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)#