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

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jenn0814Best ResponseYou've already chosen the best response.0
x4−x3−6x2+4x+80 is your answer
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
you want to use grouping to factor this polynomial
 one year ago

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

jaersynBest ResponseYou've already chosen the best response.0
how should i group them?
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
i got x^2(x^2x6)+4(x+2)
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
x^2(x3)(x+2)+4(x+2) (x+2)(x^2(x3)+4) :/
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
x^4  x^3  6x^2 + 4x + 8 = 0
 one year ago

matrickedBest ResponseYou've already chosen the best response.0
use factor and remainder theorem to solve it
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
x^2(x^2  x  6) + 4(x + 2) = 0
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
do you have to use special factors?
 one year ago

matrickedBest ResponseYou've already chosen the best response.0
see 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
 one year ago

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

cammyabboBest ResponseYou've already chosen the best response.0
no to special factors, matricked is right also
 one year ago

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

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

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

cammyabboBest ResponseYou've already chosen the best response.0
now set each equation equal to 0
 one year ago

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

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

cammyabboBest ResponseYou've already chosen the best response.0
now set each equal to zero by themselves and solve for x
 one year ago

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

cammyabboBest ResponseYou've already chosen the best response.0
you factored it out by grouping
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
it 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)
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
are you learning imaginary numbers?
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
right, but x^2x6 did
 one year ago

cammyabboBest ResponseYou've already chosen the best response.0
you further factored
 one year ago

matrickedBest ResponseYou've already chosen the best response.0
x^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
 one year ago
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