Question

If P(x) = Pormula">$x^{n}$ + P$_{n-1}$$x^{n-1}$ + · · · + P$_{0}$ is divided by (x - a), show that the remainder is P(a)

Answer 1

If $P(x) = p_nx^n + p_{n-1}x^{n-1}+\ldots+p_0$ is divided by $(x-a)$, then $P(x) = (x-a) \cdot Q(x) + R(x)$ for some polynomials $Q,R$. Moreover, $\deg R < 1$ (because $\deg (x-a) = 1$), so there exists  $\alpha \in \mathbb{R}$ such that $R(x) = \alpha$ for all $x \in \mathbb{R}$. But if we calculate $P(a)$, it turns out that $P(a) = (a-a)\cdot Q(a) + \alpha$, so $R(x) = \alpha = P(a)$. $\blacksquare$