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jaersyn
 3 years ago
find all solutions of the polynomial equation:
x^4  x^3  6x^2 + 4x + 8 = 0
jaersyn
 3 years ago
find all solutions of the polynomial equation: x^4  x^3  6x^2 + 4x + 8 = 0

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jenn0814
 3 years ago
Best ResponseYou've already chosen the best response.0x4−x3−6x2+4x+80 is your answer

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

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

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

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

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

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

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

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

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

matricked
 3 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

cammyabbo
 3 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

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

jaersyn
 3 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

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

jaersyn
 3 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?

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

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

jaersyn
 3 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)?

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

jaersyn
 3 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

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

jaersyn
 3 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)

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

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

matricked
 3 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
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