Question

Find all the zeros of the polynomial 2x³ + x² – 6x – 3, if two of its zeros are `-sqrt(3)` and `sqrt(3)`.

Answer 1

We know that, if x = a is a zero of a polynomial, then x – a is a factor of f(x).

Since `sqrt3` and `-sqrt3` are zeros of f(x).

Therefore

`(x+sqrt3)(x-sqrt3)=x^2+sqrt3x-sqrt3x-3`

= x² – 3

x² – 3 is a factor of f(x). Now, we divide f(x) = 2x³ + x² – 6x – 3 by g(x) = x² – 3 to find the other zeros of f(x).

                    2x + 1
`x^2 - 3")"overline(+ \cancel(2x^3) + x^2 - 6x - 3)`
            `+ \cancel(2x^3) - 0 - 6x`
            –                  +                
                      `+ \cancel(x^2) + 0 \cancel(-3)`
                      `+ \cancel(x^2) + 0 \cancel(-3)`
                       –               +        
                                    0

By using that division algorithm we have,

f(x) = g(x) × q(x) + r(x)

2x³ + x² – 6x – 3 = (x² – 3) × (2x + 1) + 0

2x³ + x² – 6x – 3 `= (x^2 + sqrt3)(x - sqrt3)(2x + 1)`

Hence, the zeros of the given polynomial are `-sqrt3, +sqrt3, (-1)/2`.