What are all real zeros of the polynomial? (Factor it)

Question

#g(t)= t^5-6t^3+9t#

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Answer 1 · 2017-10-14T11:56:00

#t=0 , t= -sqrt(3) , t= sqrt(3) , t= -sqrt(3) , t= sqrt(3)#

Factoring #t^5-6t^3+9t#

#t(t^4-6t^2+9)#

This has to be equated to zero. ( To find roots ):

#t(t^4-6t^2+9)=0=> t=0# ( This is one of the roots )

We now have to equate #t^4-6t^2+9# to zero.

#t^4-6t^2+9=0# ( This is just a quadratic in #t^2# )

Making the substitution #z= t^2#, we have:

#z^2-6z+9=0#

Factoring:

#(z-3)(z-3)=0#

This is the square of a binomial:

#(z-3)^2=0=> z=3#

But #z=t^2#

So:

#t^2=3=> t= +-sqrt(3)#

Since the equation is of 5th order there will be 5 roots. These can be seen to be just repeated.

Graph:

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