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