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

Mathematics
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x4−x3−6x2+4x+80 is your answer
you want to use grouping to factor this polynomial
for the first 3 terms, you can factor out an X^2

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Other answers:

how should i group them?
i got x^2(x^2-x-6)+4(x+2)
x^2(x-3)(x+2)+4(x+2) (x+2)(x^2(x-3)+4) :/
lets start over
x^4 - x^3 - 6x^2 + 4x + 8 = 0
use factor and remainder theorem to solve it
x^2(x^2 - x - 6) + 4(x + 2) = 0
do you have to use special factors?
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
no... (x^2+4)(x-3)(x+2)^2=0 now solve each equal to zero
no to special factors, matricked is right also
how'd you get there..
x=-2 and x=2 x=3
you skipped a couple steps from x^2(x^2 - x - 6) + 4(x + 2) = 0
i further factored (x^2-x-6) and grouped (x^2+4)
can you show me how, from there, you got (x^2+4)(x-3)(x+2)^2=0?
now set each equation equal to 0
the x^2 and +4 you pulled out go in their own equation
wait so from x^2(x-3)(x+2) + 4(x+2) = (x^2+4)(x+2)^2 (x-3)?
yes
now set each equal to zero by themselves and solve for x
hold on
x^2+4=0
the rest i know how, but h\where's the logic in that move, where you can take x^2+4
you factored it out by grouping
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 (x-3)
are you learning imaginary numbers?
right, but x^2-x-6 did
you further factored
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^3-3x^2+4)=0 (x+2)(x^3-2x^2-x^2+4)=0 (x+2)(x^2(x-2) - (x+2)(x-2))=0 (x+2)(x-2)(x^2-x-2)=0 (x+2)(x-2)(x^2-2x+x-2)=0 (x+2)(x-2)(x(x-2)+1(x-2))=0 or(x+2)(x-2)(x-2)(x+1)=0 now set each equal to 0 thus the roots are -1,-2,2,2
wow, nice job

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