At Leaving Cert you are only required to prove the Factor Theorem for cubics. This is a more general proof.

The Factor Theorem is an important theorem in the factorisation of polynomials. When $(x-k)$ is a factor of a polynomial $p(x)$ then $p(x)=(x-k)q(x)$ for some polynomial $q(x)$ and clearly $k$ is a root. In fact the converse is also true. Most proofs rely on the division algorithm and the remainder theorem; here a proof using strong induction on the degree of the polynomial is used. See Hungerford, T.W., (1997), Abstract Algebra: An Introduction. Brooks-Cole: U.S.A for the standard proof (this reference also describes strong induction).

Factor Theorem
$k\in\mathbb{C}$ is a root of a polynomial $p(x)$ if and only if $(x-k)$ is a factor: $p(x)=(x-k)q(x)$ where $q(x)$ is a polynomial.

Proof:
Suppose deg $p(x)=1$. Then $p(x)=ax+b$ and $p(k)=ak+b=0\Rightarrow k=-b/a$. Clearly $p(x)=(x-(-b/a))a$ and $q(x)=a$ is a polynomial (of degree 0).

Suppose for all polynomials of degree less than or equal to $n-1$, that $k$ a root implies $(x-k)$ a factor.  Let $p(x)=\sum_ia_ix^i$ be a polynomial of degree $n$ and assume $k\in\mathbb{C}$ a root. Now

$p(x)-p(k)=\sum_{i=0}^na_ix^i-\sum_{i=0}^na_ik^i$

$\underset{p(k)=0}{\Rightarrow} p(x)=\sum_{i=1}^n a_i(x^i-k^i)$

Now each $x^i-k^i$ for $i=1,\dots,n-1$ is a polynomial of degree less than or equal to $n-1$, with a root $x=k$, hence $(x-k)$ is a factor of each. Thence $x^i-k^i=(x-k)q_i(x)$ where $q_i(x)$ is a polynomial:

$p(x)=\sum_{i=1}^{n-1}a_i(x-k)q_i(x)+a_n(x^n-k^n)$

Now $x^n-k^n=(x-k)x^{n-1}+kx^{n-1}-k^n$ but $kx^{n-1}-k^n$ is a polynomial of degree $n-1$ with root $k$ and hence $(x-k)$ is a factor:

$x^n-k^n=(x-k)x^{n-1}+(x-k)h(x)$

Hence

$p(x)=(x-k)\lbrack x^{n-1}+h(x)+\sum_{i=1}^{n-1}a_iq_i(x)\rbrack$

Hence $(x-k)$ a factor $\bullet$