Remainder Theorem

When a polynomial $p(x)$ is divided by a linear factor $(x - a)$ the remainder is $P(a)$

When a linear factor is in the form $kx - b$ then it should be put in the form $k(x - \frac{b}{k})$ and the remainder is then $P(\frac{b}{k})$

Proof

Let $P(x) = (x - a) Q(x) + R$

Where $Q(x)$ is a polynomial and $R$ is the remainder when $x = a$

$P(a) = (a - a) Q(a) + R$

$P(a) = R$

$R = P(a)$

When $R = 0$

$P(x) = (x - a) Q(x)$

$x - a$ is a factor of $p(x)$

Since $p(a) = 0$

a is a root (a zero) of $p(x)$

Examples

1. Find the remainder when $x^5 + 4x^4 - 6x^2 + 3x + 2$ is divided by $x + 2$

Solution

$P(x) = x^5 + 4x^4 - 6x^2 + 3x + 2$

$x - a = x + 2$

$a = -2$

$p(-2) = (-2)^5 + 4(-2)^4 - 6(-2)^2 + 3(-2) + 2$

$p(-2) = -32 + 64 - 24 - 6 + 2$

$= 66 - 62$

$= 4$

$P(-2) = 4$

2. Find the remainder when $4x^3 - 6x^2 - 5$ is divided by $2x - 1$

Solution

$P(x) = 4x^3 - 6x^2 - 5$

$x - a = \left(x - \frac{1}{2}\right)$

$P\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 6 \times \left(\frac{1}{2}\right)^2 - 5$

$= \frac{1}{2} - \frac{3}{2} - 5$

$= 1 - 3 - 10$

$\frac{1}{2}$

$P\left(\frac{1}{2}\right) = -6$

If a is a zero of $p(x)$ then $(x - a)$ is a factor of $p(x)$ i.e. $p(x) = (x - a) Q(x)$

Proof

Let $p(x) = (x - a) Q(x) + R$

Given a is a zero of $p(x)$

Then $p(a) = 0$

$0 = (a - a) Q(a) + r$

$0 = r$

$r = 0$

$p(x) = (x - a) Q(x)$

$x - a$ is a factor of $p(x)$

Examples

Factorize completely the following polynomial function $x^4 - 5x^3 + 6x^2 + 2x - 4$

Solution

Let $p(x) = x^4 - 5x^3 + 6x^2 + 2x - 4$

$P(1) = 1 - 5 + 6 + 2 - 4$

$= 0$

$P(2) = 2^4 - 5(2)^3 + 6(2)^2 + 2 \times 2 - 4$

$= 16 - 40 + 24 + 4 - 4$

$= 0$

$(x - 1)$ and $(x - 2)$ are factors of $P(x)$

$\rightarrow P(x) = (x - 1)(x - 2) Q(x)$

$P(x) = (x^2 - 3x + 2) Q(x)$

$$\begin{array}{r} x^2 - 2x - 2 \\ \hline x^2 - 3x + 2 \big) x^4 - 5x^3 + 6x^2 + 2x - 4 \\ \phantom{x^2 - 3x + 2 \big)} -x^4 + 3x^3 - 2x^2 \\ \hline \phantom{x^2 - 3x + 2 \big)} -2x^3 + 4x^2 + 2x \\ \phantom{x^2 - 3x + 2 \big)} -2x^3 + 6x^2 - 4x \\ \hline \phantom{x^2 - 3x + 2 \big)} -2x^2 + 6x - 4 \\ \phantom{x^2 - 3x + 2 \big)} -2x^2 + 6x - 4 \\ \hline \phantom{x^2 - 3x + 2 \big)} 0 \\ \end{array}$$

$Q(x) = x^2 - 2x - 2$

$= (x^2 - 2x + 1) - 3$

$= (x-1)^2 - (\sqrt{3})^2$

$= ((x-1) + \sqrt{3})((x-1) - \sqrt{3})$

$\rightarrow P(x) = (x-1)(x-2)(x-1+\sqrt{3})(x-1-\sqrt{3})$

Synthetic division is the shortcut method to find the remainder when a polynomial function is divided by a factor $x - a$

Example

1. Use synthetic division to divide $2x^3 + x^2 - 3x + 4$ by $x + 3$

Solution

$x - a = x + 3$

So; $a = -3$

Then

$$\begin{array}{r|rrrr} 2 & 1 & -3 & 4 \\ \hline -3 & -6 & 15 & -36 \\ \hline & 2 & -5 & 12 & -32 \\ \end{array}$$ $$\begin{aligned} \text{Quotient} & \quad 2x^2 - 5x + 12 \\ \text{Remainder} & \quad -32 \\ \end{aligned}$$

Note that in the synthetic division the third row will contain the coefficients of the quotient and the remainder

2. Use synthetic division to divide $4x^3 - 6x^2 - 5$ by $2x - 1$

Solution

$x - a = 2x - 1$

$a = \frac{1}{2}$

$$\begin{array}{ccccc} \frac{1}{2} & \begin{array}{cccc} 4 & -6 & 0 & -5 \\ \hline 2 & -2 & -1 \\ \hline 4 & -4 & -2 & -6 \end{array} \end{array}$$

$Q(x) = 4x^2 - 4x - 2$

Remainder $= -6$

Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a x + a_0$

Where

$a_n, a_{n-1}, a_1, a_0$ are integral coefficients and

Let $\frac{p}{q}$ be a rational number in its lowest term

Then $\frac{p}{q}$ is a zero of $p(x)$ when $p$ is a factor of $a_0$

$q$ is a factor of $a_n$

Example

To find zero of $2x^3 - x - 3$

If $\frac{p}{q}$ is a zero of the expression

Then $p$ is a factor of $-3$ i.e. $-1, 1, -3, 3$

$q$ is a factor of $2$ i.e. $1, -1, 2, -2$

We try $-1, 1, -3, 3, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2},$ and $\frac{3}{2}$