skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Dec 19, 2016 #2 $h^3 - 81h + 154 = 0$ rational root theorem yields $h = 2$ as one solution, therefore ... $(h-2)(h^2+2h-77) = 0$ Can you determine the other two solutions? Reactions: 1 person

$h^3 - 81h + 154 = 0$ rational root theorem yields $h = 2$ as one solution, therefore ... $(h-2)(h^2+2h-77) = 0$ Can you determine the other two solutions?

P Plato MHF Helper Aug 2006 22,506 8,663 Dec 19, 2016 #4 neilkirk said: h(81-h^2) =154 Solve for h please Click to expand... Factor $h^3-81h+154$ then use the quadratic equation. SEE HERE

neilkirk said: h(81-h^2) =154 Solve for h please Click to expand... Factor $h^3-81h+154$ then use the quadratic equation. SEE HERE

N neilkirk Dec 2016 6 0 Ireland Dec 19, 2016 #5 I got it. Feel a bit of a dope now, the question even said 'one of the values is an 'integer'. Thanks for tge fast reply lads!!

I got it. Feel a bit of a dope now, the question even said 'one of the values is an 'integer'. Thanks for tge fast reply lads!!

D deesuwalka Sep 2016 84 19 India Dec 20, 2016 #6 $h(81-h^2)=154$ $81h-h^3=154$ $h^3-81h+154=0 $ Now factor out it. $(h-2)(h^2+2h-77)=0$