Question

Find all the zeroes of polynomial (2x⁴ – 11x³ + 7x² + 13x – 7), it being given that two of its zeroes are `(3 + sqrt(2))` and `(3 - sqrt(2))`.

Answer 1

The given polynomial is f(x) = 2x⁴ – 11x³ + 7x² + 13x – 7.

Since `(3 + sqrt(2))` and `(3 - sqrt(2))` are the zeroes of f(x) it follows that each one of `(x + 3 + sqrt(2))` and `(x + 3 - sqrt(2))` is a factor of f(x).

Consequently,` [(x - ( 3 + sqrt(2))] [(x - (3 - sqrt(2))]`

= `[(x - 3) - sqrt(2)] [(x - 3) + sqrt(2)]`

= `[(x - 3)^2 - 2]`

= x² – 6x + 7, which is a factor of f(x).  

On dividing f(x) by (x² – 6x + 7), we get: 

 `x^2 - 6x + 7")"overline(2x^4 - 11x^3 + 7x^2 + 13x - 7)"("2x^2 + x - 1`
                      2x⁴  – 12x³  + 14x²
                    –         +          –                           
                           x³ – 7x² + 13x – 7
                           x³ – 6x² + 7x
                         –      +        –                          
                                –x² + 6x – 7
                                –x² + 6x – 7
                               +     –     +                     
                                     x                             

f(x) = 0

⇒ 2x⁴ – 11x³ + 7x² + 13x – 7 = 0

⇒ (x² – 6x + 7) (2x² + x – 7) = 0

⇒ `(x + 3 + sqrt(2)) (x + 3 - sqrt(2)) (2x - 1) (x + 1) = 0`

⇒ `x = -3 - sqrt(2)` or `x = -3 + sqrt(2)` or `x = 1/2` or x = –1

Hence, all the zeroes are `(-3 - sqrt(2)), (-3 + sqrt(2)), 1/2` and –1.