Question

prove the remainder theorem​

Answer 1

Answer:

It is applied to factorize polynomials of each degree in an elegant manner. For example: if f(a) = a3-12a2-42 is divided by (a-3) then the quotient will be a2-9a-27 and the remainder is -123. Thus, it satisfies the remainder theorem.

Step-by-step explanation:

Answer 2

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$Theorem\: functions \:on \:an\: actual \:case\: that \:a\: polynomial\: is \:comprehensively$

$\: dividable,\: at least\: one \:time\: by\: its\: factor \:in \:order\: to \:get a\: smaller \:polynomial\: and$

$\: ‘a’\: remainder\: of \:zero. \:This \:acts \:as\: one \:of \:the\: simplest\: ways\: to \:determine \:whether$

$\: the\: value\: ‘a’\: is\: a\: root\: of\: the\: polynomial \:P(x).$

$That \:is \:when \:we \:divide \:p(x) \:by \:x-a \:we \:obtain$

$p(x) = (x-a)·q(x) + r(x),$

$as \:we \:know\: that\: Dividend = (Divisor × Quotient) + Remainder$

$But \:if \:r(x)\: is \:simply\: the\: constant \:r (remember\: when\: we\: divide \:by \:(x-a) \:the\: remainder\: is\: a \:constant)…. so\: we \:obtain\: the \:following \:solution, \:i.e$

$p(x) = (x-a)·q(x) + r$

$Observe \:what \:happens \:when\: we \:have \:x \:equal \:to \:a:-$

$p(a) = (a-a)·q(a) + r$

$p(a) = (0)·q(a) + r$

$p(a) = r$

$Hence, proved.$