Math Core

Multiplying polynomials builds bigger ones. Dividing takes them apart, and that's exactly what you need to factor a cubic or a quartic and find its zeros. The good news: polynomial division works just like the long division of numbers you learned in elementary school.

What division tells you

When you divide $437$ by $12$ , you get $36$ with remainder $5$ . That's a compact way of saying

437 = 12 \cdot 36 + 5, \qquad 0 \le 5 < 12.

Polynomials follow the same pattern. The remainder has to be "smaller" than the divisor, and for polynomials smaller means lower degree.

The division algorithm

For polynomials $P(x)$ (the dividend) and $D(x)$ (the divisor, not zero), there are unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) with

P(x) = D(x) \cdot Q(x) + R(x),

where $R(x) = 0$ or the degree of $R$ is less than the degree of $D$ . Equivalently, $\dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}$ .

If the remainder is $0$ , then $D(x)$ is a factor of $P(x)$ . When you divide by a linear divisor like $x - 3$ , the remainder is always a constant.

Polynomial long division

Each round of long division has four moves:

Divide the leading term of what's left by the leading term of the divisor. That's the next term of the quotient.
Multiply that term by the whole divisor.
Subtract the product from what's left.
Bring down the next term and repeat until the degree of what's left is less than the degree of the divisor.

Worked example: Dividing by a linear binomial

Divide $2x^3 + 3x^2 - 8x + 5$ by $x + 3$ .

step	divide	multiply by $x + 3$	subtract, what's left
1	$2x^3 \div x = 2x^2$	$2x^3 + 6x^2$	$-3x^2 - 8x + 5$
2	$-3x^2 \div x = -3x$	$-3x^2 - 9x$	$x + 5$
3	$x \div x = 1$	$x + 3$	$2$

The degree of $2$ is less than the degree of $x + 3$ , so you stop. The quotient is $2x^2 - 3x + 1$ and the remainder is $2$ :

2x^3 + 3x^2 - 8x + 5 = (x + 3)(2x^2 - 3x + 1) + 2.

Check by multiplying: $(x + 3)(2x^2 - 3x + 1) = 2x^3 + 3x^2 - 8x + 3$ , and adding $2$ gives the original. ✓

Common mistake

If the dividend skips a power of $x$ , write that term with a coefficient of $0$ before you start. For $x^4 + 2x^3 - x + 3$ , use $x^4 + 2x^3 + 0x^2 - x + 3$ . Skipping the placeholder shifts every later column and wrecks the answer. The same goes for the divisor: $x^2 + 1$ is $x^2 + 0x + 1$ .

Worked example: Dividing by a quadratic

Divide $x^4 + 2x^3 - x + 3$ by $x^2 + 1$ .

Write the dividend as $x^4 + 2x^3 + 0x^2 - x + 3$ .

step	divide	multiply by $x^2 + 1$	subtract, what's left
1	$x^4 \div x^2 = x^2$	$x^4 + x^2$	$2x^3 - x^2 - x + 3$
2	$2x^3 \div x^2 = 2x$	$2x^3 + 2x$	$-x^2 - 3x + 3$
3	$-x^2 \div x^2 = -1$	$-x^2 - 1$	$-3x + 4$

What's left, $-3x + 4$ , has degree $1$ , less than the divisor's degree $2$ . So

\frac{x^4 + 2x^3 - x + 3}{x^2 + 1} = x^2 + 2x - 1 + \frac{-3x + 4}{x^2 + 1}.

The remainder here is not a constant, which is fine: it only has to have lower degree than $x^2 + 1$ .

Synthetic division

When the divisor has the form $x - c$ , you can drop the variables and work with coefficients only. This shortcut is called synthetic division.

Write $c$ in a box on the left and the dividend's coefficients (with $0$ placeholders) in a row.
Bring the first coefficient straight down.
Multiply it by $c$ , write the product under the next coefficient, and add.
Repeat multiply-and-add across the row.

The bottom row gives the quotient's coefficients (one degree lower than the dividend) followed by the remainder.

Worked example: Synthetic division

Divide $2x^4 - 7x^3 + 5x - 1$ by $x - 3$ .

Here $c = 3$ , and the coefficients are $2, -7, 0, 5, -1$ (notice the $0$ for the missing $x^2$ term).

\begin{array}{r|rrrrr} 3 & 2 & -7 & 0 & 5 & -1 \\ & & 6 & -3 & -9 & -12 \\ \hline & 2 & -1 & -3 & -4 & -13 \end{array}

Bring down $2$ . Then $2 \cdot 3 = 6$ and $-7 + 6 = -1$ . Next $-1 \cdot 3 = -3$ and $0 + (-3) = -3$ . Then $-3 \cdot 3 = -9$ and $5 + (-9) = -4$ . Finally $-4 \cdot 3 = -12$ and $-1 + (-12) = -13$ .

The dividend has degree $4$ , so the quotient has degree $3$ :

2x^4 - 7x^3 + 5x - 1 = (x - 3)(2x^3 - x^2 - 3x - 4) - 13.

Common mistake

For a divisor like $x + 2$ , the number in the box is $c = -2$ , because $x + 2 = x - (-2)$ . Using $+2$ is the most common synthetic division mistake.

Dividing by ax − b

Synthetic division needs a leading coefficient of $1$ in the divisor. For a divisor like $2x - 1$ , write it as $2\left(x - \tfrac{1}{2}\right)$ . Do synthetic division with $c = \tfrac{1}{2}$ , then divide the quotient (but not the remainder) by $2$ .

Worked example: A divisor with leading coefficient 2

Divide $6x^3 - x^2 - 5x + 2$ by $2x - 1$ .

Use $c = \tfrac{1}{2}$ :

\begin{array}{r|rrrr} \tfrac{1}{2} & 6 & -1 & -5 & 2 \\ & & 3 & 1 & -2 \\ \hline & 6 & 2 & -4 & 0 \end{array}

So $6x^3 - x^2 - 5x + 2 = \left(x - \tfrac{1}{2}\right)(6x^2 + 2x - 4)$ . Move the factor of $2$ from the quotient into the divisor:

6x^3 - x^2 - 5x + 2 = (2x - 1)(3x^2 + x - 2).

The quotient is $3x^2 + x - 2$ and the remainder is $0$ , so $2x - 1$ is a factor. Long division gives the same result if you prefer it.

Tip

Always check a division by multiplying back: divisor times quotient, plus remainder, must equal the dividend. A faster partial check is to compare the constant terms, or to substitute $x = 1$ into both sides.

Practice

Practice 1

When $x^2 + 5x - 14$ is divided by $x - 2$ , the quotient is $x + a$ and the remainder is $0$ . Find $a$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the remainder when $4x^3 + 2x^2 - 5x + 7$ is divided by $x + 1$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Divide $x^3 - 6x^2 + 4$ by $x - 1$ .

Practice 4

Divide $x^4 - 16$ by $x - 2$ .

Practice 5

Divide $3x^3 - x^2 + 4x + 2$ by $x^2 - x + 1$ .

Practice 6

A rectangle has area $2x^3 + 5x^2 - 4x - 3$ square units and width $x + 3$ units. What is its length?

Practice 7

When a polynomial $P(x)$ is divided by $x - 4$ , the quotient is $x^2 + 3x - 1$ and the remainder is $5$ . What is the constant term of $P(x)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Divide $8x^3 + 2x^2 - 5x + 3$ by $2x + 1$ .