Find all the zeroes of polynomial(2x⁴_9x³+5x²+3x_1).if …

Given:
f(x) = (2x⁴ – 9x³ + 5x² + 3x – 1)
Zeroes = (2 + √3) and (2 – √3)
Given the zeroes, we can write the factors = (x – 2 + √3) and (x – 2 – √3)
{Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}
Multiplying these two factors, we can get another factor which is:
((x – 2) + √3)((x – 2) – √3) = (x – 2)^{2 –} (√3)²
⇒x² + 4 – 4x – 3 = x² – 4x + 1
So, dividing f(x) with (x² – 4x + 1)

f(x) = (x² – 4x + 1) (2x² – x – 1)
Solving (2x² – x – 1), we get the two remaining roots as

{tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}
where f(x) = ax² + bx + c = 0(using Quadratic Formula)

{tex}\mathrm{x}=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(2)(-1)}}{2(2)}{/tex}
{tex}\mathrm{x}=\frac{-1 \pm 3}{4}{/tex}
{tex}\Rightarrow \mathrm{x}=1,-\frac{1}{2}{/tex}
Zeros of the polynomial = {tex}1,-\frac{1}{2}, 2+\sqrt{3}, 2-\sqrt{3}{/tex}