Question

find the zeroes of polynomial p(x)=x^2-5x^3-4x^2+16x-8 if two zeroes of the polynomial are 3+√5 and 3-√5​

Answer 1

Answer:

The required zeros are

3+\sqrt{5},3-\sqrt{5},-2,1

Step-by-step explanation:

Given zeros are

3+\sqrt{5} and 3-\sqrt{5}

sum of zeros

=3+\sqrt{5}+3-\sqrt{5}

=6

product of zeros

=(3+\sqrt5)(3-\sqrt5)

=3^2-(\sqrt5)^2

=9-5

=4

Corresponding quadratic factor is

x^2-6x+4

Now,

x^4-5x^3-4x^2+16x-8=(x^2-6x+4)(x^2+px-2)

Equating coefficients of x on both sides

16=12+4p

4=4p

p=1

\therefore\:other factor is

x^2+x-2

x^2+x-2=0

\implies\:(x+2)(x-1)=0

\implies\:x=1,-2