How do you solve #x^ { 5} - 4x ^ { 4} - 2x ^ { 3} + 8x ^ { 2} - 24x + 96= 0#?

1 Answer
Cem Sentin
Jan 21, 2018

#x_1=-2i#, #x_2=2i#, #x_3=-sqrt6#, #x_4=sqrt6# and #x_5=4#

Explanation:

#x^5-4x^4-2x^3+8x^2-24x+96=0#

#x^4*(x-4)-2x^2*(x-4)-24*(x-4)=0#

#(x-4)*(x^4-2x^2-24)=0#

#(x-4)(x^2+4)(x^2-6)=0#

Hence solutions of this polynomial are #x_1=-2i#, #x_2=2i#, #x_3=-sqrt6#, #x_4=sqrt6# and #x_5=4#

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