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mathslover Group TitleBest ResponseYou've already chosen the best response.0
first of all find the factors of +20 do you know the factors of 20?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Take out the common factor of the first two terms and last two terms. 2x^3 + 10x2 + 4x + 20 = 2x^2 (x+5) + 4(x+5) = ...?
 2 years ago

ashna Group TitleBest ResponseYou've already chosen the best response.1
2x^3 + 10x^2 + 4x + 20 = 0 (2x+10)(x^2+2) = 0 2x = 10 ; x^2 = 2 x = 5 ; x = √ 2 x = 5 ; x = + or √ 2
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
@ashna It's factoring the expression, not solving equation. Moreover, please don't give direct answers to the question. You may refer to our Code of Conduct at httlp://www.opnstudy.com/codeofconduct for more information. Thanks.
 2 years ago

ashna Group TitleBest ResponseYou've already chosen the best response.1
okay .. sorry !
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@Callisto it can be done in this way also : factors of 20 = \(\large{\pm 1,\pm 2,\pm 4,\pm 5,\pm 10, \pm20}\) now given p(x) = 2x3 + 10x2 + 4x + 20 p(5)=250+25020+20=0 hence 5 is a zero of the polynomial p(x) that is (x+5) is a factor of p(x)
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
M i right in this way also ? please dont consider this as direct answer... it is my confusion it may be wrong and also the answer is not yet declared
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I haven't learnt this way. It may work, but for this question, it's not necessary to use this way, I think. @mathslover
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
ok wait lemme show my full work
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Well, I understand this method. Probably, you're doing by factor theorem. It works but it just takes a long time for this question as you have to test for many factors.
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
this is what i did
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Very right @Callisto It is factor theorem .. but yes it takes a long time
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
thanks to god that it was 20 .. if it was some other large number like 4000993 then it was next to impossible to do by this
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
For this type of problem,usually, I would see if it can be done by taking out factors/identities first. If not, then I have to use factor theorem.
 2 years ago
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