Table of Contents
Last modified on December 6th, 2024
A quintic polynomial or quintic function is a type of polynomial that has a degree of 5. It means the highest exponent of its variable, typically denoted by x, is 5. The term ‘Quintic’ is derived from a Latin word meaning ‘fifth.’
A quintic polynomial can be expressed in the standard form as:
f(x) = ax5 + bx4 + cx3 + dx2 + ex + f
Here,
a, b, c, d, e, and f are constants, and a ≠ 0
Here are a few examples of quintic polynomials:
Quintic functions are commonly used in calculus, hydrodynamics, computer graphics, optics, and spatial analysis.
A quintic polynomial (fifth-degree polynomial) has 5 roots or solutions, which may be real or complex numbers. Thus, the polynomial can have up to 5 turning points (inflection points) on its graph.
Let us graph the polynomial f(x) = x5 – 5x4 + 5x3 + 5x2 – 6x
Putting x = 0 in the given polynomial, we get
y = f(0) = (0)5 – 5(0)4 + 5(0)3 + 5(0)2 – 6(0)
⇒ y = 0
Thus, the y-intercept is (0, 0)
Putting f(x) = 0, we get
x5 – 5x4 + 5x3 + 5x2 – 6x = 0
Factoring the Quintic Polynomial
⇒ x(x4 – 5x3 + 5x2 + 5x – 6) = 0 …..(i)
Factoring the Quartic Polynomial
From (i), we get
x4 – 5x3 + 5x2 + 5x – 6 = 0
⇒ x4 – x3 – 4x3 + 4x2 + x2 – x + 6x – 6 = 0
⇒ x3(x – 1) – 4x2(x – 1) + x(x – 1) + 6(x – 1) = 0
⇒ (x – 1)(x3 – 4x2 + x + 6) = 0 …..(ii)
Factoring the Cubic Polynomial
x3 – 4x2 + x + 6 = 0
⇒ x3 + x2 – 5x2 – 5x + 6x + 6 = 0
⇒ x2(x + 1) – 5x(x + 1) + 6(x + 1) = 0
⇒ (x + 1)(x2 – 5x + 6) = 0 …..(iii)
Factoring the Quadratic Polynomial
x2 – 5x + 6 = 0
⇒ x2 – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2(x – 3) = 0
⇒ (x – 3)(x – 2) = 0 …..(iv)
Now, combining (i), (ii), (iii), and (iv), we can factor the given quintic polynomial as:
x5 – 5x4 + 5x3 + 5x2 – 6x = x(x – 1)(x + 1)(x – 3)(x – 2)
Here,
x = 0
(x – 1) = 0 ⇒ x = 1
(x + 1) = 0 ⇒ x = -1
(x – 3) = 0 ⇒ x = 3
(x – 2) = 0 ⇒ x = 2
Thus, the x-intercepts are (-1, 0), (0, 0), (1, 0), (2, 0), and (3, 0)
Now, we choose random values of x from each interval between the x- and y-intercepts. For each chosen value, we evaluate the function and complete the table by recording the corresponding function values.
x | f(x) |
-0.5 | 3.2813 |
-0.65 | 3.6308 |
1.5 | 1.4063 |
1.75 | 1.1279 |
2.25 | -1.7139 |
2.5 | -3.2813 |
If the leading coefficient is greater than zero, then
If the leading coefficient is less than zero, then
Now, identifying the sign of the leading coefficient of the given function f(x), we get
a = 1 (> 0)
Thus, as x → -∞, f(x) → -∞, and as x → ∞, f(x) → ∞
Thus, the polynomial has 5 roots and 4 distinct extrema (2 local maximums and 2 local minimums).
The derivative of a quintic function always results in a quartic function.
Let us now find the derivative of the polynomial f(x) = x5 – 5x4 + 5x3 + 5x2 – 6x
Differentiating the function with respect to x, we get
f’(x) = 5x4 – 5 ⋅ 4x3 + 5 ⋅ 3x2 + 5 ⋅ 2x – 6 (using the power rule)
⇒ f’(x) = 5x4 – 20x3 + 15x2 + 10x – 6
Calculate the first derivative of each of the following quintic polynomials:
a) f(x) = 3x5 – 2x4 + 4x2 – 6x + 1
b) g(x) = x5 – 7x3 + 5x
a) Given, f(x) = 3x5 – 2x4 + 4x2 – 6x + 1
Differentiating with respect to x,
f’(x) = 3 ⋅ 5x4 – 2 ⋅ 4x3 + 4 ⋅ 2x – 6
⇒ f’(x) = 15x4 – 8x3 + 8x – 6
b) Given, g(x) = x5 – 7x3 + 5x
Differentiating with respect to x,
g’(x) = 5x4 – 7 ⋅ 3x2 + 5
⇒ g’(x) = 5x4 – 21x2 + 5
Factor the following quintic polynomial and find its real roots:
f(x) = x5 – 4x3 + 3x
Given, f(x) = x5 – 4x2 + 3x
Putting f(x) = 0, we get
x5 – 4x2 + 3x = 0
⇒ x(x4 – 4x + 3) = 0
⇒ x(x4 – x3 + x3 – x2 + x2 – x – 3x + 3) = 0
⇒ x[x3(x – 1) + x2(x – 1) + x(x – 1) – 3(x – 1)] = 0
⇒ x(x – 1)(x3 + x2 + x – 3) = 0
⇒ x(x – 1)(x3 – x2 + 2x2 – 2x + 3x – 3) = 0
⇒ x(x – 1)[x2(x – 1) + 2x(x -1) + 3(x – 1)] = 0
⇒ x(x – 1)(x – 1)(x2 + 2x + 3) = 0
Here,
x2 + 2x + 3 = 0 gives imaginary roots
x = 0 and (x – 1) = 0 ⇒ x = 1
Thus, x5 – 4x2 + 3x = x(x – 1)(x – 1)(x2 + 2x + 3)
The real roots are x = 0 and 1
Graph the quintic polynomial f(x) = x5 – 5x and answer the following:
a) Find the x-intercepts
b) Determine the turning points and inflection points
c) Identify the end behavior
a) X-Intercepts:
x5 – 5x = 0
⇒ x(x4 – 5) = 0
⇒ x = 0 and x4 – 5 = 0
⇒ x = 0 and x4 = 5
Solving x4 = 5, we get
The real x-intercepts are approximately
x = 0, -1.495, 1.495
b) Turning Points or Inflection Points:
The real turning points occur at x = 1
Here is an inflection point at x = 0
c) End Behavior:
Since the leading term is x5:
as x → -∞, f(x) → -∞
as x → ∞, f(x) → ∞
Problem: Identifying QUINTIC BINOMIAL
What is a quintic binomial? Which of the following polynomials is a quintic binomial?
a) f(x) = x5
b) g(x) = x5 + x3 – 2x + 3
c) h(x) = x5 + 2
d) m(x) = 5x5 – 3x4 + 7x2 + 15
A quintic binomial is a type of polynomial that consists of two terms, and the leading term is of degree 5 (meaning the highest exponent is 5).
a) Given, f(x) = x5 is of a single term with the degree 5
Thus, it is not a quintic binomial.
b) Given, g(x) = x5 + x3 – 2x + 3 consists of 4 terms with the degree 5
Thus, it is not a quintic binomial.
c) Given, h(x) = x5 + 2 is of two terms with the degree 5
Thus, it is a quintic binomial.
d) Given, m(x) = 5x5 – 3x4 + 7x2 + 2x – 15 consists of 5 terms with the degree 5
Thus, it is not a quintic binomial.
Last modified on December 6th, 2024