Question

Find all the zeroes of (x⁴ + x³ – 23x² – 3x + 60), if it is given that two of its zeroes are `sqrt(3)` and `-sqrt(3)`.

Answer 1

Let f(x) = x⁴ + x³ – 23x² – 3x + 60

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

Consequently, `(x - sqrt(3)) (x + sqrt(3)) = (x^2 - 3)` is a factor of f(x).

On dividing f(x) by (x² – 3), we get:

`x^2 - 3")"overline(x^4 + x^3 - 23x^2 - 3x + 60)"("x^2 + x - 20`
            x⁴            – 3x²                     
            –                +                        
           x³ – 20x² – 3x + 60
           x³             – 3x
           –                 +                       
                 –20x² + 60
                 –20x² + 60
                 +         –                        
                             x                       

f(x) = 0

⇒ (x² + x – 20) (x² – 3) = 0

⇒ (x² + 5x – 4x – 20) (x² – 3)

⇒ [x(x + 5) – 4(x + 5)] (x² – 3)

⇒ `(x - 4) (x + 5) (x - sqrt(3)) (x + sqrt(3)) = 0`

⇒ x = 4 or x = –5 or x = `sqrt(3)` or x = `-sqrt(3)`

Hence, all the zeroes are `sqrt(3), -sqrt(3), 4` and –5.