Question

If P(x) = P
`_(n)`
`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`