Theorem: Remainder of division of a polynomial $f(x)$ by a linear factor $(x-a)$ is $f(a)$
Prerequisites:
Euclidean Polynomials Division (proof)
Proof:
Any polynomial can be written according to the Euclidean polynomial division as:
$\qquad\quad f(x) = q(x)g(x) + r(x)\qquad\text{where }\;\; deg\big(r(x)\big) < deg\big(g(x)\big)\qquad\qquad\ldots (1)$
Here $q(x)$, $g(x)$ and $r(x)$ are the quotient, divisor and remainder respectively and $deg(.)$ represents the degree of the respective polynomial.
Since $g(x) = (x-a)$, therefore $deg\big(g(x)\big) = 1$
Hence, $deg\big(r(x)\big) = 0$
$\Rightarrow \quad\;\; r(x) = r =$ constant
$\therefore \qquad\! r(x)$ is independent of $x$.
Putting $x=a$ and $g(x) = (x-a)$ in equation $(1)$,
$\qquad\quad f(a) = q(a)(a-a) + r\\
\Rightarrow\;\quad\; f(a) = r$
Hence the result
Recommended:
Factor Theorem
Factor representation of polynomials
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