Prove remainder theorem
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Posted by Mritunjay Rai 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
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.
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