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

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