A polynomial where the leading coefficient is $1$; that is, the coefficient of the term with the highest power is $1$ is said to be a monic polynomial. In engineering, many of the polynomials used will be monic.

Given a monic polynomial $p(x)$ and a monic linear polynomial $x - x_0$, it is always possible to write $p(x) = (x - x_0)q(x) + r$ where if $p(x)$ was of degree $n$, then $q(x)$ is of degree $n - 1$ and $r$ is a constant. Thus, our problem is:

Problem 3.A: Find the quotient and remainder when dividing a polynomial by $x - x_0$

If the remainder is zero, we say that $x - x_0$ is a factor of the polynomial and we may write $p(x) = (x - x_0)q(x)$.

For example, given the polynomial $p(x) = x^3 - 4x^2 - 17x + 60$, we note that:

• $p(x) = (x - 0)(x^2 - 4x - 17) + 60 = x(x^2 - 4x - 17) + 60$
• $p(x) = (x - 1)(x^2 - 3x - 20) + 40$
• $p(x) = (x - 2)(x^2 - 2x - 21) + 18$
• $p(x) = (x - 3)(x^2 - x - 20)$
• $p(x) = (x - 4)(x^2 - 17) - 4$

The remainder is the function evaluated at the point $x_0$, so in this case, we see that $p(0) = 60$, $p(1) = 40$, $p(2) = 18$, $p(3) = 0$ and $p(4) = -4$. In the fourth case, when the remainder is zero, we observe that the point $x = 3$ is a root of the given polynomial.

The algorithm for polynomial division is much more straight-forward than long division. If we are dividing a polynomial by $x - x_0$:

Algorithm 3.A: Polynomial long division by $x - x_0$

Let the remainder $r(x)$ initially be assigned $p(x)$.
Let the quotient $q(x)$ initially be the zero polynomial.
While the remainder is not a constant:

• Assume the highest power in the remainder is $x^k$ and its coefficient is $a_k$.
• Add $a_k x^{k - 1}$ onto the quotient $q(x)$, and
• subtract $a_k x^{k - 1}(x - x_0)$ from the remainder. The degree of the remainder is now $k - 1$ or less.

This is special case of the long polynomial division algorithm. In engineering, it is often most useful to divide a polynomial by its leading coefficient, as both the original polynomial and the polynomial with leading coefficient $1$ both have the same roots.

In the following examples, we will see how the polynomial division algorithm works by dividing the above polynomial by three of the above five terms:

$r(x)$	$q(x)$
$x^3 - 4x^2 - 17x + 60$	$0$
$(x^3 - 4x^2 - 17x + 60) - x^2(x - 1) = -3x^2 - 17x + 60$	$x^2$
$(-3x^2 - 17x + 60) - (-3x)(x - 1) = -20x + 60$	$x^2 - 3x$
$(-20x + 60) - (-20)(x - 1) = 40$	$x^2 - 3x - 20$

Thus, $x^3 - 4x^2 - 17x + 60 = (x - 1)(x^2 - 3x - 20) + 40$. Thus, we note that $x = 1$ is not root of the given polynomial; specifically, $p(1) = 40$.

Next, let us divide the same polynomial by $x - 2$:

$r(x)$	$q(x)$
$x^3 - 4x^2 - 17x + 60$	$0$
$(x^3 - 4x^2 - 17x + 60) - x^2(x - 2) = -2x^2 - 17x + 60$	$x^2$
$(-2x^2 - 17x + 60) - (-2x)(x - 2) = -21x + 60$	$x^2 - 2x$
$(-21x + 60) - (-21)(x - 2) = 18$	$x^2 - 2x - 21$

Thus, $x^3 - 4x^2 - 17x + 60 = (x - 2)(x^2 - 2x - 21) + 18$. Thus, we note that $x = 2$ is not root of the given polynomial; specifically, $p(2) = 18$.

We will now repeat the procedure by dividing the same polynomial by a known root, $x - 3$, which yields

$r(x)$	$q(x)$
$x^3 - 4x^2 - 17x + 60$	$0$
$(x^3 - 4x^2 - 17x + 60) - x^2(x - 3) = -x^2 - 17x + 60$	$x^2$
$(-x^2 - 17x + 60) - (-x)(x - 3) = -20x + 60$	$x^2 - x$
$(-20x + 60) - (-20)(x - 3) = 0$	$x^2 - x - 20$

Thus, $p(x) = (x - 3)(x^2 - x - 20)$ and $x = 3$ is a root of the polynomial $p$.

Polynomial quotient and remainder

The ratio of two polynomials $\frac{p(x)}{d(x)}$ can always be written as $q(x) + \frac{r(x)}{d(x)}$ where $q$ and $r$ are polynomials where $\deg(q) + \deg(d) = \deg(p)$ and $\deg(r) < \deg(d)$. Describe an algorithm for finding the quotient $q$ and the remainder $r$.

Problem

1. Argue that if the polynomial $p(x)$ is monic, then the quotient, too, must be monic.

2. Describe an algorithm for dividing a monic polynomial by a monic quadratic polynomial $x^2 + b_1 x + b_0$.