Synthetic division

Long division by $x - a$ writes the same letters again and again. Every row repeats $x^3$ , $x^2$ , $x$ , and every subtraction changes signs. Synthetic division (also called Horner's method) strips all of that away and works with the coefficients alone. It gives exactly the same quotient and remainder as long division, in a fraction of the space, and it makes factorising cubics and quartics much faster.

The method

To divide $f(x)$ by $x - a$ :

Write the coefficients of $f(x)$ in a row, in descending powers. Put $0$ for any missing power.
Write $a$ (the number that makes $x - a$ zero) at the left.
Bring the first coefficient straight down to the bottom row.
Multiply that bottom number by $a$ and write the product under the next coefficient. Add the column and write the sum in the bottom row.
Repeat step 4 until the last column.

The last number in the bottom row is the remainder. The other bottom numbers are the coefficients of the quotient, which has degree one less than $f(x)$ .

Notice that we add each column. Long division subtracts; synthetic division avoids that by using $a$ instead of $-a$ .

Example 1

Find the quotient and remainder when $4x^3 - 3x + 9$ is divided by $x + 3$ .

Here $x + 3 = x - (-3)$ , so $a = -3$ . The coefficients are $4, 0, -3, 9$ (the $x^2$ term is missing).

$a = -3$	$x^3$	$x^2$	$x$	constant
coefficients	$4$	$0$	$-3$	$9$
products		$-12$	$36$	$-99$
bottom row	$4$	$-12$	$33$	$-90$

How each column was made:

Bring down $4$ .
$4 \times (-3) = -12$ ; then $0 + (-12) = -12$ .
$-12 \times (-3) = 36$ ; then $-3 + 36 = 33$ .
$33 \times (-3) = -99$ ; then $9 + (-99) = -90$ .

$\text{Quotient} = 4x^2 - 12x + 33, \qquad \text{Remainder} = -90$

Check with the remainder theorem: $f(-3) = 4(-27) - 3(-3) + 9 = -108 + 9 + 9 = -90$ . It agrees.

Factorising with synthetic division

Combine it with the factor theorem: find a zero by trial, then use synthetic division to remove its factor. A remainder of $0$ in the table confirms the factor.

Example 2

Factorise $2x^3 - 11x^2 + 17x - 6$ using synthetic division.

Let $f(x) = 2x^3 - 11x^2 + 17x - 6$ .

$f(1) = 2 - 11 + 17 - 6 = 2 \neq 0$

$f(-1) = -2 - 11 - 17 - 6 = -36 \neq 0$

$f(2) = 16 - 44 + 34 - 6 = 50 - 50 = 0$

So $x - 2$ is a factor. Divide with $a = 2$ :

$a = 2$	$x^3$	$x^2$	$x$	constant
coefficients	$2$	$-11$	$17$	$-6$
products		$4$	$-14$	$6$
bottom row	$2$	$-7$	$3$	$0$

The remainder is $0$ , and the quotient is $2x^2 - 7x + 3$ . Splitting with product $6$ and sum $-7$ :

$2x^2 - x - 6x + 3 = x(2x - 1) - 3(2x - 1) = (2x - 1)(x - 3)$

$2x^3 - 11x^2 + 17x - 6 = (x - 2)(2x - 1)(x - 3)$

Example 3

Factorise $3x^4 - 10x^3 + 5x^2 + 10x - 8$ .

The coefficients add to $3 - 10 + 5 + 10 - 8 = 0$ , so $f(1) = 0$ and $x - 1$ is a factor.

$a = 1$	$x^4$	$x^3$	$x^2$	$x$	constant
coefficients	$3$	$-10$	$5$	$10$	$-8$
products		$3$	$-7$	$-2$	$8$
bottom row	$3$	$-7$	$-2$	$8$	$0$

The quotient is $3x^3 - 7x^2 - 2x + 8$ . For it, the even-power coefficients give $-7 + 8 = 1$ and the odd-power coefficients give $3 - 2 = 1$ , so $x + 1$ is a factor. Divide the quotient by $x + 1$ , that is $a = -1$ :

$a = -1$	$x^3$	$x^2$	$x$	constant
coefficients	$3$	$-7$	$-2$	$8$
products		$-3$	$10$	$-8$
bottom row	$3$	$-10$	$8$	$0$

That leaves $3x^2 - 10x + 8 = 3x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2)$ .

$3x^4 - 10x^3 + 5x^2 + 10x - 8 = (x - 1)(x + 1)(x - 2)(3x - 4)$

Dividing by ax − b

Synthetic division is built for $x - a$ , but a small adjustment handles a divisor like $2x - 3$ . Since $\displaystyle 2x - 3 = 2\left(x - \frac{3}{2}\right)$ , divide by $\displaystyle x - \frac{3}{2}$ first, then divide the quotient (not the remainder) by $2$ .

Example 4

Find the quotient and remainder when $4x^3 - 3x + 9$ is divided by $2x - 3$ .

Use $\displaystyle a = \frac{3}{2}$ with coefficients $4, 0, -3, 9$ .

$\displaystyle a = \frac{3}{2}$	$x^3$	$x^2$	$x$	constant
coefficients	$4$	$0$	$-3$	$9$
products		$6$	$9$	$9$
bottom row	$4$	$6$	$6$	$18$

So $\displaystyle 4x^3 - 3x + 9 = \left(x - \frac{3}{2}\right)(4x^2 + 6x + 6) + 18 = (2x - 3)(2x^2 + 3x + 3) + 18$ .

$\text{Quotient} = 2x^2 + 3x + 3, \qquad \text{Remainder} = 18$

The remainder theorem agrees: $\displaystyle f\left(\frac{3}{2}\right) = 4\left(\frac{27}{8}\right) - \frac{9}{2} + 9 = \frac{27}{2} - \frac{9}{2} + 9 = 18$ .

Practice

Use synthetic division in every question.

Find the quotient and remainder when $x^4 + 8x^3 + x - 5$ is divided by $x + 2$ .
Find the quotient and remainder when $x^3 - 6x^2 + 2x - 4$ is divided by $x - 3$ .
Find the quotient and remainder when $2x^4 - x^3 + 3x - 7$ is divided by $x - 2$ .
Find the quotient and remainder when $4x^3 - 12x^2 + 14x - 3$ is divided by $\displaystyle x - \frac{1}{2}$ .
Find the quotient and remainder when $6x^3 + 5x^2 - 4$ is divided by $2x - 1$ .
Factorise $x^3 - 2x^2 - 5x + 6$ .
Factorise $2x^3 - 7x^2 - 10x + 24$ .
Factorise $2x^3 - x^2 - 15x + 18$ .

Answers

Show answers

Coefficients $1, 8, 0, 1, -5$ with $a = -2$ ; bottom row $1, 6, -12, 25, -55$ . Quotient $x^3 + 6x^2 - 12x + 25$ , remainder $-55$ .
Bottom row $1, -3, -7, -25$ . Quotient $x^2 - 3x - 7$ , remainder $-25$ .
Coefficients $2, -1, 0, 3, -7$ with $a = 2$ ; bottom row $2, 3, 6, 15, 23$ . Quotient $2x^3 + 3x^2 + 6x + 15$ , remainder $23$ .
Bottom row $\displaystyle 4, -10, 9, \frac{3}{2}$ . Quotient $4x^2 - 10x + 9$ , remainder $\displaystyle \frac{3}{2}$ .
With $\displaystyle a = \frac{1}{2}$ on $6, 5, 0, -4$ the bottom row is $6, 8, 4, -2$ . Divide the quotient by $2$ : quotient $3x^2 + 4x + 2$ , remainder $-2$ .
Coefficients add to $0$ , so $a = 1$ ; bottom row $1, -1, -6, 0$ ; $x^2 - x - 6 = (x - 3)(x + 2)$ . Answer $(x - 1)(x - 3)(x + 2)$ .
$f(4) = 128 - 112 - 40 + 24 = 0$ ; bottom row $2, 1, -6, 0$ ; $2x^2 + x - 6 = (2x - 3)(x + 2)$ . Answer $(x - 4)(2x - 3)(x + 2)$ .
$f(2) = 16 - 4 - 30 + 18 = 0$ ; bottom row $2, 3, -9, 0$ ; $2x^2 + 3x - 9 = (2x - 3)(x + 3)$ . Answer $(x - 2)(x + 3)(2x - 3)$ .

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The method#

Exam­ple 1#

Fac­toris­ing with syn­thetic divi­sion#

Exam­ple 2#

Exam­ple 3#

Divid­ing by ax − b#

Exam­ple 4#

Prac­tice#

Answers#