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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