Factor theorem and remainder theorem (HL)

See also polynomial division, and polynomials

Recall from polynomial division that

p(x) = d(x)q(x) + r(x)

\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}

$p, d, q, r$ are polynomials. $p$ is divided by divisor $d$ to get some quotient $q$ and remainder $r$ . The degree of $r$ is always smaller than $d$ .

Factor theorem

$x - x_1$ is a factor of polynomial $p$ if and only if $p(x_1) = 0$ . Nothing surprising here.

Remainder theorem

The remainder is of a lower degree than the divisor. So if we divide $p(x)$ by $x - x_1$ , the remainder has to be a constant, namely the remainder is $p(x_1)$ . That is, if we have

p(x) = (x - x_1)q(x) + r(x)

then evaluating at $x = x_1$ yields

p(x_1) = r(x_1)

Example: (1969 American High School Mathematics Examination #34) The remainder $R$ obtained by dividing $x^{100}$ by ${x^2 - 3x + 2}$ is a polynomial of degree less than $2$ . Find $R$ .

The idea is to use remainder theorem to write a system of equations on the coefficients of $R$ .

$x^2 - 3x+2$ factors into $(x-1)(x-2)$ with roots $1, 2$ . The polynomial can be written as

p(x) = x^{100} = q(x)(x-1)(x-2) + ax + b

Then evaluating $p(1)$ and $p(2)$ we obtain the system of equations

\begin{align*} 1 &= a + b \\ 2^{100} &= 2a + b \end{align*}

Subtracting the first equation from the second yields ${a = 2^{100} - 1}$ then it follows that ${b = 2 - 2^{100}}$ .

Hence, ${R = (2^{100} - 1)x + (2 - 2^{100}) \qed}$ or with some simplification ${R = 2^{100}(x - 1) - (x - 2)}$ .

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