Obtain all the zeros if the polynomial x⁴ – 3x³ – x² + 9x – 6 if two of its zeros are √3 and –√3.

Let p(x) = x⁴ – 3x³ – x² + 9x – 6

p(x) is a polynomial with degree 4 hence it will have 4 zeroes

As √3 and –√3 are zeroes of p(x) then (x – √3) and (x – (–√3)) are factors of p(x)

As both (x – √3) and (x + √3) are factors of p(x) then their product (x – √3)(x + √3) will also be factor of p(x)

Using identity (a + b)(a – b) = a² – b²

⇒ (x – √3)(x + √3) = x² – 3

Dividing p(x) by x² – 3

Using dividend = divisor × quotient + remainder

⇒ p(x) = (x² – 3)(x² – 3x + 2) + 0

Now remaining two zeroes we will get when (x² – 3x + 2) = 0

⇒ x² – 3x + 2 = 0

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 1)(x – 2) = 0

⇒ x – 1 = 0 and x – 2 = 0

⇒ x = 1 and x = 2

All four zeroes of x⁴ – 3x³ – x² + 9x – 6 are x = 1, x = 2, x = √3 and x = –√3