How to write polynomial equations based on the roots.

Question

There are three questions.

The first has the root 6-√7

The second has the roots -2, 1+√7

The third has the roots -6, 0, 3, -√5

Answer 1

First case: Root = a = (6 - √7)

Associated factor is: (x – a)

In order to make the polynomial coefficients integers add another root equal to the conjugate if the given root contains a radical.

Second root: b = 6 + √7

The associated factor is: (x – b)

Functional equation:

(x – a)(x – b) = x² - (a + b)x + ab

a + b = (6 - √7) + (6 + √7) = 12

ab = (6 - √7)(6 + √7) = 36 – 7 = 29

Therefore, function is:

f(x) = 0 = x² -12x +20

Second case:

a = -2; associated factor is: (x + 2)

b = (1 + √7); associated factor is: (x - b)

Add conjugate for third root

c = (1 - √7) ; associated factor is: (x - c)

Portion of functional equation using the conjugates:

(x – b)(x – c) = x² - (b + c)x + bc

b + c = (1 + √7) + (1 - √7) = 2

bc = (1 + √7)(1 - √7) = 1 – 7 = -6

(x – b)(x – c) = x² - 2x – 6

Functional equation:

f(x) = (x – a)(x – b)(x – c)

f(x) = (x + 2)( x² - 2x – 6)

f(x) = x³ – 10x -12

Third case: roots = -6, 0 , 3, -√5

a = -6; associated factor is: (x + 6)

b = 0; associated factor is: (x)

c = 3; associated factor is: (x – 3)

d = -√5; associated factor is: (x + √5)

Add conjugate for fifth root

E = √5; associated factor is: (x - √5)

(x + √5)(x - √5) = x² - 5

Functional equation:

f(x) = (x + 6)(x)(x - 3)(x² - 5)

f(x) = (x)(x² + 3x - 18)(x² - 5)

f(x) = (x³ + 3x² - 18x)(x² - 5)

f(x) = x⁵ + 3x⁴ - 23x³ - 15x² + 90x

Answer 2

Any polynomial with a root of the for a + sqrt(b) must also have the root a - sqrt(b)

I will do the first one: (Write the equation in the form (x-r₁)(x-r₂)... = 0

(x -6+sqrt(7))(x-6-sqrt(7)) = 0 or x² - 12x +36 - 7 = x² -12x + 29 = 0

The second equation will be a cubic and the 3rd a quintic equation.

Please consider a tutor. Take care.