The Division Identity

You divided $2x^3 - 3x^2 + x - 5$ by $x - 2$ last lesson and got quotient $2x^2 + x + 3$, remainder $1$. Write that as a single equation with no division sign in it. Now substitute $x = 2$ into your equation. What happens to the right-hand side, and why is that interesting?

Every polynomial division can be written without a division sign at all:

Last lesson's division gave quotient $2x^2 + x + 3$ and remainder $1$ when $2x^3 - 3x^2 + x - 5$ was divided by $x - 2$. Written as one equation with no division sign:

$$2x^3 - 3x^2 + x - 5 = (x-2)(2x^2 + x + 3) + 1$$

In general, for any polynomials $P$ and $A$ with $A$ not the zero polynomial, there are unique polynomials $Q$ and $R$ with

$$P(x) = A(x)\,Q(x) + R(x), \qquad \deg R < \deg A$$

This is the same statement as the numerical one from last lesson, $47 = 5 \times 9 + 2$, with "$\deg R < \deg A$" replacing "$0 \leq r < d$".

The gain is that this is an identity: it holds for every value of $x$, not just for some. That means you may substitute anything you like into it, and whatever comes out is true. Almost everything in the next three lessons is a consequence of exercising that freedom well.

Two degree facts follow immediately, and both are useful as checks.

The remainder. $\deg R < \deg A$, always. So:

dividing by a linear polynomial such as $x - 3$ leaves a remainder of degree less than $1$, which means a constant; dividing by a quadratic leaves a remainder of the form $ax + b$ or a constant; dividing by a cubic permits a quadratic remainder.

The quotient. Taking degrees in $P = AQ + R$: the term $AQ$ has degree $\deg A + \deg Q$, and $R$ has strictly lower degree than $A$, hence lower than $AQ$. So the degree of the right side is $\deg A + \deg Q$, giving

$$\deg Q = \deg P - \deg A$$

A cubic divided by a linear polynomial has a quadratic quotient; a quartic divided by a quadratic has a quadratic quotient. If your quotient has the wrong degree, the division went wrong, and you know it before checking a single coefficient.

Here is where the identity earns its keep. Take

$$P(x) = (x - 2)Q(x) + R$$

where $R$ is a constant because the divisor is linear. Substitute $x = 2$:

$$P(2) = (2-2)Q(2) + R = 0 \times Q(2) + R = R$$

The entire quotient disappeared, whatever it was, because it was multiplied by zero. So the remainder equals $P(2)$, and you never needed to know $Q$ at all.

Return to the hook. What is the remainder when $x^{100} - 1$ is divided by $x - 1$? Write $x^{100} - 1 = (x-1)Q(x) + R$ and put $x = 1$:

$$1^{100} - 1 = 0 \times Q(1) + R \quad \Longrightarrow \quad R = 0$$

The remainder is zero, so $x - 1$ is a factor of $x^{100} - 1$. That took one line. Choose the substitution that makes the divisor vanish, and the quotient never has to be computed.

The identity also works in reverse: knowing something about the remainder tells you about the polynomial.

Suppose $P(x) = x^3 + kx^2 - 5x + 2$ leaves a remainder of $8$ when divided by $x - 1$. Write the identity and substitute $x = 1$:

$$P(1) = 8 \quad \Longrightarrow \quad 1 + k - 5 + 2 = 8 \quad \Longrightarrow \quad k = 10$$

When more than one unknown is involved, more than one substitution is needed, and the result is a pair of simultaneous equations. If $P(x) = x^3 + ax^2 + bx - 6$ leaves remainder $0$ on division by $x - 1$ and remainder $-12$ on division by $x + 1$:

$$P(1) = 0: \quad 1 + a + b - 6 = 0 \quad \Longrightarrow \quad a + b = 5$$

$$P(-1) = -12: \quad -1 + a - b - 6 = -12 \quad \Longrightarrow \quad a - b = -5$$

Adding gives $2a = 0$, so $a = 0$ and $b = 5$.

With a divisor of degree $2$ the remainder is $cx + d$ rather than a constant, and the same technique produces equations in $c$ and $d$ by substituting each root of the divisor.

Dividing the identity through by $A(x)$ gives a form you will meet again in senior courses:

$$\frac{P(x)}{A(x)} = Q(x) + \frac{R(x)}{A(x)}$$

So $\dfrac{2x^3 - 3x^2 + x - 5}{x-2} = 2x^2 + x + 3 + \dfrac{1}{x-2}$.

Read it as: a polynomial part plus a genuinely fractional part that cannot be simplified further, because the numerator now has lower degree than the denominator. It is the exact analogue of writing $\tfrac{47}{5} = 9 + \tfrac{2}{5}$.

One caution. The identity $P = AQ + R$ holds for every real $x$, including the values where $A(x) = 0$, because it contains no division. The fraction form does not: at $x = 2$ the right-hand side above is undefined. That is precisely why the substitution trick in the previous card uses the identity and never the fraction.

Five items on the division identity. Work each one, then reveal the answer.

1. 1 A degree-$5$ polynomial is divided by a degree-$2$ polynomial. What is the degree of the quotient?

   Show answer
   Quotient degree is dividend degree minus divisor degree.$3$

2. 2 What form must the remainder take on division by $x^2 + 1$?

   Show answer
   Degree less than $2$.$cx + d$ (possibly with $c = 0$)

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

   Show answer
   Substitute $x = 1$: $1 - 2 + 5$.$4$

4. 4 Find the remainder when $x^3 + 8$ is divided by $x + 2$.

   Show answer
   Substitute $x = -2$: $-8 + 8$.$0$

5. 5 If $P(x) = (x-5)Q(x) + 11$, what is $P(5)$?

   Show answer
   The first term vanishes at $x = 5$.$11$