## Factoring out a root

Let $$P(x)$$ be a polynomial of degree $$n \geq 1$$ with root $$c$$, i.e. $$P(c) = 0$$. Prove that $$P(x) = (x-c)Q(x)$$, where $$Q$$ is some polynomial of degree $$n-1$$.
Try to factor out $$x - c$$ from the polynomial $$P(x) - P(c) = P(x) - 0$$.
Assume that $$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$$. Then $P(x) = P(x) - P(c) = a_n(x^n - c^n) + a_{n-1}(x^{n-1} - c^{n-1}) + \cdots + a_1 (x - c) + 0.$ From each of the parenthesized expressions $$x^k - c^k$$ we can factor out $$x-c$$ because $x^k - c^k = (x-c)(x^{k-1} + x^{k-2} c + x^{k-3}c^2 + \cdots + xc^{k-2} + c^{k-1}).$ So we can factor out $$(x-c)$$ from the entire polynomial $$P(x)$$.