x^3 = -2x^2 + 3x ??

Question

Can someone please solve for x?

Thank you so much!

Answer 1

x^3 + 2x^2 -3x=0
take x common
x(x^2 + 2x-3)=0
x(x-1)(x+3)=0
hence solution is 0,1,-3 for a detailed discussion IM me at Jaagrut_live

Answer 2

x^3=-2x^2+3x =>x^3+2x^2-3x=0 =>x^3+3x^2-x^2-3x=0 =>x^2(x+3)-x(x+3)=0 =>(x+3)(x^2-x)=0 therefore,either x+3=0 or x^2-x=0 . If x+3=0,then x=-3 ;if x^2-x=0 ,then x^2=x or x=19dividing both sides by x. The problem may also be solved in the following way x^3+2x^2-3x=0 => x(x^2+2x-3)=0 =>x(x^2+3x-x-3)=0 => x{x(x+3)-1(x+3)} => x(x+3)(x-1)=0 Therefore,either x=0 or x+3=0 or x-1=0 .Therefore x=0 or -3 or 1. ans.

Answer 3

This is a quadratic so you need to move the right over to the left, setting the whole thing to 0.

x^3 + 2x^2 + 3x =0

Factor out an x

X(x^2 + 2x + 3) =)

Factor the polynomial

X( + )( + )

then solve each of the 3 parts for x. The first x counts! (value is 0)

Answer 4

x^3 + 2x^2 - 3x = 0
x(x^2 + 2x - 3) = 0
x(x-1)(x+3)=0
x { -3, 0, 1}

Answer 5

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