Finitely generated module equal to free module of rank $n$? Not just the quotient of free module.

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(For commutative algebra)

I have a Ring defined as the polynomial Ring over Field $K$: $R = K[x]$. I need to show that the finitely generated module $M$ is equal to a free module of rank $n$ for finite $n$. I can use the surjective function $f\colon R^n \to M$ to show that $M = R^n/\ker(f)$, but I am not sure how to get to the final result from here (that $M=R^n$).