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

## Solution :

We have : $$2 \pm \sqrt{3}$$ are two zeroes of the polynomial

p(x) = $$x^4 – 6x^3 – 26x^2 + 138x – 35$$

Let x = $$2 \pm \sqrt{3}$$,  So, x – 2 = $$\pm \sqrt{3}$$

Squaring, we get

$$x^2 – 4x + 4$$ = 3,   i.e.  $$x^2 – 4x + 1$$ = 0

Let us divide p(x) by $$x^2 – 4x + 1$$ to obtain other zeroes.

$$\therefore$$  p(x) = $$x^4 – 6x^3 – 26x^2 + 138x – 35$$

= ($$x^2 – 4x + 1$$)($$x^2 – 2x – 35$$)

= ($$x^2 – 4x + 1$$)($$x^2 – 7x + 5x – 35$$)

= ($$x^2 – 4x + 1$$)(x + 5)(x – 7)

So, (x+ 5) and (x – 7) are the other factors of p(x).

$$\therefore$$    -5  and 7 are other zeroes of the given polynomial.