How do you find the real solutions of the polynomial $x^{3} + 6 = 2 x^{2} + 5 x$ $x^3+6=2x^2+5x$ ?

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Answer 1 · 2017-11-16T12:11:30

Solution: $x = 1, x = 3, x = - 2$ $x=1 ,x=3 ,x=-2$

$x^{3} + 6 = 2 x^{2} + 5 x or x^{3} + 6 - 2 x^{2} - 5 x = 0$ $x^3+6=2x^2+5x or x^3+6-2x^2-5x =0$ or

$x^{3} - 2 x^{2} - 5 x + 6 = 0 or x^{3} - x^{2} - x^{2} + x - 6 x + 6 = 0$ $x^3-2x^2-5x +6 =0 or x^3-x^2-x^2+x-6x+6 =0$ or

$x^{2} (x - 1) - x (x - 1) - 6 (x - 1) = 0$ $x^2(x-1)-x(x-1)-6(x-1) =0$ or

$(x - 1) (x^{2} - x - 6) = 0 or (x - 1) (x^{2} - 3 x + 2 x - 6) = 0$ $(x-1)(x^2-x-6)=0 or (x-1) ( x^2-3x+2x-6)=0$ or

$(x - 1) {x (x - 3) + 2 (x - 3)} = 0$ $(x-1) { x(x-3)+2(x-3)}=0$ or

$(x - 1) {(x - 3) (x + 2)} = 0$ $(x-1) {(x-3)(x+2)}=0$ or

$(x - 1) (x - 3) (x + 2) = 0$ $(x-1)(x-3)(x+2)=0$

$x-1=0 :. x=1 or x-3=0 :.x=3$ or

$x+2=0:. x =-2$

Solution: $x=1 ,x=3 ,x=-2$ [Ans]

How do you find the real solutions of the polynomial x^3+6=2x^2+5xx3+6=2x2+5xx^3+6=2x^2+5x?