1$, $b=-5$, and $c=6$.

Method A: Factoring

We look for two numbers that multiply to $c=6$ and add up to $b=-5$. Let these numbers be $m$ and $n$. We need $m \times n = 6$ and $m + n = -5$. After checking pairs of factors of 6 (like (1,6), (2,3), (-1,-6), (-2,-3)), we find that the numbers $-2$ and $-3$ satisfy both conditions: $(-2) \times (-3) = 6$ and $(-2) + (-3) = -5$.

So, we can factor the quadratic expression:

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

According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for $x$:

$x - 2 = 0 \implies x = 2$

$x - 3 = 0 \implies x = 3$

The zeros of the polynomial $P(x) = x^2 - 5x + 6$ are $\textbf{x = 2}$ and $\textbf{x = 3}$.

Check: $P(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$. $P(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$. Both values are indeed zeros.


Method B: Using the Quadratic Formula

For the equation $x^2 - 5x + 6 = 0$, we have $a=1$, $b=-5$, and $c=6$. Substitute these values into the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} $$

Simplify the expression:

$$ x = \frac{5 \pm \sqrt{25 - 24}}{2} $$ $$ x = \frac{5 \pm \sqrt{1}}{2} $$ $$ x = \frac{5 \pm 1}{2} $$

This gives two possible values for $x$:

$x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3$

$x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2$

The zeros are $x = 3$ and $x = 2$. This matches the result obtained by factoring.

Number of Zeros

The number of zeros a polynomial has is related to its degree. A fundamental result in algebra, the Fundamental Theorem of Algebra, states that a polynomial of degree $n \ge 1$ with complex coefficients has exactly $n$ complex roots (zeros), when counted with multiplicity (meaning if a root appears multiple times in factoring, it is counted that many times).

When we are specifically looking for real zeros of a polynomial with real coefficients:

Finding the zeros of polynomials of degree 3 or higher can be more challenging and often involves techniques like the Rational Root Theorem, synthetic division (using potential rational roots to find factors), or numerical methods when exact analytical solutions are difficult or impossible to find.



Geometrical Representation of a Polynomial (Graphing Zeroes)

Connecting Algebra and Geometry

Mathematics often benefits from visual representations. Algebraic expressions, especially polynomials, can be represented geometrically using graphs. For a polynomial in one variable, say $P(x)$, we can associate it with a function $y = P(x)$. By assigning various values to $x$, calculating the corresponding value of $y = P(x)$, and plotting these $(x, y)$ pairs on a coordinate plane, we can sketch the graph of the polynomial. This graph is typically a smooth, continuous curve.

The graph of a polynomial provides a powerful visual tool for understanding its properties. In particular, there is a direct and important connection between the zeros of a polynomial and its graph.


Geometrical Meaning of Zeros

Recall that the zeros of a polynomial $P(x)$ are the values of $x$ for which $P(x) = 0$. When we plot the graph of $y = P(x)$, the y-coordinate of any point on the graph is given by the value of the polynomial $P(x)$ for the corresponding x-coordinate. The points on the coordinate plane where the y-coordinate is zero are the points that lie on the x-axis.

Therefore, the real zeros of a polynomial $P(x)$ are the x-coordinates of the points where the graph of $y = P(x)$ intersects or touches the x-axis. These points are also known as the x-intercepts of the graph.

The number of distinct real zeros of a polynomial corresponds to the number of distinct points where its graph intersects or touches the x-axis. A polynomial of degree $n$ can have at most $n$ distinct real zeros, and its graph can intersect or touch the x-axis at most $n$ times.


Visualizing Zeros Through Graphs

Let's examine the graphs of different degrees of polynomials to see how their real zeros are represented geometrically.

1. Graph of a Linear Polynomial ($y = ax + b$, $a \neq 0$)

The graph of a linear polynomial is always a straight line. As long as the leading coefficient $a$ is not zero, the line is not horizontal and must intersect the x-axis at exactly one point.

The zero of the linear polynomial $P(x) = ax + b$ is found by solving $ax+b=0$, which gives $x = -b/a$. On the graph of $y = ax + b$, this value $x = -b/a$ is the x-coordinate of the single point where the line crosses the x-axis (the x-intercept).

Graph of a straight line showing one x-intercept at x = -b/a.

The image shows a straight line graph. The point where it crosses the x-axis (the x-intercept) is the location of the polynomial's zero. The x-coordinate of this point is the value of the zero.

2. Graph of a Quadratic Polynomial ($y = ax^2 + bx + c$, $a \neq 0$)

The graph of a quadratic polynomial is a parabola. A parabola can intersect or touch the x-axis in three possible ways, corresponding to the number of distinct real zeros:

3. Graph of a Cubic Polynomial ($y = ax^3 + bx^2 + cx + d$, $a \neq 0$)

The graph of a cubic polynomial is a more complex curve than a parabola. A cubic polynomial (degree 3) must cross the x-axis at least once (because odd-degree polynomials with real coefficients always have at least one real root). It can intersect or touch the x-axis at one, two, or three distinct points. Thus, a cubic polynomial can have one, two, or three distinct real zeros.

Examples of cubic graphs showing 1, 2, and 3 x-intercepts.

The image shows examples of cubic graphs demonstrating these possibilities: a graph crossing once (one real zero), a graph touching and crossing (two distinct real zeros, one with multiplicity 2), and a graph crossing three times (three distinct real zeros).

4. General Polynomials

For a general polynomial $P(x)$ of degree $n$, the graph of $y = P(x)$ can have at most $n$ x-intercepts. The x-coordinates of these intercepts are the real zeros of the polynomial. The behavior of the graph (crossing vs. touching) at each x-intercept indicates the multiplicity of the corresponding zero.

Visualizing the graph of a polynomial is a powerful way to understand the concept of zeros and their relationship to the polynomial equation $P(x) = 0$. The graph provides a visual summary of where the polynomial's value is zero.



Relationship Between Zeroes and Coefficients of a Polynomial

One of the most elegant and useful concepts in polynomial algebra is the direct connection between the roots (zeros) of a polynomial and its coefficients. These connections are formally described by Vieta's formulas, named after the French mathematician François Viète.

The basis for these relationships lies in the Factor Theorem, which states that if $c$ is a zero of a polynomial $P(x)$, then $(x-c)$ is a factor of $P(x)$. If a polynomial of degree $n$ has zeros $\alpha_1, \alpha_2, \ldots, \alpha_n$, then it can be written in its factored form as $P(x) = a_n(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$, where $a_n$ is the leading coefficient. By expanding this factored form and comparing the coefficients of the corresponding powers of $x$ with the standard form of the polynomial, we can derive these relationships.


Relationship for a Linear Polynomial (Degree 1)

Consider a linear polynomial in one variable, $P(x) = ax + b$, where $a$ and $b$ are real coefficients and $a \neq 0$.

A linear polynomial of degree 1 has exactly one complex root, which in this case is always a real root. Let this unique zero be $\alpha$. By definition, $P(\alpha) = 0$.

Setting the polynomial equal to zero to find the root:

$$ ax + b = 0 $$

Subtract $b$ from both sides:

$$ ax = -b $$

Divide both sides by $a$ (since $a \neq 0$):

$$ x = -\frac{b}{a} $$

So, the unique zero is $\alpha = -\frac{b}{a}$.

This gives the relationship between the single zero and the coefficients of a linear polynomial:

$\text{Zero} = -\frac{\text{Constant term}}{\text{Coefficient of } x}$

[Zero of Linear Polynomial]

For a linear polynomial, the sum of the zeros is just the zero itself, and the product of the zeros is also just the zero itself (this relationship becomes more meaningful for polynomials of degree 2 or higher).


Relationship for a Quadratic Polynomial (Degree 2)

Consider a quadratic polynomial in one variable, $P(x) = ax^2 + bx + c$, where $a, b, c$ are real coefficients and $a \neq 0$. A quadratic polynomial of degree 2 has exactly two complex roots (zeros), counting multiplicity. Let the two zeros be denoted by $\alpha$ and $\beta$. These roots can be real or complex.

If $\alpha$ and $\beta$ are the zeros of $P(x)$, then $(x - \alpha)$ and $(x - \beta)$ are factors of $P(x)$. Therefore, $P(x)$ can be written in factored form as $P(x) = k(x - \alpha)(x - \beta)$ for some constant $k$.

The leading term of $P(x) = ax^2 + bx + c$ is $ax^2$, with coefficient $a$.

The leading term of $k(x - \alpha)(x - \beta)$ is $k \times x \times x = kx^2$, with coefficient $k$.

Comparing the leading coefficients, we find that $k = a$.

So, we have the identity:

$$ ax^2 + bx + c = a(x - \alpha)(x - \beta) $$

Now, let's expand the right side of this equation:

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

Group the terms containing $x$:

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

Distribute the constant $a$ into the terms inside the parentheses:

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

Now, we equate the coefficients of corresponding powers of $x$ from the standard form $ax^2 + bx + c$ and the expanded factored form $ax^2 - a(\alpha + \beta)x + a(\alpha \beta)$:

These two relationships for quadratic polynomials are fundamental Vieta's formulas for degree 2.

Example 1. Find the sum and product of the zeros of the polynomial $P(x) = 5x^2 - 7x - 10$ without explicitly finding the zeros.

Answer:

The given polynomial is $P(x) = 5x^2 - 7x - 10$. This is in the standard quadratic form $ax^2 + bx + c$.

Comparing the given polynomial with the standard form, we identify the coefficients:

$a = 5$

$b = -7$

$c = -10$

Let the zeros of the polynomial be $\alpha$ and $\beta$.

Using the relationship for the sum of zeros:

$$ \alpha + \beta = -\frac{b}{a} $$

Substitute the values of $a$ and $b$:

$$ \alpha + \beta = -\frac{(-7)}{5} = \frac{7}{5} $$

Using the relationship for the product of zeros:

$$ \alpha \times \beta = \frac{c}{a} $$

Substitute the values of $a$ and $c$:

$$ \alpha \times \beta = \frac{-10}{5} = -2 $$

Answer: The sum of the zeros of $5x^2 - 7x - 10$ is $\textbf{\frac{7}{5}}$, and the product of the zeros is $\textbf{-2}$.

Verification (Optional):

To verify this, we could find the zeros using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(5)(-10)}}{2(5)} $$ $$ x = \frac{7 \pm \sqrt{49 + 200}}{10} $$ $$ x = \frac{7 \pm \sqrt{249}}{10} $$

The two zeros are $\alpha = \frac{7 + \sqrt{249}}{10}$ and $\beta = \frac{7 - \sqrt{249}}{10}$.

Sum of zeros: $\alpha + \beta = \frac{7 + \sqrt{249}}{10} + \frac{7 - \sqrt{249}}{10} = \frac{(7 + \sqrt{249}) + (7 - \sqrt{249})}{10} = \frac{7 + \sqrt{249} + 7 - \sqrt{249}}{10} = \frac{14}{10} = \frac{7}{5}$. (Matches the formula).

Product of zeros: $\alpha \times \beta = \left(\frac{7 + \sqrt{249}}{10}\right) \times \left(\frac{7 - \sqrt{249}}{10}\right)$. This is in the form $(A+B)(A-B)=A^2-B^2$.

$$ \alpha \times \beta = \frac{(7)^2 - (\sqrt{249})^2}{10^2} = \frac{49 - 249}{100} = \frac{-200}{100} = -2 $$

The calculated sum and product match the values obtained from the coefficient relationships.


Relationship for a Cubic Polynomial (Degree 3)

Consider a cubic polynomial in one variable, $P(x) = ax^3 + bx^2 + cx + d$, where $a, b, c, d$ are real coefficients and $a \neq 0$. A cubic polynomial of degree 3 has exactly three complex roots, counting multiplicity. Let these three zeros be denoted by $\alpha, \beta,$ and $\gamma$.

Similar to the quadratic case, if $\alpha, \beta, \gamma$ are the zeros, then $(x-\alpha)$, $(x-\beta)$, and $(x-\gamma)$ are factors of $P(x)$. Thus, $P(x) = a(x-\alpha)(x-\beta)(x-\gamma)$ since $a$ is the leading coefficient.

Let's expand the right side:

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

Now, distribute $x$ and $-\alpha$ from the first factor:

$$ = a [ x(x^2 - (\beta + \gamma)x + \beta \gamma) - \alpha(x^2 - (\beta + \gamma)x + \beta \gamma) ] $$ $$ = a [ (x^3 - (\beta + \gamma)x^2 + \beta \gamma x) - (\alpha x^2 - \alpha(\beta + \gamma)x + \alpha\beta\gamma) ] $$

Distribute the negative sign in the second parenthesis and $\alpha$:

$$ = a [ x^3 - (\beta + \gamma)x^2 + \beta \gamma x - \alpha x^2 + \alpha\beta x + \alpha\gamma x - \alpha\beta\gamma ] $$

Group terms by powers of $x$ ($x^3$, $x^2$, $x$, constant):

$$ = a [ x^3 + (-\beta - \gamma - \alpha)x^2 + (\beta \gamma + \alpha\beta + \alpha\gamma)x - \alpha\beta\gamma ] $$

Rewrite the coefficients and distribute the outer $a$:

$$ = a x^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \beta\gamma + \gamma\alpha)x - a(\alpha\beta\gamma) $$

Comparing this to the standard form $ax^3 + bx^2 + cx + d$, we equate the coefficients of corresponding powers of $x$:

These are the fundamental Vieta's formulas for a cubic polynomial.

Example 2. For the polynomial $P(x) = x^3 + 4x^2 + x - 6$, find the sum of zeros, the sum of the products of zeros taken two at a time, and the product of zeros.

Answer:

The given polynomial is $P(x) = x^3 + 4x^2 + x - 6$. This is in the standard cubic form $ax^3 + bx^2 + cx + d$.

Comparing coefficients:

$a = 1$

$b = 4$

$c = 1$

$d = -6$

Let the zeros be $\alpha, \beta, \gamma$.

Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{4}{1} = \textbf{-4}$.

Sum of products of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{1}{1} = \textbf{1}$.

Product of zeros: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{(-6)}{1} = \textbf{6}$.

Answer: The sum of the zeros is $-4$, the sum of the products of the zeros taken two at a time is $1$, and the product of the zeros is $6$.

Verification (Optional):

Factoring (or using the Rational Root Theorem and synthetic division, which will be covered later) reveals that the zeros of $x^3 + 4x^2 + x - 6$ are $x=1, x=-2, x=-3$. Let's check the formulas:

Sum: $1 + (-2) + (-3) = 1 - 2 - 3 = 1 - 5 = -4$. (Matches $-\frac{b}{a}$)

Pairwise Products: $(1)(-2) + (-2)(-3) + (-3)(1) = -2 + 6 - 3 = 4 - 3 = 1$. (Matches $\frac{c}{a}$)

Product: $(1)(-2)(-3) = -2 \times -3 = 6$. (Matches $-\frac{d}{a}$)

Example 3. If the sum of the zeros of the polynomial $P(x) = 2x^3 - kx^2 + 5x + 9$ is $3$, find the value of $k$.

Answer:

The given polynomial is $P(x) = 2x^3 - kx^2 + 5x + 9$. This is a cubic polynomial in the standard form $ax^3 + bx^2 + cx + d$.

Comparing coefficients:

$a = 2$

$b = -k$

$c = 5$

$d = 9$

Let the zeros of the polynomial be $\alpha, \beta, \gamma$. We are given that the sum of the zeros is $3$.

$$ \alpha + \beta + \gamma = 3 $$

Using the relationship for the sum of zeros of a cubic polynomial, we know that:

$$ \alpha + \beta + \gamma = -\frac{b}{a} $$

Substitute the expression for the sum of zeros and the values of $a$ and $b$ from the polynomial:

$$ 3 = -\frac{(-k)}{2} $$

Simplify the right side:

$$ 3 = \frac{k}{2} $$

Multiply both sides by $2$ to solve for $k$:

$$ 3 \times 2 = \frac{k}{2} \times 2 $$ $$ 6 = k $$

So, $k = 6$.

Answer: The value of $k$ is $\textbf{6}$. The polynomial is $2x^3 - 6x^2 + 5x + 9$.

Generalization: Vieta's Formulas

The relationships observed for quadratic and cubic polynomials are specific cases of a general set of formulas known as Vieta's formulas, which apply to polynomials of any degree.

For a general polynomial of degree $n$ in one variable $x$, written in standard form:

$$ P(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots + a_1 x^1 + a_0 $$

where $a_n \neq 0$, and its $n$ complex roots (zeros) are $\alpha_1, \alpha_2, \ldots, \alpha_n$ (counting multiplicity), Vieta's formulas describe the sums and products of these roots in terms of the coefficients:

Vieta's formulas are incredibly useful for checking polynomial roots, finding roots when some are already known, constructing polynomials with specified roots, and proving various theorems about polynomials.