So, I'm working on my polynomial functions exercises, and all goes well, until I reach this one. Somehow, somewhere, I seem to always be doing something wrong and not get to the desired result. My course says it should be $2$. It's not always the most reliable book, but as I myself can't think of a better solution, I'm going to assume it's right and ask you guys for some help. Just don't hold me accountable if it isn't.
This is the exercise:
Solve the equation. Follow the following steps:
1. Put all factors on the left and equate to $0$.
2. Solve the brackets, reduce and sort the factors.
3. Factor to the second degree using known methods (if impossible, solve graphically).
4. Equate each factor to $0$ separately and solve the equation.
I'm sorry if anything in my formulation is unclear[I've translated the mathematical lingo from Dutch].
And then here's the specific exercise:
$$\frac{(x - 2)^2}{2}-\frac{(x + 1)}{3} = (x - 3)^3$$
I tried solving like this:
Step 1:
$$\frac{(x - 2)^2}{2} - \frac{(x + 1)}{3} - (x + 3)^3 = 0$$
Step 2.1:
$$\frac{(x² - 4x + 4)}{2} - \frac{x}{3} - \frac{1}{3} - (x³ + 9x² + 27x + 27) = 0 $$
That's where I get lost. Though I'm not entirely sure of the first two steps. My solution for step $2.2$ would be:
$$\frac{x^2}{2} - 2x + 2 - \frac{x}{3} - \frac{1}{3} - x³ - 9x² - 27x - 27 = 0 $$
But after this, I have no idea what to do.
Please help me out here! My textbook doesn't offer explanations to solutions, so any help is thoroughly appreciated.
Thank you!
Move everything from one side to the other
$$\frac12 (x-2)^2 - \frac13 (x+1) = (x-3)^3 $$
which implies that $$\frac12 (x-2)^2 - \frac13 (x+1) - (x-3)^3 = 0 $$ (You made a small mistake by changing the sign within the bracket of the last term.)
Your method description says that you can solve it graphically. If you plot the function $$f(x) := \frac12 (x-2)^2 - \frac13 (x+1) - (x-3)^3$$ you will note that there is a root around $x\approx 2$. Indeed, we can test this and if you calculate it you will find that $f(2) = 0$.
Given that $x = 2$ is a root. We know that $x-2$ is a factor of $f(x)$. Using polynomial long division we can find that $$f(x) = -\frac16(x-2)(6x^2-45x+86). $$
Therefore, one root is $x = 2$ and the last two can be found by solving $$6x^2-45x+86 = 0$$ which can be solved by the quadratic formula. (The roots of the last quadratic will not be real numbers. Therefore, there is only one real root, namely $x=2$.)