How do you find the real solutions of the polynomial #2x^5+3x^4=18x+27#?

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Answer 1 · 2017-02-15T10:55:40

The solutions are #S={-3/2, sqrt3, -sqrt3}#

Ler's rewrite the polynomial and factorise

#2x^5+3x^4=18x+27#

#x^4(2x+3)=9(2x+3)#

#x^4(2x+3)-9(2x+3)=0#

#(2x+3)(x^4-9)=0#

#(2x+3)(x^2-3)(x^2+3)=0#

#(2x+3)(x-sqrt3)(x+sqrt3)(x^2+3)=0#

Therefore,

the solutions are

#2x+3=0#, #=>#, #x=-3/2#

#x-sqrt3=0#, #=>#, #x=sqrt3#

#x+sqrt3=0#, #=>#, #x=-sqrt3#

#x^2+3=0#, no real solutions