Question

Obtain all the zeros of the polynomial x⁴ + x³ – 14x² – 2x + 24 if two of its zeros are `sqrt(2)` and `-sqrt(2)`.

Answer 1

Given: The polynomial f(x) = x⁴ + x³ – 14x² – 2x + 24 and two zeros x = `sqrt(2)` and x = `-sqrt(2)`.

Step-wise calculation:

1. If `±sqrt(2)` are zeros then `(x - sqrt(2))(x + sqrt(2)) = x^2 - 2` is a factor of f(x).

2. Divide f(x) by x² – 2.

Assume quotient ax² + bx + c.

Expanding (x² – 2)(ax² + bx + c) and matching coefficients with f(x) gives:

a = 1 (x⁴ term)

b = 1 (x³ term)

c – 2a = –14

⇒ c – 2 = –14

⇒ c = –12

Constant term check: –2c = 24 ⇒ c = –12 (consistent). 

So, the quotient is x² + x – 12.

3. Factor the quotient:

x² + x – 12 = (x + 4)(x – 3)

4. Therefore, the remaining zeros are x = –4 and x = 3.

The four zeros are 3, –4, `sqrt(2)` and `-sqrt(2)`.