How do you factor completely $h^3-27h+10$ ?

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Answer 1 · 2017-05-09T12:54:02

$h^3-27h+10=(h-5)(h^2+5h-2)=(h-5)(h+5/2+sqrt33/2)(h+5/2-sqrt33/2)$

As the function is $h^3-27h+10$ , one of the zeros could be factor of $10$ i.e. $+-2$ or $+-5$ . As is seen $5$ is a zero of the function as

$(5)^3-27(5)+10=125-135+10=0$

and hence $(h-5)$ is a factor of $h^3-27h+10$

Dividing it by $(h-5)$ , we get

$h^2(h-5)+5h(h-5)-2(h-5)=(h-5)(h^2+5h-2)$

As discriminant of $h^2+5h-2$ is $5^2-4xx1xx(-2)=33$ , which is not a perfect square and hence we can only have additional irrational factor.

$h^2+5h-2=(h^2+2xx5/2xxh+(5/2)^2)-(5/2)^2-2$

= $(h+5/2)^2-33/4=(h+5/2)^2-(sqrt33/2)^2$

= $(h+5/2+sqrt33/2)(h+5/2-sqrt33/2)$

Hence $h^3-27h+10=(h-5)(h+5/2+sqrt33/2)(h+5/2-sqrt33/2)$

How do you factor completely h^3-27h+10h^3-27h+10?