Question

Find all zeros of $x^4-6x^3-26x^2+138x-35$, if two zeros are $2+\sqrt{3}$ and $2-\sqrt{3}$

Answer 1

Given,

$f\left(x\right)=x^4-6x^3-26x^2+138x-35$

$=(x+5)\frac{x^4-6x^3-26x^2+138x-35}{x+5}$

$=(x+5)(x^3-11x^2+29x-7)$

$=(x+5)(x-7)(x^2-4x+1)$

$x^2-4x+1=0$

solving the above quadratic equation, we get, $x=2+\sqrt{3},x=2-\sqrt{3}$

$=(x+5)(x-7)[x-(2\pm \sqrt{3}\left)\right]$

$x=-5,7,2+\sqrt{3},2-\sqrt{3}$