Question

Write a cubic polynomial whose zeroes are 2+ root 5 /2 , 2 - root 5 /2 and 4Please answer urgently. And with proper steps to do it

Answer 1

Here,
given, three zeros of cubical equation
(2 +√5)/2 , ( 2 -√5)/2 and 4 .

sum of roots = (2 +√5 + 2-√5)/2 + 4 = 6

sum of products of two consecutive roots = ( 2+√5)(2-√5)/4 + (2+√5)/2 ×4 + (2-√5)/2 ×4 = -1/4 + 4 +2√5 +4 -2√5
= 8 -1/4 = 31/4

product of roots = (2+√5)(2-√5)×4/4 = -1

now, equation :-
x³ -(sum of roots )x² + (sum of products of two roots )x - product of roots =0

x³ - 6x² + 31/4x +1 = 0

Answer 2

Answer:Given- Zeroes of the polynomial 2+√5/2, 2-√5/2 and 4.

To Find: Cubic polynomial of the given zeroes.

Step-by-step explanation: Sum of the roots= alpha+beta+Gama(represent alpha, beta and gama with their respective symbols)

So, 2+√5/2+ 2-√5/2 +4

= 2+√5/2 +2-√5/2+4=6(-√5&+√5 get cancelled, 2+2=4 and 4/2=2+4=6)

Sum of product of 2 consecutive roots=(alpha × beta)(beta × Gama)(Gama × alpha)

=2+√5(2-√5)/2×2 +2+√5×4/2 +2-√5/2(4)

=2^2-√5^2/4 +4+2√5+4-2√5

(2^2 and -√5^2 implies 2square and 5square,+2√5&-2√5 are cancelled and we get -1/4 and 8)

=-1/4+8= 31/4

Product of zeroes= alpha×beta×gama

=2+√5/2×2-√5/2×4

=2^2-√5^2/2×4

=-1/4×4

=-1

Therefore (top 1 dot and below 2 dots to represent therefore mathematically) the cubic polynomial is x^3-(sum of roots)x^2+(sum of 2roots)x-product of roots=0

=x^3- 6x^2+31/5x+1=0