Question

Find all real solutions by factoring the equation: -2x^2 + 2 = 3x^3 - 3x

Answer 1

Step 1:

Write the equation

`-2x^2\text{ + 2 = }3x^3\text{ - 3x}`

Step 2:

`3x^3+2x^2\text{ - 3x - 2 = 0}`

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

`\begin{gathered} 3x^3+2x^2\text{ - 3x - 2 = 0} \\ 3*(-1)^3+2*(-1)^2\text{ -3}*\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}`

Step 4

`\text{Divide }3x^3+2x^2\text{ - 3x - 2 by x + 1 to find the other factors.}`
`\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}`

Step 5:

Hence

`\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}`

Final answer

Equate each factor to zero to find the values of x.

`\begin{gathered} \text{x + 1 = 0, x = -1} \\ \text{x - 1 = 0 , x = 1} \\ 3x\text{ + 2 = 0 , x = }(-2)/(3) \end{gathered}`

x = -1 , x = 1, x = -2/3