If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes.
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 - 4x + 4 = 3,
⇒ x2 - 4x + 1= 0
Now, dividing p(x) by x2 - 4x + 1
∴ p(x) = x4 - 6x3 - 26x2 + 138x - 35
= (x2 - 4x + 1) (x2 - 2x - 35)
= (x2 - 4x + 1) (x2 - 7x + 5x - 35)
= (x2 - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2 - 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.
