Question 812219
{{{16h^5-25h^3+9h=0}}}
{{{(16h^4-25h^2+9)h=0}}} --> {{{highlight(x=0)}}} is one solution
The other solutions, if there are any, must come from
{{{16h^4-25h^2+9=0}}}
I know that I can factor {{{16y^2-25y+9=(16y-9)(y-1)}}} , so
{{{16h^4-25h^2+9=(16h^2-9)(h^2-1)}}}
and that can be applied to transform the equation:
{{{16h^4-25h^2+9=0}}}
{{{(16h^2-9)(h^2-1)=0}}}
{{{(4h+3)(4h-3)(h+1)(h-1)=0}}}
So, the other solutions are
{{{highlight(h=-3/4)}}} , {{{highlight(h=3/4)}}} ,
{{{highlight(h=-1)}}} , and {{{highlight(h=1)}}} .
 
NOTE: You could also say that from {{{16h^4-25h^2+9=0}}} ,
we do the change of variable {{{y=h^2}}}
to get the quadratic equation {{{16y^2-25y+9=0}}} ,
with solutions {{{y=1}}} and {{{y=9/16}}}
and returning from {{{y}}} to {{{h}}} we get
{{{h^2=1}}} --> {{{system(h=1, "or" ,h=-1)}}}
and
{{{h^2=9/16}}} --> {{{system(h=3/4, "or" ,h=-3/4)}}} .