Question

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

Answer 1

Given: 

p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 and the two zeroes, √2 and – √2 

So, 

the polynomial is (x + √2) (x – √2) = x² – 2. 

Let us divide p(x) by (x² – 2) 

Here, 2x⁴ – 3x³ – 3x² + 6x – 2 = (x² – 2) (2x² – 3x + 1) 

= (x² – 2) [(2x² – (2 + 1) x + 1] 

= (x² – 2) (2x² – 2x – x + 1) 

= (x² – 2) [(2x (x – 1) –1(x – 1)] 

= (x² – 2) (2x – 1) (x – 1)

The other two zeroes are \(\frac{1}2\) and 1.