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plz help :) list the possible zeros for the polynomial.. h(x)=2x^3+5x^2-31x-15 plz help :)
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try factoring it first :)
oh. i see :( let me think :)
thank you!! :)
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
x (x (2 x+5)-31)-15 --- this is factor? :O
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it's a ...kind of factor. not the kind you're thinking of ;)
No need to graph, if you know Newton-raphson or method of bisection.
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