Let `f(x) = x^(4) - 6x^(3)-26x^(2)+138x-35.`
Let ` alpha = (2+sqrt3) and beta =(2-sqrt3).` Then,
` (alpha+beta) = 4 and alpha beta = (4-3) = 1.`
So, the quadratic polynomial whose roots are `alpha and beta` is given by
`x^(2) -(alpha+beta) x + alpha beta=(x^(2) -4x+1).`
`:. (x^(2)-4x+1) ` is a factor of f(x).
On dividing f(x) by `(x^(2)-4x+1)`, we get
`:. f(x)=(x^(2)-4x+1)(x^(2)-2x-35).`
`:. ` the other two zeros of f(x) are given by `(x^(2)-2x-35) = 0`.
Now, `x^(2) -2x-35 = 0 rArr x^(2)-7x+5x-35=0`
`rArr x(x-7)+5(x-7)=0`
` rArr (x-7)(x+5) =0 `
` rArr x -7 = 0 or x + 5 = 0`
` rArr x = 7 or x =- 5.`
Hence, the other two zeros of f(x) are 7 and -5.
