• Remainder Theorem

Remainder Theorem

Remainder Theorem is a way to solve the remainder of a divided polynomial. It was introduced by Etienne Bezout.

First Degree Divisor

If a polynomial in an unknown quantity '$x$' is divided by a first degree expression in the same variable $(x-k)$, the remainder will be equal to the sum obtained in the polynomial when $x=k$.

$$ \text{remainder} = f(x)_{x \to k} $$

Example:

1. What is the remainder of the polynomial x³ + 4x² - 3x + 8 when divided by (x-5)?
Solution:
$$ \text{Let } P(x) = x^3 + 4x^2 - 3x + 8$$ $$\begin{align} \text{Divisor: } & x-k = x - 5 \\ & k = 5 \\ \end{align}$$ @$k=5$, (substitute k to the polynomial)
$$\begin{align} remainder = P(k) & = P(5) \\ & = (5)^3 + 4(5)^2 -3(5) +8 \\ & = 218 \\ \end{align}$$


2. Given f(x) = (x+3)(x-4)+4, when f(x) is divided by (x-k), the remainder is k. Find k from the following choices:
• 4
• 2
• 6
• 8
Solution:
$$\begin{align} \text{Let }P(x) & = (x+3)(x-4)+4 \\ & = x^2 + 3x - 4x - 12 + 4\\ & = x^2 -x - 8 \\ \end{align}$$ $$\begin{align} \text{Divisor: } & x-k \\ & x = k \\ \end{align}$$ $$\begin{align} remainder = k & = P(k) \\ k & = k^2 - k - 8 \\ & k^2 - 2k - 8 = 0 \\ & k = (-2,4)\\ \end{align}$$ Therefore, $k = 4$.

n^th Degree Divisor

For remainders from divisors in n^th degree, see table below.

Divisor	Remainder
$x-k$	constant
$ax^2+bx+c$	$ax+b$
$ax^3+bx^2+cx+d$	$ax^2+bx+c$
$n^{th} \text{ degree}$	$(n-1)^{th} \text{ degree}$

Example:

1. When the polynomial P(x) = x³ + 3x² - 2Ax + 3, where A is a constant, is divided by (x²+1) we get a remainder equal to 5x. Find A.
Solution:
$$ P(x) = x^3 + 3x^2 - 2Ax + 3$$ $$\begin{align} \text{Divisor: } & x^2 + 1 \\ & x^2 + 1 = 0 & \text{equate to zero} \\ & x = \sqrt{-1} \\ & x = \pm i \\ \end{align}$$ @$x = +i$, (substitute to the polynomial)
$$\begin{align} remainder = 5x & = P(x) = P(i) \\ 5(i) & = (i)^3 + 3(i)^2 - 2A(i) + 3 \\ 5i & = -i + 3(-1) - 2 A i + 3 \\ A & = \frac{6i}{-2i} \\ A & = -3 \\ \end{align}$$
@$x = -i$, (substitute to the polynomial)
$$\begin{align} remainder = 5x & = P(x) = P(-i) \\ 5(-i) & = (-i)^3 + 3(-i)^2 - 2A(-i) + 3 \\ -5i & = i + 3(-1) + 2 A i + 3 \\ A & = \frac{-6i}{2i} \\ A & = - 3 \\ \end{align}$$
Therefore, $A=-3$.


2. Let P(x) be a polynomial such that when P(x) is divided by (x-19), the remainder is 99, and when P(x) is divided by (x-99) the remainder is 19. What is the remainder when P(x) is divided by (x-19)(x-99)?
Solution:
From the problem, we know the following: $$\begin{align} & \frac{P(x)}{x-19} \text{ remainder = } 99, k = 19, P(k) = P(19) = 99 \\ & \frac{P(x)}{x-99} \text{ remainder = } 19, k = 99, P(k) =P(99) = 19 \\ \end{align}$$
$$\begin{align} \text{Divisor: } & (x-19)(x-99) &\text{Divisor is a polynomial in 2nd degree.}\\ & x = (19, 99) \\ \end{align}$$ @$x = 19$, (substitute to the remainder $ax+b$)
$$\begin{align} remainder = ax+b & = P(x) = P(19) \\ a(19) + b & = 99\\ 19a + b & = 99 & \text{Eq. 1}\\ \end{align}$$
@$x = 99$, (substitute to the remainder $ax+b$)
$$\begin{align} remainder = ax+b & = P(x) = P(99) \\ a(99) + b & = 19 \\ 99a + b & = 19 & \text{Eq. 2}\\\ \end{align}$$
By substitution and elimination between Eq. 1 & 2, $$\begin{align} 19a + b & = 99\\ 99a + b & = 19\\ \hline a = -1 \\ b = 118\\ \end{align}$$
Substituting a and b to the remainder.
$$\begin{align} remainder & = ax+b \\ & = (-1)x+118\\ & = -x + 118\\ \end{align}$$