Question 908446
{{{ 8y^3 + 40y^2 - 16y }}}
Factor out {{{ 8y }}} from each term
{{{ 8y*( y^2 + 5y - 2 ) }}}
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I have 2 factors now, but I can 
make it into 3 factors by solving
the quadratic
I'm not sure if that's needed, 
but anyway:
{{{ y^2 + 5y = 2 }}}
{{{ y^2 + 5y + (5/2)^2 = 2 + (5/2)^2 }}}
{{{ y^2 + 5y + 25/4 = 8/4 + 25/4 }}}
{{{ ( y + 5/2 )^2 = 33/4 }}}
You can finish if you need to.
I'm guessing that it's not required here