# 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.

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Think of it as this: $x^4-5x^3+5x^2+11x-\frac{2}3x^3+\frac{10}3x^2-\frac{10}3x-\frac{22}3=x^4-5x^3+5x‌​^2+11x-\frac{2}3x^3+\frac{10}3x^2-\frac{10}3x-\frac{22}3$ – EpicGuy Oct 21 '13 at 7:33
$f(x_0)=0\iff3f(x_0)=0$. – Michael Hoppe Oct 21 '13 at 10:54

In general, if you know that $p(x)$ is a polynomial with zeroes $\alpha_1, \dots, \alpha_n$, and you know that there are no other zeroes, then $p(x) = a(x-\alpha_1)\dots(x-\alpha_n)$ for some $a \neq 0$. This follows from the Factor Theorem. As the questions asks for a polynomial, both of the polynomials you have written are correct solutions, but it is incorrect to say that they are the same polynomial; they just have the same zeroes.
It would be better to write $f(x) = x^4 - 5x^3 + ...$, and then define
$$g(x) = 3f(x) = 3x^4 - 17x^3 + ...$$