A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
find all solutions of the polynomial equation:
x^4  x^3  6x^2 + 4x + 8 = 0
anonymous
 4 years ago
find all solutions of the polynomial equation: x^4  x^3  6x^2 + 4x + 8 = 0

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x4−x3−6x2+4x+80 is your answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you want to use grouping to factor this polynomial

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for the first 3 terms, you can factor out an X^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how should i group them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got x^2(x^2x6)+4(x+2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^2(x3)(x+2)+4(x+2) (x+2)(x^2(x3)+4) :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^4  x^3  6x^2 + 4x + 8 = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use factor and remainder theorem to solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^2(x^2  x  6) + 4(x + 2) = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you have to use special factors?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see by plugging x=2 the expression x^4  x^3  6x^2 + 4x + 8 becomes 0 hence x+2 is one of the factor for getting other factors divide x^4  x^3  6x^2 + 4x + 8 by x+2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no... (x^2+4)(x3)(x+2)^2=0 now solve each equal to zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no to special factors, matricked is right also

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how'd you get there..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you skipped a couple steps from x^2(x^2  x  6) + 4(x + 2) = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i further factored (x^2x6) and grouped (x^2+4)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you show me how, from there, you got (x^2+4)(x3)(x+2)^2=0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now set each equation equal to 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the x^2 and +4 you pulled out go in their own equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait so from x^2(x3)(x+2) + 4(x+2) = (x^2+4)(x+2)^2 (x3)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now set each equal to zero by themselves and solve for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the rest i know how, but h\where's the logic in that move, where you can take x^2+4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you factored it out by grouping

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is my understanding that in order to factor by grouping both pieces need to have common factors like, "4(x+2)" doesn't have a (x3)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you learning imaginary numbers?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right, but x^2x6 did

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^4  x^3  6x^2 + 4x + 8=0 x^4 +2x^3 3x^3  6x^2 + 4x + 8=0 X^3(x+2) 3x^2(x+2) +4(x+2)=0 (x+2)(x^33x^2+4)=0 (x+2)(x^32x^2x^2+4)=0 (x+2)(x^2(x2)  (x+2)(x2))=0 (x+2)(x2)(x^2x2)=0 (x+2)(x2)(x^22x+x2)=0 (x+2)(x2)(x(x2)+1(x2))=0 or(x+2)(x2)(x2)(x+1)=0 now set each equal to 0 thus the roots are 1,2,2,2
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.