Table of Contents
Last modified on November 9th, 2024
A polynomial function is a type of mathematical function that involves a sum of terms, each consisting of a variable (usually denoted by x) raised to a whole-number exponent and multiplied by a constant coefficient.
Here are a few examples of polynomial functions:
A polynomial function in standard form is written as:
f(x) = anxn + an – 1xn – 1 + … + a1x + a0
Here,
For example,
In the polynomial f(x) = 3x2 – 2x + 1,
A polynomial function can be classified either based on the number of terms or its degree, such as:
Polynomials can be categorized by the number of terms they contain. The types are:
A polynomial with a single term is called a monomial. It has the general form: f(x) = axn; here, a ≠ 0. An example is 2x
A polynomial with exactly two terms is called a binomial. The general form is axm + byn; here, a and b are constants, and x and y are non-negative integers. An example is g(x) = 2x + 7
A polynomial function with three terms is called a trinomial. For example, h(x) = x2 + 2x + 7
The degree of a polynomial function is defined as the highest exponent of the variable in the function. Accordingly, polynomials can be classified into the following types:
A function of the form f(x) = 0 (here, ∀ an = 0 and n ∈ ℕ) is called a zero polynomial.
A function of the form f(x) = a or f(x) = ax0 is called a constant polynomial. An example is f(x) = 7
A function of the form f(x) = ax + b is called a linear polynomial. An example is f(x) = 2x + 7
A function of the form f(x) = ax2 + bx + c is called a quadratic polynomial. An example is f(x) = x2 + 2x + 7
A function of the form f(x) = ax3 + bx2 + cx + d is called a cubic polynomial. An example is f(x) = 3x3 + x2 + 2x + 7
A function of the form f(x) = ax4 + bx3 + cx2 + dx + e is called a quartic polynomial. An example is f(x) = x4 – 4x3 + 3x2 + 2x – 8
A function of the form f(x) = ax5 + bx4 + cx3 + dx2 + ex + f is called a quintic polynomial. An example is f(x) = -x5 + x4 – 4x3 + 3x2 + 2x – 8
The graph of a polynomial function is smooth and continuous, with no sharp corners or breaks. The degree of the polynomial determines the number of turning points (where the graph changes direction) and the general shape of the graph.
To graph a polynomial function, we first create a table of values with some selected values of x and the corresponding values of f(x). Then, we plot these points and connect them with a smooth curve.
The constant polynomial function f(x) = a contains no variable x. Its graph is a horizontal line at y = a, which means it remains constant regardless of the value of x.
Let us graph y = 2
x | -2 | -1 | 0 | 1 | 2 |
f(x) | 2 | 2 | 2 | 2 | 2 |
After plotting the points, we get a horizontal line y = 2
Here, the graph has no turning points or roots (unless the constant is zero, forming the zero polynomial function, in which case the graph is the x-axis).
The graph of a linear polynomial function f(x) = ax + b forms a straight line. Here, a represents the slope, and b represents the y-intercept. The graph can have at most one root.
Plotting the linear polynomial y = 2x + 7, we get
x | -5 | -4 | -2 | -1 | 0 |
f(x) | -3 | -1 | 3 | 5 | 7 |
The graph of a quadratic polynomial function f(x) = ax2 + bx + c forms a parabolic curve. Here, a determines the direction of the parabola, and c represents the y-intercept.
Here, the graph can have zero, one, or two roots.
The graph of y = x2 + 5x + 6 is shown below.
x | -5 | -4 | -2 | -1 | 0 |
f(x) | 6 | 2 | 0 | 2 | 6 |
Higher polynomial functions are of the form f(x) = anxn + an – 1xn – 1 + … + a2x2 + a1x + a0. If the polynomial is of degree n, the graph of the function can be intersected by a straight line at up to n points. The constant term, a0, represents the y-intercept.
For example,
Plotting y = x4 + 2x3 – x – 2, we get
x | -2 | -1 | -0.5 | 0 | 1 |
f(x) | 0 | -2 | -1.6875 | -2 | 0 |
The zeros or roots (also known as the x-intercepts) of a polynomial function f(x) are values that satisfy the equation f(x) = 0. To find the zeros, the function is set equal to 0, and then, the equation is solved for the variable.
Let us find the zeros of the polynomial f(x) = x – 5
Let f(x) = 0
x – 5 = 0
Solving for x
⇒ x = 5
Thus, x = 5 is the root of f(x) = x – 5
Let us find the zeros of the polynomial f(x) = x2 + 5x + 6
Let f(x) = 0
x2 + 5x + 6 = 0 …..(i)
As we know, the standard form of quadratic polynomial is f(x) = ax2 + bx + c …..(ii)
Comparing (i) and (ii), we have
a = 1, b = 5, and c = 6
Solving for x
Here, to solve for the variable x, we can use the factoring method or the quadratic formula.
Using the quadratic formula, we get
x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
⇒ x = ${\dfrac{-5\pm \sqrt{5^{2}-4\times 1\times 6}}{2\times 1}}$
⇒ x = -2 and x = -3
Thus, x = -2 and x = -3 are the roots of f(x) = x2 + 5x + 6
Note: The irrational roots and complex roots of a polynomial function always occur in pairs.
Now, let us find the zeros of the cubic polynomial function f(x) = x3 – 2x2 – x + 2
Let f(x) = 0
x3 – 2x2 – x + 2 = 0
Solving for x
⇒ (x3 – 2x2) + (-x + 2) = 0
⇒ x2(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(x2 – 1) = 0
⇒ (x – 2)(x + 1)(x – 1) = 0
⇒ x = 2, x = -1, and x = 1
Thus, x = -1, 1, and 2 are the roots of f(x) = x3 – 2x2 – x + 2
Determine which of the following is a polynomial function.
a) f(x) = ${x^{2}+\dfrac{3}{2x}+7}$
b) g(x) = ${x^{2}+\dfrac{3}{2}x+7}$
c) h(x) = ${x^{-2}+\dfrac{3}{2}x+7}$
d) k(x) = ${\sqrt{x}+\dfrac{3}{2}x+7}$
e) p(x) = ${\dfrac{3}{x-1}}$
a) Here, f(x) = ${x^{2}+\dfrac{3}{2x}+7}$ is not a polynomial function since it has a negative exponent for the variable x
b) Here, g(x) = ${x^{2}+\dfrac{3}{2}x+7}$ is a polynomial function.
c) Here, h(x) = ${x^{-2}+\dfrac{3}{2}x+7}$ is not a polynomial function since it has a negative exponent for the variable x
d) Here, k(x) = ${\sqrt{x}+\dfrac{3}{2}x+7}$ is not a polynomial function since it has a fractional exponent for the variable x
e) Here, p(x) = ${\dfrac{3}{x-1}}$ is not a polynomial since it has a negative exponent for the variable x
Determine the degree of the polynomial function y = 5x + 2x5 – 3x3 – x4 + 1
Given, y = 5x + 2x5 – 3x3 – x4 + 1
⇒ y = 2x5 – x4 – 3x3 + 5x + 1
Here, the variable is x, and the highest exponent of x is 5
Thus, the degree of the polynomial is 5
Find the zeros of the following polynomials:
a) f(x) = x2 – 7x + 12
b) g(x) = x3 – 16x
c) h(x) = x4 – 5x2 + 4
a) Given, f(x) = x2 – 7x + 12
Here, x2 – 7x + 12 = 0
⇒ x2 – 4x – 3x + 12 = 0
⇒ x(x – 4) – 3(x – 4) = 0
⇒ (x – 4)(x – 3) = 0
⇒ x = 4 and 3
Thus, x = 3 and 4 are the roots of f(x) = x2 – 7x + 12
b) Given, g(x) = x3 – 16x
Here, x3 – 16x = 0
⇒ x(x2 – 16) = 0
⇒ x(x + 4)(x – 4) = 0
⇒ x = 0, -4, and 4
Thus, x = -4, 0, and 4 are the roots of g(x) = x3 – 16x
c) Given, h(x) = x4 – 5x2 + 4
Here, x4 – 5x2 + 4 = 0
⇒ x4 – 4x2 – x2 + 4 = 0
⇒ x2(x2 – 4) – 1(x2 – 4) = 0
⇒ (x2 – 4)(x2 – 1) = 0
⇒ (x – 4)(x + 4)(x – 1)(x + 1) = 0
⇒ x = -4, 4, 1, and -1Thus, x = -4, -1, 1, and 4 are the roots of h(x) = x4 – 5x2 + 4
Last modified on November 9th, 2024