Division and the Four Names

You know that $47 = 5 \times 9 + 2$. Which of those four numbers could you change to $7$ and still have a sensible division statement, and which could not? Say why for the one that cannot.

Every division, of numbers or of polynomials, produces the same four objects: the dividend being divided, the divisor doing the dividing, and the quotient and remainder that come out. Learn them here, on a number small enough to see all at once.

Divide $47$ by $5$. The answer is $9$ with $2$ left over, and every part of that sentence has a name.

$$\underbrace{47}_{\text{dividend}} = \underbrace{5}_{\text{divisor}} \times \underbrace{9}_{\text{quotient}} + \underbrace{2}_{\text{remainder}}$$

Two conditions make this the right answer rather than merely a true statement. The quotient must be a whole number, and the remainder must satisfy

$$0 \leq \text{remainder} < \text{divisor}$$

Without the second condition the statement is useless, because $47 = 5 \times 8 + 7$ is also true, and $47 = 5 \times 0 + 47$, and infinitely many others. The remainder being smaller than the divisor is what makes the answer unique.

Hold on to that, because the polynomial version replaces "smaller than" with "of lower degree than", and it does exactly the same job.

Long division of numbers works through place value: hundreds first, then tens, then units. Polynomial long division works through powers: $x^3$ first, then $x^2$, then $x$, then the constant. The four steps are identical.

Divide. Take the leading term of what is left and divide it by the leading term of the divisor. This gives one term of the quotient.

Multiply. Multiply the whole divisor by that new quotient term.

Subtract. Subtract that product from what you have.

Bring down. Bring down the next term and repeat.

Stop when what remains has a lower degree than the divisor, because at that point the first step is no longer possible: you cannot divide $x$ by $x^2$ and get a polynomial term.

Divide $2x^3 - 3x^2 + x - 5$ by $x - 2$.

Step 1. $2x^3 \div x = 2x^2$. Multiply: $2x^2(x-2) = 2x^3 - 4x^2$. Subtract:

$$(2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2$$

Step 2. Bring down the $+x$. Now $x^2 \div x = x$. Multiply: $x(x-2) = x^2 - 2x$. Subtract:

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

Step 3. Bring down the $-5$. Now $3x \div x = 3$. Multiply: $3(x-2) = 3x - 6$. Subtract:

$$(3x - 5) - (3x - 6) = 1$$

Stop. What is left, $1$, has degree $0$, which is lower than the divisor's degree of $1$. So the quotient is $2x^2 + x + 3$ and the remainder is $1$:

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

That last line is the check, and it is worth doing: expand the right side and you must recover the dividend exactly.

Divide $x^3 - 8$ by $x - 2$ and the first difficulty arrives immediately: there is no $x^2$ term and no $x$ term to bring down.

The fix is the one from Lesson 1. A missing term is a term with coefficient zero, so write it in:

$$x^3 - 8 = x^3 + 0x^2 + 0x - 8$$

Now the division proceeds normally. $x^3 \div x = x^2$; multiply to get $x^3 - 2x^2$; subtract to leave $2x^2$; bring down the $0x$; and so on. The answer is

$$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$$

with remainder zero, which says $x - 2$ is a factor of $x^3 - 8$.

Skipping the placeholders is the most reliable way to get a wrong answer here. The columns are doing real work: they are the polynomial equivalent of place value, and a drifted column subtracts an $x^2$ term from an $x$ term without complaining.

Nothing about the algorithm assumes the divisor is linear. Divide $x^4 + 3x^2 - 2x + 1$ by $x^2 + 1$:

$x^4 \div x^2 = x^2$. Multiply: $x^2(x^2+1) = x^4 + x^2$. Subtract from $x^4 + 0x^3 + 3x^2$, giving $2x^2$, and bring down the rest.

$2x^2 \div x^2 = 2$. Multiply: $2(x^2+1) = 2x^2 + 2$. Subtract from $2x^2 - 2x + 1$:

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

Stop, because $-2x - 1$ has degree $1$, lower than the divisor's degree $2$. So

$$x^4 + 3x^2 - 2x + 1 = (x^2+1)(x^2 + 2) + (-2x - 1)$$

The one change worth noticing: with a quadratic divisor the remainder may be linear, not just a number. The rule is always "lower degree than the divisor", so a divisor of degree $2$ permits a remainder of degree $1$ or $0$. Expecting a bare number here is a genuine trap.

Five items on the algorithm and the vocabulary. Work each one, then reveal the answer.

1. 1 Write the division statement for $58$ divided by $7$.

   Show answer
   $7 \times 8 = 56$, leaving $2$.$58 = 7 \times 8 + 2$

2. 2 Divide $x^2 + 5x + 6$ by $x + 2$.

   Show answer
   $x^2 \div x = x$; the division is exact.Quotient $x + 3$, remainder $0$

3. 3 Divide $x^2 + 3x - 1$ by $x - 1$.

   Show answer
   Quotient $x + 4$, then $(4x - 1) - (4x - 4) = 3$.Quotient $x + 4$, remainder $3$

4. 4 How should $x^3 + 1$ be written before dividing by $x + 1$?

   Show answer
   Every missing power needs a placeholder.$x^3 + 0x^2 + 0x + 1$

5. 5 A cubic is divided by a quadratic. What are the possible degrees of the remainder?

   Show answer
   Lower than the divisor's degree of $2$.$1$ or $0$ (or the remainder is zero)