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evy15
 3 years ago
Find all zeros 5x^520x^440x^316x^245x+180=0
evy15
 3 years ago
Find all zeros 5x^520x^440x^316x^245x+180=0

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Asad0000
 3 years ago
Best ResponseYou've already chosen the best response.1What do you mean by "Find all zeros" ?

wio
 3 years ago
Best ResponseYou've already chosen the best response.0Use rational root theorem to find first zero. Then use synthetic division to factor it out.

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0hi @evy15 are you in numerical analysis? the root are 2.07316, 1.2355, 5.5573 and two complex roots check it for yourself if they are true

matricked
 3 years ago
Best ResponseYou've already chosen the best response.0@evy though the equation is 5x^520x^440x^316x^245x+180=0 correct bu t may be i feel u have typed in 16X^2 instead of 160x^2

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0can someone please help me and explain to me how to begin this problem

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0Have you done numerical methods in your course? This is not a problem that can be solved by factorization, or rearrangements. I suggest you triple check the question.

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0its correct can you tell me how to begin

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0It can be solved numerically, or graphically. Are these expected methods to use in your course?

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0yes the thing is i missed a day in class and now i dont know how to work this out

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0Are you working on factorization, Descartes rule of signs, etc.? I suggest you check your course outline to see what you're expected to know, or check with a friend to see what has been done that day. The prof would also be pleased to tell you what has been covered that day. This problem requires numerical methods which are techniques completely different from your previous problems.

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0can u help me and explain it to me

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0i have tried to use the book but i keep getting it wrong

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0What is the name of your math course?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0Then there is probably another typo in the question.

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0there isn't, i already checked

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0I believe you, but your prof made a typo! Check in the index of your textbook to see if Newton's method is in there. What is the title of your textbook?

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0Holt McDougal Larson Algebra 2

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0Can you find "Newton's method" in the index?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0On which chapter are you working?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0Is it paper homework or online?

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0its a take home test and i couldnt find it

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0As matricked pointed out, IF 16x^2 had been +160x^2, then there are rational roots obtainable by grouping. 5x^520x^440x^3+160x^245x+180 =5(x^54x^48x^3+32x^29x+36) \(= 5( x^4(x4) 8x^2(x4) 9(x4)) \) \(= 5(x4)(x^48x^29) \) \(= 5(x4)(x^29)(x^2+1) \) \(= 5(x4)(x+3)(x3)(x^+1) \)

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0*last factor is \( (x^2+1) \)

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0ok do you solve for them now

evy15
 3 years ago
Best ResponseYou've already chosen the best response.0hold on a second, im sorry
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