Answer :
Step 1:
Write the equation
[tex]-2x^2\text{ + 2 = }3x^3\text{ - 3x}[/tex]Step 2:
[tex]3x^3+2x^2\text{ - 3x - 2 = 0}[/tex]Next, find the first zero by trial and error. choose value of x that when you substitute will result in zero.
x = -1 will result into zero
Let check
[tex]\begin{gathered} 3x^3+2x^2\text{ - 3x - 2 = 0} \\ 3\times(-1)^3+2\times(-1)^2\text{ -3}\times\text{ (-1) - 2} \\ -3\text{ + 2 + 3 -2 = 0} \\ \text{Hence -1 is zero and x = -1 or x + 1 is a factor.} \end{gathered}[/tex]Step 4
[tex]\text{Divide }3x^3+2x^2\text{ - 3x - 2 by x + 1 to find the other factors.}[/tex][tex]\begin{gathered} \text{ 3x}^2\text{ - x -2} \\ \text{ x+ 1 }\sqrt[]{3x^3+2x^2\text{ - 3x - 2 }} \\ \text{ -(3x}^3+3x^2)\text{ } \\ \text{ -x}^2\text{ - 3x - 2} \\ \text{ -(-x}^2\text{ - x)} \\ \text{ -2x - 2} \\ \text{ -2x - 2} \\ \text{ 0} \end{gathered}[/tex]Step 5:
Hence
[tex]\begin{gathered} 3x^3+2x^2\text{ - 3x - 2 = 0} \\ (x+1)(3x^2-\text{ x - 2) = 0} \\ (x+1)(3x^2-\text{ 3x+ 2x - 2) = 0} \\ (x+1)\lbrack3x(x^{}-\text{ 1)+2(x - 1)\rbrack = 0} \\ (x+1)(x-1)(3x+2) \end{gathered}[/tex]Final answer
Equate each factor to zero to find the values of x.
[tex]\begin{gathered} \text{x + 1 = 0, x = -1} \\ \text{x - 1 = 0 , x = 1} \\ 3x\text{ + 2 = 0 , x = }\frac{-2}{3} \end{gathered}[/tex]x = -1 , x = 1, x = -2/3
