If two zeroes of the polynomial 138x - 35 are 2 √3, find other zeroes

The given equation is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x4-6x3-26x2+138x-35 Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get, x2-4x+1, this is a factor of a given polynomial f(x).

Now, on dividing f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x4-6x3-26x2+138x-35

= (x2-4x+1)(x2 –2x−35)

On factorisation(x2–2x−35) we get,

x2–(7−5)x −35 

= x2– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, the zeroes are given by: x= −5 and x = 7.

Hence, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.