2 + √3 and 2 — √3 are the zeros of p(x) = x⁴ — 6x³ — 26x² + 138x — 35. Find the remaining zeros of p(x).

Given, 2 + √3 and 2–√3 are zeros of p(x).

⇒ and are the factors of p(x).

Thus = (x² – 4x + 4 –3) = (x² – 4x + 1) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8

and divisor polynomial = s(x) = x² – 2

⇒ p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8 = (x² – 4x + 1)(x² – 2x — 35)

On factorising x² – 2x — 35, we get

x² – 2x — 35 = x² –7x + 5x— 35

= x (x – 7) + 5 (x – 7)

= (x – 7) (x + 5)

Hence the other two zeros of p(x) are 7 and –5.