## Factoring out a root

### Task number: 3064

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 \(n-1\).

#### Hint

Try to factor out \(x - c\) from the polynomial \(P(x) - P(c) = P(x) - 0\).

#### Resolution

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 \(P(x)\).