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

How do you find the real solutions of the polynomial 2x^5+3x^4=18x+272x^5+3x^4=18x+27?