FIND THE NTH DEGREE POLYNOMIAL WITH THE GIVEN CONDITION 

Find an nth degree polynomial function with real coefficients satisfying the given conditions.

Problem 1 :

n = 3; 3 and i are zeros; f(2) = 25 

Solution:

The given zeroes are 3 and i. Since one of the roots is in the form of complex number, its conjugate cane be considered as other root.

i and -i are two roots.

The roots are x = 3, i and -i

Using the complex roots, we can find the quadratic function. By multiplying the quadratic function with linear function, we will get cubic polynomial which is required.

α = i and β = -i

Quadratic function with roots α and β :

x² -  (α+β)x + αβ = 0

α+β = i + (-i) ==> 0

αβ = i(-i) ==> -i²

= -(-1)

αβ = 1

x² -  (0)x + 1 = 0

x² + 1 = 0

Linear factor is (x - 3)

By multiplying quadratic factor and linear factor, we will get the polynomial with the highest exponent of 3.

f(x) = a (x - 3)(x² + 1) -----(1)

Given condition is f(2) = 25

When x = 2, y = 25

25 = a (2 - 3)(2² + 1)

25 = a(-1)(5)

-5a = 25

a = -5

By applying the value of a in (1), we get

f(x) = -5 (x - 3)(x² + 1)

f(x) = -5 (x³ + x - 3x² - 3)

f(x) = -5x³ + 15x² - 5x + 15

So, the required polynomial is

f(x) = -5x³ + 15x² - 5x + 15

Problem 2 :

n = 3; -4 and i are zeros; f(-3) = 60

Solution:

The zeroes are -4, i and -i. Since one of the root is complex number, its conjugate is taken as another root.

α = i and β = -i

Quadratic function with roots α and β :

x² -  (α+β)x + αβ = 0

α+β = i + (-i) ==> 0

αβ = i(-i) ==> -i²

= -(-1)

αβ = 1

x² -  (0)x + 1 = 0

x² + 1 = 0

f(x) = a(x + 4)(x² + 1)

Given condition is f(-3) = 60

60 = a(-3 + 4)((-3)² + 1)

60 = a(1)(10)

a = 60/10

a = 6

f(x) = 6(x + 4)(x² + 1)

= 6(x³ + x + 4x² + 4)

= 6x³ + 6x + 24x² + 24

f(x) = 6x³ + 24x² + 6x + 24

Problem 3 :

n = 4; 3, 1/3, and 1 + 2i are zeros; f(1) = 48

Solution :

The given roots are 3, 1/3, 1 + 2i and 1 - 2i

Creating quadratic function with complex roots :

α = 1 + 2i and β = 1 - 2i

α + β = 1 + 2i + 1 - 2i ==> 2

αβ = (1 + 2i)(1 - 2i)

= 1 - (2i)²

= 1 - 4i²

= 1 - 4(-1)

= 1 + 4

= 5

x² -  (α+β)x + αβ = 0

x² -  2x + 5 = 0

Linear factor is (x - 1/3)

f(x) = a (x - 3) (x - 1/3) (x² -  2x + 5) ----(1)

Given condition is f(1) = 48

Applying the value of a in (1)

Problem 4 :

n = 3; -1 and -3 + 2i are zeros; leading coefficient is 1

Solution:

The zeroes are -1, -3 + 2i and -3 - 2i

Creating quadratic function with complex roots :

α = -3 + 2i and β = -3 - 2i

α + β = -3 + 2i + (-3) - 2i ==> -6

αβ = (-3 + 2i)(-3 - 2i)

= 9 - 4i²

= 9 + 4

= 13

The quadratic function will be

x² - (-6)x + 13 = 0

x² + 6x + 13 = 0

Linear factor is (x + 1) 

f(x) = a(x + 1)(x² + 6x + 13) ----(1)

Leading coefficient is 1. So, a = 1

f(x) = (x + 1)(x² + 6x + 13)

f(x) = x³ + 7x² + 19x + 13

So, the required polynomial is 

f(x) = x³ + 7x² + 19x + 13

Problem 5 :

n = 4; 2i, 5, and -5 are zeros; leading coefficient is 1.

Solution:

Quadratic function with the roots 2i and -2i

α + β = 2i + (-2ii ==> 0

αβ = (2i)(- 2i)

= - 4i²

= 4

The quadratic function with the roots 2i and -2i is

 (x² + 0x + 4)

 (x² + 4)

The quadratic function with the roots 5 and -5 is

α + β = 5 + (-5) ==> 0

αβ = (5)(- 5)

= - 25

The quadratic function with the roots 5 and -5 is

 (x² + 0x + (-25)

= (x² - 25)

= (x² + 4)(x² - 25)

= x⁴ - 25x² + 4x² - 100

= x⁴ - 21x² - 100

So, the required polynomial is  x⁴ - 21x² - 100.