If two zeroes of the polynomial x^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 ± `sqrt3` , find other zeroes

#### Solution 1

Given that 2 + `sqrt3` and 2 - `sqrt3` are zeroes of the given polynomial

Therefore, `(x - 2-sqrt3)(x-2+sqrt3)` = x^{2} + 4 - 4x - 3

= x^{2} - 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing by x^{4} - 6x^{3} - 26x^{2} + 138x - 35 by x^{2} - 4x + 1

Clearly x^{4} - 6x^{3} - 26x^{2} + 138x - 35 = (x^{2} - 4x +1)(x^{2} -2x -35)

It can be observed that (x^{2} - 2x - 35) is also a factor of the given polynomial.

And (x^{2} - 2x - 35) = (x -7)(x + 5)

Therefore, the value of the polynomial is also zero when x - 7 = 0 or x + 5 = 0

or x = 7 or -5

Hence, 7 and - 5 are also zeroes of this polynomial

#### Solution 2

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35.

Let x = 2±√3

So, x-2 = ±√3

On squaring, we get x^{2} - 4x + 4 = 3,

^{2}- 4x + 1

^{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(x - 7) + 5 (x - 7)]

= (x

^{2}- 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).

∴ - 5 and 7 are other zeroes of the given polynomial.