Last modified on December 6th, 2024

chapter outline

 

Quintic Polynomial

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.’ 

Formula

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:

  • f(x) = x5 + 3x3 – 9x2 + 5
  • g(x) = 9x5 + 2x3 – x2 + 5
  • h(x) = x5 – x4 + 4x – 18
  • k(x) = x5 + 2
  • m(x) = 2x5 + x3 – 3x + 7

Quintic functions are commonly used in calculus, hydrodynamics, computer graphics, optics, and spatial analysis.

Degree and Roots

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.

Graphing

Let us graph the polynomial f(x) = x5 – 5x4 + 5x3 + 5x2 – 6x

Finding the Y-Intercept

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)

Identifying the X-Intercepts

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)

Choosing Random Values of x and Completing the Table

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.

xf(x)
-0.53.2813
-0.653.6308
1.51.4063
1.751.1279
2.25-1.7139
2.5-3.2813

Determining the End Behavior of the Function

If the leading coefficient is greater than zero, then

  • as x → -∞, f(x) → -∞ 
  • as x → ∞, f(x) → ∞ 

If the leading coefficient is less than zero, then 

  • as x → -∞, f(x) → ∞
  • as x → ∞, f(x) → -∞

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) → ∞

Plotting the Points

Quintic Polynomial Graph

Thus, the polynomial has 5 roots and 4 distinct extrema (2 local maximums and 2 local minimums).

Finding Derivatives

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

Solved Examples

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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