Question

find all the zeroes of the polynomial (x^4-5x^3-4x^2+16x-8), if the two of its zeroes are (3+√5)​

Answer 1

Answer:

The required zeros are

3+√5,3-√5-2,1

Step-by-step explanation:

Given zeros are

3+√5 and 3-√5

sum of zeros

=3+√5+3-√5

=6

product of zeros

=(3+√5)(3-√5)=

=3²-(√5)²

=9−5

=4

Corresponding quadratic factor is

x²-6x+4x

Now,

x⁴-5x³-4x²+16x-8=(x²−6x+4)(x² +px−2)

Equating coefficients of x on both sides

16=12+4p

4=4p

p=1

∴ other factor is

x²+x-2

x²+x-2=0

(x+2)(x-1)=0

x=1,-2