Question

If two zeros of the polynomial $f\left(x\right)=x^4-6x^3-26x^2+138x-35$ are $2\pm \sqrt{3}$, find ohter zeros. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are two zeros of $f\left(x\right)$

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

$=(x-2{)}^2-(\sqrt{3}{)}^2=x^2-4x+1$

is a factor of $f\left(x\right)$

Let us now divide $f\left(x\right)$ by $x^2-4x+1$

We have,



$\therefore f\left(x\right)=(x^2-4x+1)(x^2-2x-35)$

Hence, other two zeros of $f\left(x\right)$ are the zeros of the polynomial $x^2-2x-35$

We have,

$x^2-2x-35=x^2-7x+5x-35=(x-7)(x+5)$

Hence, other two zeros of $f\left(x\right)$ are 7 and -5.