Worksheet 04 - Definition of Polynomial Functions

Section 04: Polynomial Functions

Definition of a Polynomial Function

A function \(f\) whose equation can be written in the form \[f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_{0}\] is called a polynomial function of degree n. Here, \(a_{0}, a_{1}, \ldots, a_{n}\) are real numbers, \(a_{n} \neq 0\), and \(n\) is a natural number.

The numbers \(a_{0}, a_{1}, \ldots, a_{n}\) are called coefficients.

Counter-example: A function with the equation

\[f(x) = \frac{x + 0.5x^{2}}{x^{2} - 2x + 1}\]

is not a polynomial function (it is a rational function).


Worked Examples

Example 1

A function with the equation \[f(x) = 7x^{4} - \sqrt{5}x + 1\] is a polynomial of degree 4.

Coefficients: \(a_{4} = 7\), \(a_{3} = 0\), \(a_{2} = 0\), \(a_{1} = -\sqrt{5}\), \(a_{0} = 1\).

Example 2

A function with the equation \[g(x) = x^{2} - 4x - x^{2}\] is a polynomial of degree 1 (simplifies to: \(g(x) = -4x\)).

Coefficient: \(a_{1} = -4\), \(a_{0} = 0\).

Example 3

A function with the equation \[h(x) = \frac{x^{2}}{x + 1}\] is not a polynomial function, because it cannot be written in the form \[h(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_{0}\]


Practice Problems

Instructions: For each function below, determine:

  1. Is it a polynomial function? (Yes/No)
  2. If yes, what is its degree?
  3. If yes, list all non-zero coefficients

Problem 1

\(f(x) = -4x^{5} - 4\)

Yes, this is a polynomial function.

Degree: 5

Coefficients: \(a_{5} = -4\), \(a_{0} = -4\)

(Note: \(a_{4} = a_{3} = a_{2} = a_{1} = 0\))


Problem 2

\(g(x) = x^{20} + 5x^{5}\)

Yes, this is a polynomial function.

Degree: 20

Coefficients: \(a_{20} = 1\), \(a_{5} = 5\)

(All other coefficients are 0)


Problem 3

\(h(x) = 2^{x} - 3x\)

No, this is not a polynomial function.

Reason: The term \(2^{x}\) is an exponential function (the variable \(x\) is in the exponent), which cannot be written as a power of \(x\) with a constant exponent.


Problem 4

\(i(x) = x^{-2} + 4x\)

No, this is not a polynomial function.

Reason: The term \(x^{-2} = \frac{1}{x^{2}}\) has a negative exponent. Polynomial functions require non-negative integer exponents only.


Problem 5

\(j(x) = \frac{4}{x} + x\)

No, this is not a polynomial function.

Reason: The term \(\frac{4}{x} = 4x^{-1}\) has a negative exponent. This is a rational function, not a polynomial.


Problem 6

\(k(x) = 100\)

Yes, this is a polynomial function.

Degree: 0

Coefficient: \(a_{0} = 100\)

This is a constant function, which is a polynomial of degree 0.


Problem 7

\(l(x) = (x - 1)(x - 3)\)

Yes, this is a polynomial function.

First, expand the product: \[l(x) = (x - 1)(x - 3) = x^{2} - 3x - x + 3 = x^{2} - 4x + 3\]

Degree: 2

Coefficients: \(a_{2} = 1\), \(a_{1} = -4\), \(a_{0} = 3\)


Problem 8

\(m(x) = \sqrt{2}x^{2} - x + 1\)

Yes, this is a polynomial function.

Degree: 2

Coefficients: \(a_{2} = \sqrt{2}\), \(a_{1} = -1\), \(a_{0} = 1\)

Note: The coefficient \(\sqrt{2}\) is a real number (even though it’s irrational), so this is still a valid polynomial function. The key requirement is that coefficients are real numbers, not that they are rational.


Additional Practice Problems

For each function below, determine if it is a polynomial function and state its degree if applicable.

Problem 9

\(n(x) = 3x^{4} - 2x^{3} + x^{2} - 7\)

Yes, this is a polynomial function.

Degree: 4

Coefficients: \(a_{4} = 3\), \(a_{3} = -2\), \(a_{2} = 1\), \(a_{1} = 0\), \(a_{0} = -7\)


Problem 10

\(p(x) = \frac{x^{3} + 2x}{x}\)

Yes, this is a polynomial function (after simplification).

First, simplify: \[p(x) = \frac{x^{3} + 2x}{x} = \frac{x(x^{2} + 2)}{x} = x^{2} + 2\]

(Note: This simplification is valid for \(x \neq 0\))

Degree: 2

Coefficients: \(a_{2} = 1\), \(a_{1} = 0\), \(a_{0} = 2\)

Important: While the original form looks like a rational function, after simplification it becomes a polynomial. However, the domain excludes \(x = 0\) in the original form.


Problem 11

\(q(x) = x^{3/2} + 5x - 1\)

No, this is not a polynomial function.

Reason: The term \(x^{3/2} = x^{1.5} = \sqrt{x^{3}}\) has a fractional exponent. Polynomial functions require integer (whole number) exponents only.


Challenge Problems

Challenge 1

\(s(x) = (2x + 1)^{3}\)

Yes, this is a polynomial function.

Expand using the binomial theorem or repeated multiplication: \[(2x + 1)^{3} = (2x)^{3} + 3(2x)^{2}(1) + 3(2x)(1)^{2} + (1)^{3}\] \[= 8x^{3} + 12x^{2} + 6x + 1\]

Degree: 3

Coefficients: \(a_{3} = 8\), \(a_{2} = 12\), \(a_{1} = 6\), \(a_{0} = 1\)


Challenge 2

\(t(x) = \sqrt{x^{4} + 2x^{2} + 1}\)

It depends! Let’s analyze this carefully.

Notice that: \[x^{4} + 2x^{2} + 1 = (x^{2})^{2} + 2(x^{2})(1) + 1^{2} = (x^{2} + 1)^{2}\]

Therefore: \[t(x) = \sqrt{(x^{2} + 1)^{2}} = |x^{2} + 1|\]

Since \(x^{2} \geq 0\) for all real \(x\), we have \(x^{2} + 1 \geq 1 > 0\), so: \[t(x) = x^{2} + 1\]

Yes, this simplifies to a polynomial function.

Degree: 2

Coefficients: \(a_{2} = 1\), \(a_{1} = 0\), \(a_{0} = 1\)