Question

If two zeros of the polynomial p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 are `sqrt(2)` and `-sqrt(2)`, find its other two zeros.

Answer 1

Given: Two zeros of p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 are `sqrt(2)` and `-sqrt(2)`.

Step-wise calculation:

1. If `±sqrt(2)` are zeros then x² – 2 is a factor.

So, write p(x) = (x² – 2)(ax² + bx + c).

2. Expand the product:

(x² – 2)(ax² + bx + c) = ax⁴ + bx³ + (c – 2a)x² − 2bx − 2c

3. Match coefficients with p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2:

a = 2

b = –3

c – 2a = –3

⇒ c – 4 = –3 

⇒ c = 1

4. Thus, the other quadratic factor is 2x² – 3x + 1.

Solve 2x² – 3x + 1 = 0:

`x = (3 ± sqrt(9 - 8))/4` 

= `(3 ± 1)/4` 

⇒ x = 1 or x = `1/2`

The other two zeros are 1 and `1/2`.