Question

If two zeroes of the polynomial $x^4-6x^3-26x^2+138x-35$ are $2\pm \sqrt{3}$, find other zeroes.

Answer 1

$\because 2\pm \sqrt{3}$ are the zeroes of the given $p\left(x\right)$

$\therefore [x-(2+\sqrt{3}\left)\right][x-(2-\sqrt{3}\left)\right]$

$\therefore [x-2+\sqrt{3}][x-2-\sqrt{3}]$

$=x^2-4x+1=0\Rightarrow$ factor of

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

Now,

we divide the $p\left(x\right)$ by $x^2-41+1$

Refer image,

we split

$x^2-2x-35$

$\Rightarrow x^2-7x+5x-35$

$\Rightarrow x(x-7)+5(x-7)$

$\therefore x=-5$ and $x=7$

$\therefore$ zeroes of the given $p\left(x\right)$ are

$(2+\sqrt{3}),2-\sqrt{3},-5$ and $7$