How do you solve #x^ { 5} - 4x ^ { 4} - 3x ^ { 3} - 12x ^ { 2} - 4x + 16= 0#?

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Answer 1 · 2018-05-02T15:45:08

#(1): "The Solution Set "[sub RR]={+-2,4}#.

#(2): "The Solution Set "[sub CC]={+-2,4, +-i}#.

#ul(x^5-4x^4)-ul(3x^3-12x^2)-ul(4x+16)=0#.

#:. x^4(x-4)-3x^2(x-4)-4(x-4)=0#.

#:. (x-4)(x^4-3x^2-4)=0#.

#:. (x-4){ul(x^4-4x^2)+ul(x^2-4)}=0#.

#:. (x-4){(x^2-4)(x^2+1)}=0#.

#:. (x-4)(x-2)(x+2)(x^2+1)=0#.

Therefore, #"the Solution Set "[sub RR]={+-2,4}#.

Further, in #CC, x^2+1=0 rArr x^2=-1 rArr x=+-i#.

So, #"the Solution Set "[sub CC]={+-2,4, +-i}#.