Given that x √ 5 is a factor of the cubic polynomial x3 3 √ 5 x 2 13x 3 √ 5

Solution : Let `f(x) = x^(3)-3sqrt(5)x^(2)+13x - 3sqrt(5)` and given that, `(x-sqrt(5))` is a one of the factor of `f(x)`. Now, using division algorithm,
`{:(" "ul(" "x^(2)-2sqrt(5)x+3)),( {:x-sqrt(5)) " "x^(3)-3sqrt(5)x^(2)+13x-3sqrt(5)),(" "ul(underset(-)(x^(3))underset(+)-sqrt(5)x^(2)" ")),(" "-2sqrt(5)x^(2)+13x-3sqrt(5)),(" "ul(underset(+)(-2sqrt(5)x^(2))underset(-)(+)10x" ") ),(" "3x-3sqrt(5)),(" "ul(underset(-)(3x)underset(+)(-)3sqrt(5))),(" "xx):}`
`:. x^(3)-3sqrt(5) +13x - 3sqrt(5)=(x^(2)-2sqrt(5)+3)xx (x-sqrt(5))` [`:'` dividend = divisor `xx` quotient + remainder]
`=(x-sqrt(5))[x^(2)-{(sqrt(5)+sqrt(2))+(sqrt(5)-sqrt(2))} x+3]` " "[by splitting the middle term]
`= (x-sqrt(5))[x^(2)-(sqrt(5)+sqrt(2))x-(sqrt(5)-sqrt(2))x +(sqrt(5)+sqrt(2))(sqrt(5)-sqrt(2))]`
`[ :' 3 = (sqrt(5)+sqrt(2))(sqrt(5)-sqrt(2))]`
`= (x-sqrt(5))[x{x-(sqrt(5)+sqrt(2))} =(sqrt(5)-sqrt(2)){x-sqrt(5)+sqrt(2))}]`
`=(x-sqrt(5)){x-(sqrt(5)+sqrt(2))} {x-(sqrt(5)-sqrt(2))}`
Hence, all the zeroes of polynomial are `sqrt(5), (sqrt(5)+sqrt(2))` and `(sqrt(5)-sqrt(2))`.

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