Polynomial Division Using Synthetic Division

In dividing the polynomial $x^2-5x+6$x2−5x+6 by the linear polynomial $x-1$x−1, we would normally use polynomial division procedures as shown in the diagram here:

Hence, calling $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 and $D\left(x\right)=x-1$D(x)=x−1 we would write:

$\frac{P\left(x\right)}{D\left(x\right)}=\left(x-4\right)+\frac{2}{x-1}$P(x)D(x)​=(x−4)+2x−1​

We say that the quotient is $\left(x-4\right)$(x−4) and the remainder is $2$2.

Synthetic division

To save time and space, and minimise error, an efficient method of polynomial division emerges, known as synthetic division. While only applicable to linear divisors, the method derives from a more general algorithm known as Horner’s method of synthetic division that caters for larger degree divisors. It is called synthetic division because it changes subtractions into additions, by changing the sign of the constant term in the divisor polynomial .

To see how it works we again divide $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 by the monic polynomial $D\left(x\right)=x-1$D(x)=x−1 .

In the working we need only be concerned with the coefficients of $x^2-5x+6$x2−5x+6 , namely $1$1, $-5$−5  and $6$6 , and the re-signed constant term $1$1 in the divisor polynomial $x-1$x−1 . The new working is shown here: 

After placing the coefficients of $P\left(x\right)$P(x) in the top row, and the re-signed constant  $1$1 just below the symbol $P\left(x\right)$P(x) , we proceed by progressively determining the other five numbers in the white spaces in the table as follows.

Add the numbers in the second column (there is only one in this first instance) and place the result in the sum row (this is the $1$1 shown in red). Then multiply the same re-signed constant $+1$+1 by that sum and place the result (shown as a blue number) in the cell in the third column. Again, add the two numbers in the third column (the green $-4$−4). 

Now multiply the re-signed constant $+1$+1 by the green $-4$−4, and place in the fourth column (shown in purple). Again add the two numbers in the fourth column so that $6+-4=2$6+−4=2 .

Remarkably, the bottom sum row now contains the coefficients of the quotient polynomial. Hence we have  $Q\left(x\right)=x-4$Q(x)=x−4  with remainder $2$2. Thus:

$P\left(x\right)=\left(x-1\right)\left(x-4\right)+2$P(x)=(x−1)(x−4)+2 

 

As a second example that illustrates the algorithm’s efficiency we divide the polynomial $P\left(x\right)=x^5-32$P(x)=x5−32 by the linear polynomial $D\left(x\right)=x-2$D(x)=x−2. The table is shown here.

Note that the coefficients in the top row include four zeros because $P\left(x\right)=x^5+0x^4+0x^3+0x^2+0x-32$P(x)=x5+0x4+0x3+0x2+0x−32.

Here, using the same technique, the result shows that $D\left(x\right)=\left(x-2\right)$D(x)=(x−2) is a factor of $P\left(x\right)=x^5-32$P(x)=x5−32, so that: 

 $P\left(x\right)=\left(x-2\right)\left(x^4+2x^3+4x^2+8x+16\right)$P(x)=(x−2)(x4+2x3+4x2+8x+16) 

Synthetic division works for linear divisors of the form $D\left(x\right)=x-a$D(x)=x−a, but the method can also be used for linear divisors of the form $D\left(x\right)=ax-b$D(x)=ax−b. In such cases the constant to be used is the number $\frac{b}{a}$ba​, and the addition and multiplication procedure is exactly the same as shown above. However, the coefficients (but not the remainder) that are determined in the sum row must all be divided by $a$a.

As an example suppose $P\left(x\right)=2x^4-3x^3-2x^2+2x+5$P(x)=2x4−3x3−2x2+2x+5 and $D\left(x\right)=2x+1$D(x)=2x+1. The table becomes:

 

$P\left(x\right)$P(x)	$2$2	$-3$−3	$-2$−2	$2$2	$5$5
$-\frac{1}{2}$−12​		$-1$−1	$2$2	$0$0	$-1$−1
Sum Row	$2$2	$-4$−4	$0$0	$2$2	$4$4

Check that the re-signed constant becomes a negative number.

The coefficients $2$2, $-4$−4, $0$0 and $2$2 are halved so that $Q\left(x\right)=x^3-2x^2+0x+1$Q(x)=x3−2x2+0x+1 and the remainder $4$4 is as determined by the procedure. Thus:

$P\left(x\right)=\left(2x+1\right)\left(x^3-2x^2+1\right)+4$P(x)=(2x+1)(x3−2x2+1)+4

 

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3