Question

find all the zeros of the polynomial x4-5x3-4x2+16x-8 if two of its zeroes are 3+ root5 and 3- root5

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

$x^2+x-2$

$x^2+x-2=0$

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

$\implies\:x=1,-2$