I assume you meant to say
[tex]f(x)=2x^4+3x^3-5x^2-9x-3[/tex]
Given that x = √3 and x = -√3 are roots of f(x), this means that both x - √3 and x + √3, and hence their product x ² - 3, divides f(x) exactly and leaves no remainder.
Carry out the division:
[tex]\dfrac{2x^4+3x^3-5x^2-9x-3}{x^2-3} = 2x^2+3x+1[/tex]
To compute the quotient:
* 2x ⁴ = 2x ² • x ², and 2x ² (x ² - 3) = 2x ⁴ - 6x ²
Subtract this from the numerator to get a first remainder of
(2x ⁴ + 3x ³ - 5x ² - 9x - 3) - (2x ⁴ - 6x ²) = 3x ³ + x ² - 9x - 3
* 3x ³ = 3x • x ², and 3x (x ² - 3) = 3x ³ - 9x
Subtract this from the remainder to get a new remainder of
(3x ³ + x ² - 9x - 3) - (3x ³ - 9x) = x ² - 3
This last remainder is exactly divisible by x ² - 3, so we're left with 1. Putting everything together gives us the quotient,
2x ² + 3x + 1
Factoring this result is easy:
2x ² + 3x + 1 = (2x + 1) (x + 1)
which has roots at x = -1/2 and x = -1, and these re the remaining zeroes of f(x).
