Proof of the Polynomial Remainder Theorem

Revision as of 16:59, 30 April 2019 by Zhang2018 (talk | contribs) (Synopsis: Written below is a brief description of the polynomial remainder theorem. The theorem has a wide range of application in subjects spanning from Algebra to)
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The remainder theorem states when a polynomial denoted as $f(x)$ is divided by $x-a$ for some value of $x$, whether real or unreal, the remainder of $\frac{f(x)}{x-a}=f(a)$ Written below is the proof of the polynomial remainder theorem.

All polynomials can be written in the form $f(x)=d(x)\cdot\q(x)+r(x)$ (Error compiling LaTeX. ! Undefined control sequence.), where $d(x)$ is the divisor of the function/polynomial $f(x)$, $q(x)$ is the quotient. amd $r(x)$ is the remainder. Because the $deg r=0$ or $deg r\less\deg d$ (Error compiling LaTeX. ! Undefined control sequence.) and the $deg d=1$, degrees must be whole numbers, and so $deg r=0$. So to speak, $r(x)$ is a constant. We denote this constant $b$.

Knowing this, we can write

$f(x)=d(x)\cdot\q(x)+b$ (Error compiling LaTeX. ! Undefined control sequence.)

$f(x)=(x-a)\cdot\q(x)+b$ (Error compiling LaTeX. ! Undefined control sequence.)

$f(a)=(a-a)\cdot\q(a)+b$ (Error compiling LaTeX. ! Undefined control sequence.)

$f(a)=b$

We have hereby proven when the quantity $x-a$ is divided into a polynomial $f(x)$ of any degree, the value of $f(a)=b$, where b is the remainder. The remainder must be a constant because $deg r=0$.

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