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If P(x) = P[tex]_{n}[/tex][tex]x^{n}[/tex] + P[tex]_{n-1}[/tex][tex]x^{n-1}[/tex] + · · · + P[tex]_{0}[/tex] is divided by (x - a), show that the remainder is P(a)

Sagot :

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

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