Factoring Polynomials through Constructive Laziness and Synthetic Division
by Jerry Tobin
Synthetic division is presented in many algebra textbooks as a method of dividing a polynomial by a
binomial of the form (x – a). As presented, this process which reduces the amount of writing by
establishing a discipline to maintain columns according to powers of x. This is similar to the use of
columns to denote powers of 10 for simple number arithmetic such as addition, subtraction (which is
addition), multiplication, and long division. Learning synthetic division also tends to encourage
students to be what I define as “constructively lazy” later in this explanation. Synthetic division can
be used much more strongly for factoring polynomials and tying together the concepts of roots of
polynomials (x-intercepts for graphing), the remainder theorem for long division by (x – a), the
application of Descartes’ rule of signs, and the occurrence of complex conjugate pairs. The method
explained below illustrates that synthetic division can be performed iteratively on a shorter
polynomial each time a zero is found and that the coefficients to the left of the zero actually provide
the shorter polynomial which is also a factor of the dividend polynomial.
To start developing the shortened version of synthetic division and finding roots of a polynomial of
degree 2 or greater, consider an easily-factored quadratic equation such as x2 – 5x + 6 = 0.
This factors as: (x - 3) (x - 2) = 0
Case 1:
x-3=0
x=+3
Case 2:
x -2=0
x=+2
For long division, we would write:
1x  2
2
x

5
x6
(x - 3)
x 2 - 3x
- 2x + 6
-2x + 6
0
Note: In this process, we change the sign on two lines to get the term x2 to change and later the term
-2x to change so they will subtract and leave zeroes in their respective columns.
There is no remainder, so we have successfully factored the polynomial to (x - 3) (x - 2) as expected.
This process has extra writing. The process would be unchanged if the x from x – 3 was left out as
well as the entries for powers of x contained in the dividend. Instead, if the column positions are
maintained for each power of x, only the coefficients need to be written. (Note, if a power of x is
missing, a zero must appear in the column as a placeholder. This is no different than what is done
for long division of numbers for missing powers of 10, and a later example will show a missing
power of x).
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So, for the development of synthetic division, consider this same division done without writing x but
maintaining the columnar positions for the respective powers.
This would give:
1 2
-3 1  5  6
1 -3
-2+6
-2 + 6
0
Note: We subtract this line to get the term, - 3, to change.
Note: We subtract this line again to get the term, + 6, to change.
Now, to continue to illustrate synthetic division, instead of changing the sign for each line because
of the number in the divisor, consider changing the sign of the number in the divisor instead. With
this change, the division would no longer require repeatedly subtracting after each line is entered.
Instead, addition can be done repeatedly, and the chance for error from forgetting which sign(s) to
change reduced.
To show the process with the sign changed as discussed above, consider:
1 2
+3 1  5  6
1+3
-2 + 6
-2 - 6
0
But, this still has extra writing as indicated by the + 3, - 2 + 6, and -6 in this color.
1 2
+3 1  5  6
1+3
-2+6
-2 - 6
0
If what was done is considered carefully, in this process, the 1 from first term of the dividend was
brought down and multiplied by the +3 from the divisor and their product added to the -5 from the
dividend coefficient in the next column to obtain the -2 in that column. This process was repeated
by multiplying the resulting -2 by the +3 in the divisor and then adding in the last column to obtain
the zero remainder.
This process still has redundancy (and therefore extra writing). It could be represented by two
lines without writing the numbers below the division symbol except those which are shown in black
and this color. The division process could now be written as it is shown below.
This gives:
1 2
3 1 56
1
-2
0
Synthetic Division – J. Tobin
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The next (and even constructively lazier) method is to only write the numbers in the bottom
two lines. For this method, the constant from the divisor is written on the line of the results.
The constant from (x – a), which is 3 in this case and is the value for x at which the polynomial
is being evaluated, is written on the line for the resultant to the left, and the coefficients
resulting from the division process are written on the same line with this constant.
(Constructive laziness is defined by this author as doing the least one can to get a proper
result. Examples include knowing multiple methods by which a particular problem can be
solved and then selecting the easiest method on a case-by-case method or applying
mechanical advantage, such as with a lever or pulley, to elevate a weight. Shorthand
notation used by mathematicians also exemplifies constructive laziness.)
An example of shortened synthetic division dividing x2 – 5x + 6 by (x – 3) is as follows:
| 1 -5
3| 1 -2
+ 6
+ 0
This line shows the coefficients of the dividend.
This line shows constant from the divisor to the left of the vertical
line. To the right are the coefficients of a shorter polynomial obtained
by long division followed by the remainder for the division process.
Basic Method Description
To review this process step-by-step, consider dividing x2 - 5x + 6 by (x – 3). First, write the
coefficients of the polynomial in the correct columns (watch out for missing powers which requires
insertion of 0 in the respective columns).
Then, change the sign of the – 3 to + 3. Write this constant on the second line on the left.
|1 -5
+ 6
3|
Now, the division process can begin. Bring down the first coefficient (1) from the dividend (this
also becomes the first coefficient of the quotient) and multiply this number by the constant (3) on the
left, then add this product to the second coefficient and write the sum (note: 1*3+(-5) = -2 ) in the
second position on the second line.
| 1 -5
3| 1 -2
+ 6
Now, multiply the constant (3) on the left by the next number on the second line (-2) and add this
product to the next coefficient from the dividend. Record the sum (as before, 3*(-2) + 6 = 0) in the
last position. This is the remainder for the division process.
| 1 -5
3| 1 -2
+ 6
+ 0
Note: The dividend divided by the divisor is the polynomial using the coefficients on the right in the
second line with the remainder divided by the divisor. For this example, this gives
x2  5x  6
0
 x2
=x–2
x 3
x 3
This tells us that x2 - 5x + 6 = (x - 3) (x – 2) because the remainder was zero.
Synthetic Division – J. Tobin
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One of the valuable aspects of this highly-abbreviated method, besides saving writing, is that
repeated efforts can be done using the original dividend until a zero is found, thus indicating a factor
of the dividend polynomial. Due to the remainder theorem for long division, each effort for this will
provide a value for the polynomial at x= (the constant from the divisor) and therefore provide a
point on the graph of the dividend polynomial versus x.
When the remainder is zero, the other coefficients on the right are the coefficients for a shorter
polynomial which is also a factor of the original polynomial. For repeated factors, this method will
enable the user to find them if the other factors are rational numbers.
To understand this, consider factoring x3 - x2 - 5x – 3. Note that this can be written as
1x3 - 1x2 - 5x – 3 x0.
From Descartes’ rule of signs, f (x) = x3 - x2 - 5x – 3 has only one sign change, so exactly one
positive real root exists. Similarly, f (-x) = - x3 - x2 + 5x – 3, which has two sign changes and
tells us that two or zero negative real roots exist.
The value of x which would be the easiest to try, if it is a rational number, would be from the
factors of the coefficient of the x0 term, which is 3, divided by the coefficient of the highest
power of x which is (1). Thus, 3/1 = 3 which factors as 3 = (1) * (3). So the possible integers
which would work are - 3, - 1, 1, or 3.
If we try x = - 3, the synthetic division will look like the following:
| 1 -1 - 5 - 3
- 3 | 1 -4 + 7 - 24
This has revealed neither a factor of x3 - x2 - 5x – 3 nor a zero for y = x3 - x2 - 5x – 3.
However, it does give a point for the graph of y = x3 - x2 - 5x – 3 at (x , y) = ( + 3 , -24)
according to the remainder theorem for long division of a polynomial by a binomial.
(To check this result, consider ( -3)3 –( -3)2 -5( -3) – 3 = -27 - 9 + 15 – 3 = -39 + 15 = -24.)
Now, if we try x = + 3, we get:
| 1 -1
+3| 1 +2
-5 - 3
+1
0
This shows that 3 is a zero for y(x) = x3 - x2 - 5x – 3 (or y(3) = 0). It also can be read that
y(x) = [x – (+3)] * (1 x2 + 2x +1) or y = (x - 3) (1x2 + 2x + 1) = (x – 3) (x2 + 2x + 1)
Please note that factoring y(x) by a linear binomial results in a new factor of one less degree (in
this case, we can read the blue coefficients with descending powers from 2 to 0). This is
important to understand the process for finding additional roots with less effort.
Synthetic Division – J. Tobin
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Now the factoring process can be continued by doing synthetic division on the shorter
polynomial, 1 x2 + 2x +1. Finding a root of the polynomial with a zero remainder will always
provide a shorter polynomial which is also often factorable. Most textbooks leave the
student to perform synthetic division again on the original polynomial, but this will not
reveal repeated roots and isn’t as constructively lazy as performing synthetic division on
a shorter polynomial. The only major consideration for this is that a non-zero remainder will
not be a value for y on the graph of y versus the original dividend polynomial (x3 - x2 - 5x – 3
in this example).
Now, dividing the coefficient of the new x0 term by the coefficient of the new highest power
of x gives 1/1 = 1 = (1) * (1). So, the possible integer zeroes remaining are x = +1 and x = +1.
For x2 + 2x + 1= (x + 1) (x + 1), we need to consider x + 1 = 0, or x= -1. Also, for this
example, Descartes’ rule of signs told us that one positive real root exists and it has been found
(+ 3_.
Using this information and continuing with synthetic division on the shorter polynomial gives:
| 1 +2
+1
-1| 1 +1
| 0
(Note that a line (|) has been inserted before the zero to separate the zero
from the other coefficients on that line. This separates the zero root from the shorter
polynomial which is also a factor of the original long polynomial into which x + 1 has been
divided.
By the remaining coefficients to the left of the 0, we see that the last polynomial factor will be
2
a repetition of (x+1). Thus, 1 x + 2x +1 = (x +1) *(x + 1). Referring back to the original
polynomial, x3 - x2 - 5x – 3, we have from the two synthetic divisions that
x3 - x2 - 5x – 3 = (x – 3) (x +1) (x + 1)
The entire factoring process with the least possible writing can be illustrated by:
| 1 -1
+3| 1 +2
-5 - 3
+ 1 | 0 The 1st factor found is (x – 3) or x3 - x2 - 5x–3 = (x – 3)( x2 + 2x +1)
-1| 1 +1
| 0
-1| 1 | 0
The 2nd factor found is (x + 1) or x3 - x2 - 5x – 3= (x -3)(x + 1) (x + 1)
Confirms that the 3rd factor is (x + 1).
Thus, the three roots are all real and are found at x = + 3. – 1, and – 1. This agrees with the
expectation from Descartes’ rule of signs.
It is important to emphasize that this shortened method identifies repeated roots (no
matter how many times they repeat).
Synthetic Division – J. Tobin
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By inserting the vertical line before each zero ( | in this paper), the next synthetic division
process is clearly identified for the shortened polynomial from the last iteration.
| 1 -1
+3| 1 +2
-5 - 3
+ 1 | 0 The 1st factor found is (x – 3) or x3 - x2 - 5x–3 = (x – 3)( x2 + 2x +1)
-1 | 1 +1
| 0
-1 | 1 | 0
The 2nd factor found is (x – 1) or x3 - x2 - 5x–3= (x -3)(x + 1) (x + 1)
Confirms that he third factor is (x + 1) or x3 - x2 - 5x–3= (x -3)(x + 1) (x + 1).
If the roots are chosen in a different order, the result will be the same (of course):
| 1 -1
-1 | 1 - 2
-5 - 3
- 3 | 0 The 1st factor found is (x + 1) or x3 - x2 - 5x–3 = (x + 1)( x2 - 2x - 3)
+3| 1 +1
| 0
-1 | 1 | 0
The 2nd factor found is (x – 3) or x3 - x2 - 5x - 3 = (x + 1)(x - 3)(x + 1)
Confirms that the 3rd factor is (x + 1) or x3 - x2 - 5x - 3 = (x + 1)(x - 3)(x + 1)
Also, this method does not require the coefficient of the highest power of x to be one.
Consider the following:
2x3 – 5x2 + 8x - 20 .
To factor this, consider 20 = (20)(1) = (10)(2) = (5)(4). Also, 2 = (2)(1).
So the numbers to try each of are + 1, 2, 4, 5, 10 or 20 divided by 2 or 1. Thus, the available
numbers for rational solutions are + ½, 1, 2, 5/2, 5, 10 , and 20. (Hint: Don’t use the largest
number because it will almost never be a factor, and recall both Descartes’ rule of signs and the
intermediate value theorem to choose the next number to try after tries which do not locate
zeroes.) Synthetic division as described only works for (1x = a).
For this example, use x = 5/2. This gives:
| 2 - 5 + 8 - 20
+ 5/2 | 2 + 0 + 8 | 0
As above, the second “|” has been inserted to indicate that
there is no remainder and that the shorter polynomial in this color can now be used for further
factoring.
This result indicates that 2x3 – 5x2 + 8x – 20 = (x - 5/2) (2x2 + 8).
The shorter polynomial from this process has a common multiple of 2, so this becomes
2x3 – 5x2 + 8x – 20 = (x - 5/2) (2) (x2 + 4) = (2x – 5) (x2 + 4).
Synthetic Division – J. Tobin
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The quadratic factor obtained from this process, x2 + 4, has imaginary roots of + 2i and - 2i.
This is not problematic because synthetic division can be done for x = 2i with the dividend of
1x2 + 0x + 4. Remember that for any missing powers of x, zeroes must be used a placeholders
in the columns for the dividend polynomial. Thus, proceeding with synthetic division, we
have:
| 1 + 0 + 4
- 2i | 1 - 2i | 0 x2 + 4 = (x + 2i) (1x – 2i)
+ 2i | 1
Confirms that x2 + 4 = (x + 2i) (1x – 2i)
| 0
This shows that x2 + 4 = (x + 2i) (x-2i) and that the synthetic division process works for
roots which are integers, rational numbers, or imaginary numbers.
Actually, the synthetic division process works for all numbers.
As another example showing the use of zeroes as placeholders in the respective columns for
missing powers of x, consider factoring x4 – 81. The polynomial can be factored as
x4 – 81 = (x2 – 9)( x2 +9) = (x + 3 (x – 3)( x2 +9)
Synthetic division of this will again illustrate the use of zeroes for missing powers of x.
| 1
+3 | 1
0
3
0
9
0 - 81
27 | 0 The 1st factor is (x – 3), so x4 –81 = (x – 3) (x3+ 3x2 + 9x +27)
-3 | 1
0
9
|0
- 3i | 1
+ 3i | 1
The 2nd factor is (x + 3), so x4 –81 = (x – 3)(x + 3)(x2 + 9)
4
The third factor is (x + 3i), so x –81 = (x – 3)(x + 3) (x + 3i) (1x – 3i)
- 3 |0
The fourth factor is (x - 3i), so x4 – 81 = (x – 3) (x + 3) (x + 3i) (x - 3i),
|0
and the zeroes of the graph of y = x4 – 81 are at x = + 3. and x = - 3.
The two imaginary roots show a complex conjugate solution pair (as they always occur in
conjugate pairs when factoring polynomials), indicating possible local minima or maxima in
the graph of y = x4 – 81 with a region resembling a parabola locally but not touching the xaxis. (For this function, only one minimum exists at x = 0, but it is not found by this process).
One more example shows how to factor 9x4 – 85x2 +36. Thus
| 9
+3 | 9
0 - 85
27
-4
0
- 12
- 3 | 9
0
-4
| 0
+ 2/3 | 9
6
| 0
+ 36
| 0 The 1st factor is (x – 3)
The 2nd factor is (x + 3)
The 3rd factor is (x – 2/3) and
9x4 – 85x2 +36 = (x – 3) (x + 3) (x – 2/3) (9x +6)
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Any important point to recognize is that the remaining coefficients in the division process (in
the latest shortened factor polynomial of (9x + 6) have a common multiple of three. Therefore,
the fourth factor in this example is 3 which gives
9x4 – 85x2 +36 = (x – 3) (x + 3) (x – 2/3) (3) (3x +2) or
9x4 – 85x2 +36 = (x – 3) (x + 3) (3x – 2) (3x +2)
for which the roots are x = +3, -3, + 2/3, and –2/3.
Repeating the synthetic division and continuing to seek the roots without factoring out the 3
gives:
| 9
+3 | 9
0 - 85
27
-4
- 3 | 9
0
-4
+ 2/3 | 9
6
| 0
- 2/3 | 9 | 0
0 + 36
- 12 | 0 One root is x = +3
| 0
A second root is x = -3
A third root is x = + 2/3
The fourth and final root is x = - 2/3
So, the zeroes of the graph of y = 9x4 – 85x2 +36 are at x = +3, -3, + 2/3, and - 2/3.
This example shows the importance of factoring out common multiples when they appear on a
line for division. Otherwise, the process would seem to give factors which will not multiply
together to give the actual dividend. But the root will always be accurately identified as long
as no errors are committed in the process.
Conclusion
This method of synthetic division:
o expands upon the concept of decreasing the writing for division of polynomials by a
binomial and, for cases where the factors of the polynomial can be found easily,
requires only a minimum of writing to see the factors;
o reinforces the concept of being constructively lazy by enabling the person using the
process to try to factor a shorter (and therefore simpler) polynomial each time a factor
is found;
o enables the person using the process to find repeated roots which cannot be found by
conventional synthetic division; and
o enables the person using the process to confirm complex conjugate pairs when the
factors contain the form (ax2 + bx + c) such that b2 – 4ac < 0.
But the main value for algebra students, once they have mastered the process, is the ease
and speed with which factors (and coincidentally roots) can be found for polynomials
with rational roots, real or complex.
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