Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

plz help :) list the possible zeros for the polynomial.. h(x)=2x^3+5x^2-31x-15 plz help :)

OpenStudy (anonymous):

try factoring it first :)

OpenStudy (anonymous):

oh. i see :( let me think :)

OpenStudy (anonymous):

thank you!! :)

OpenStudy (anonymous):

well...it is factorable: x (x (2 x+5)-31)-15. --> this gives you all the zeroes ;) but you're right., it's not easily factorable. graph it :). graphing calculator. otherwise there's nothing...really easy to see besides one

OpenStudy (anonymous):

x (x (2 x+5)-31)-15 --- this is factor? :O

OpenStudy (anonymous):

it's a ...kind of factor. not the kind you're thinking of ;)

OpenStudy (anonymous):

No need to graph, if you know Newton-raphson or method of bisection.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!