Getting rid of the denominator of a polynomial I'm tutoring a high school precalculus student; our current topic is the roots of higher order polynomials.

The problem we're solving is:

Find a polynomial with the roots $\frac23$, -1, and $(3 + \sqrt{2}i)$

I started the solution by explaining that since one of the roots is a complex number, that root's complex conjugate must also be a root. So the polynomial (well, one of many possible answers) must be:

$(x-\frac 23)(x+1)(x-(3+\sqrt{2}i))(x-(3-\sqrt{i}))$

When I multiplied all these terms, I get:

$f(x) = x^4-5x^3+5x^2+11x- \frac 23x^3+ \frac{10}3x^2- \frac{10}3x- \frac {22}3$

Now, I put as the next step, "multiply all terms by 3 to get rid of the denominator."

This does in fact arrive at the correct answer:

$f(x)=3x^4-17x^3+25x^2+23x-22$

This solution matches the instructor's solution. What my college-educated brain can't figure out is: why does this work? I've got an equation and I'm multiplying everything on the right-hand side by 3 without doing the same to the left hand side. How does this maintain equality between the left and right hand sides of the equation?

I realize this is remedial math here; I've been doing this algebra for a while today and it's currently 1AM so hopefully I'm just overlooking something every high school kid knows because I'm cross eyed and tired. Still, if someone could point out the obvious, I'd be grateful.
They aren't the same function, so it's not correct to call them the same letter. However, the new polynomial will still have the exact same roots as the old polynomial, so it's one of the many other possible answers for the question.

It would be better to write $f(x) = x^4 - 5x^3 + ...$, and then define

$$g(x) = 3f(x) = 3x^4 - 17x^3 + ...$$
I know this is an old question, but since I'd found myself here, so too may others, so...

Your problem asks for "a" solution, not "the" solution.  As user61527 has said, your solutions really aren't the same function, but they are multiples of some function.  Since you are seeking an equation whilst knowing the zero points, every solution will equal zero so any multiple thereof will also equal zero.  Both your original solution (with fractions) and your simplified one are therefore correct and, while being different functions, satisfy the instructions to find "a" solution.