Which are the solutions of the equation 3x^3-6x^2=x-2?

Expert Answers
hala718 eNotes educator| Certified Educator

3x^3-6x^2=x-2

Let us move all terms to the left side:

==> 3x^3-6x^2-x+2 =0

==> (3x^3-6x^2)-(x-2)=0

Now factor 3x^2

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

Now factor (x-2):

==> (x-2)(3x^2-1)=0

==> (x-2)=0 ==> x=2

and 3x^2-1=0

==> x= 1/sqrt3  and x= -1/sqrt3

The the solution is x= {2, 1/sqrt3, -1/sqrt3}

neela | Student

To solve 3x^3-6x^2 = x-2.

Solution:

3x^3-6x^2 = x-2.

3x^2 (x-2) = x-2. Or

3x^2(x-2) - 1(x-2) = 0.

(x-2)(3x^2-1) = 0. Or

x-2 = 0 Or 3x^2 -1 = 0

So x-2 = 0 gives: x=2.

3x^2-1 = 0 gives: 3x^2= 1. Or x^2 = 1/3 . Or x  = sqrt(1/3) or x = -sqrt(1/3)

giorgiana1976 | Student

To solve the equation, we'll move the terms from the right side to the left side and, after that, we'll group the first 2 terms together and the next 2 terms together, so that we can factorize:

(3x^3 - 6x^2) - (x - 2) = 0

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

We'll factorize again:

(x - 2)(3x^2 - 1) = 0

We'll put each factor as being 0.

x-2 = 0

x1 = 2

3x^2 - 1 = 0

The expression is a difference of squares:

3x^2 - 1 = (xsqrt3 - 1)(xsqrt3 + 1) = 0

xsqrt3-1 = 0

xsqrt3 = 1

x2 = 1/sqrt3

x2 = sqrt3/3

xsqrt3+1 = 0

xsqrt3 = -1

x3 = -1/sqrt3

x3 = -sqrt3/3

The solutions of the equation are: {-sqrt3/3,sqrt3/3,2}