Question

Find all the zeros of 2x⁴ – 3x³ – 3x² + 6x – 2 if it is given that two of its zeros are 1 and `1/2`.

Answer 1

Given:

Polynomial: (P(x) = 2x⁴ – 3x³ – 3x² + 6x – 2

Two zeros: x = 1 and x = 12

Step-wise calculation:

1) Use the factor theorem

If 1 is a zero, then (x – 1) is a factor.

If 12 is a zero, then (x – 12) is a factor, i.e. (2x – 1) is a factor.

So, (x – 1)(2x – 1) = 2x² – 3x + 1 is a factor of P(x).

2) Divide P(x) by (2x² – 3x + 1)

Assume P(x) = (2x² – 3x + 1)(ax² + bx + c) 

Expand: (2x² – 3x + 1)(ax² + bx + c)

= 2ax⁴ + (2b – 3a)x³ + (2c – 3b + a)x² + (–3c + b)x + c 

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

2a = 2 ⇒ a = 1

2b – 3a = –3 

⇒ 2b – 3 = –3 

⇒ b = 0

2c – 3b + a = –3 

⇒ 2c – 0 + 1 = –3 

⇒ 2c = –4 

⇒ c = –2

Check linear and constant terms:

–3c + b = –3(–2) + 0 = 6

c = –2

So, P(x) = (2x² – 3x + 1)(x² – 2).

3) Factor completely and find zeros

Factor the quadratic:

2x² – 3x + 1 = (2x – 1)(x – 1) 

And x² – 2 = 0 ⇒ x = `±sqrt(2)`

Thus, `P(x) = (x - 1)(2x - 1)(x - sqrt(2))(x + sqrt(2))`

All zeros of 2x⁴ – 3x³ – 3x² + 6x – 2 are `1, 1/2, sqrt(2), -sqrt(2)`.