Tile Patterns of Regular Polygons

A regular polygon is a shape with all congruent sides and all congruent angles.  For example, an equilateral triangle is a regular polygon because it has 3 congruent sides and 3 congruent angles, and a square is a regular polygon because it has 4 congruent sides and 4 congruent angles.

Regular Tessellations

Some regular polygons can be used to make a tile pattern (or a tessellation), and others cannot.  In order for a regular polygon to be tileable, its interior angle must divide evenly into 360°, and the interior angle of a regular polygon with n sides is θ = ¹/_n(n – 2)180°.  A regular triangle, which has n = 3 sides, has an interior angle θ = ¹/₃(3 – 2)180° = 60°, which divides evenly into 360°, which means it can be used to make a tile pattern.  A square, which has n = 4 sides, has an interior angle θ = ¹/₄(4 – 2)180° = 90°, which divides evenly into 360°, which means it also can be used to make a tile pattern.  However, a regular pentagon, which has n = 5 sides, has an interior angle θ = ¹/₅(5 – 2)180° = 108°, which does not divide evenly into 360°, which means it cannot be used to make a tile pattern.  A regular hexagon, which has n = 6 sides, has an interior angle θ = ¹/₆(6 – 2)180° = 120°, which divides evenly into 360°, which means it can be used to make a tile pattern. 

Interior Angles of Regular Polygons

Since the next biggest factor of 360° after 120° is 180°, which is too big to be an interior angle of a regular polygon, there are no other regular polygons that can be used to make a tile pattern.  Therefore, the equilateral triangle, the square, and the hexagon are the only regular polygons that can be used to make a tile pattern.

Regular Polygons That Can Be Tiled

Semi-Regular Tessellations

If the requirement that all the regular polygons in the tile pattern must have the same number of sides is removed, then there are a few more tile patterns that can be obtained called semi-regular tessellations.  Semi-regular tessellations (or Archimedean tessellations) are tile patterns that contain two or more regular polygons with the same order around each vertex.  For example, at my house growing up we had a brick driveway consisting of octagons and squares.  This was a semi-regular tessellation because it was a tile pattern containing regular polygons in which each vertex was surrounded by 2 regular octagons and 1 square.

Brick Driveway Pattern Consisting of

Octagons and Squares

One way to find all the possible semi-regular tessellations is to examine which combinations of regular polygons fit snugly around a single vertex.  In order for this to happen, each interior angle, which for a regular polygon with n sides is θ = ¹/_n(n – 2)180°, must add up to 360°.  If there are 3 regular polygons around one vertex with sides n₁, n₂, and n₃; then ¹/_n1(n₁ – 2)180° + ¹/_n2(n₂ – 2)180° + ¹/_n3(n₃ – 2)180° = 360°.  After distributing, the equation becomes 180° – ^360°/_n1 + 180° – ^360°/_n2 + 180° – ^360°/_n3 = 360°; after combining like terms, the equation becomes -^360°/_n1 – ^360°/_n2 – ^360°/_n3 = -180°; and after dividing by -360°, the equation simplifies to

¹/_n1 + ¹/_n2 + ¹/_n3 = ¹/₂

This holds true for the known three hexagons example (¹/₆ + ¹/₆ + ¹/₆ = ¹/₂ or (6, 6, 6)), and for the brick driveway example consisting of 1 square and 2 octagons (¹/₄ + ¹/₈ + ¹/₈ = ¹/₂ or (4, 8, 8)).  Using trial and error and the fact that n is an integer and n ≥ 3, the only possible solutions are (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12), (4, 5, 20), (4, 6, 12), (4, 8, 8), (5, 5, 10), and (6, 6, 6).

All Combinations of 3 Polygons Fitting

Snugly around a Single Vertex

However, even though all of these solutions represent all the possible ways for 3 regular polygons to fit snugly around a single vertex, not all of them can be used to make a tessellation.  One final requirement for a tessellation is that the number of different sides must divide evenly into the third side.  For example, the solution (5, 5, 10), which represents a pentagon, pentagon, and decagon, fits snugly around a single vertex but cannot be tessellated.  The first pentagon, with 5 sides, would need to alternately share sides with the other pentagon and decagon, 2 different shapes, but 2 does not divide evenly into 5, and so these 3 shapes cannot be tessellated.  (See here for more details.)

Attempt at a (5, 5, 10) Tessellation Results in Some Gaps

This requirement eliminates all the above solutions except for (3, 12, 12), (4, 6, 12), (4, 8, 8), and (6, 6, 6).

(3, 12, 12)

(4, 6, 12)

(4, 8, 8)

The same logic can be applied to find semi-regular tessellations with 4 or more regular polygons.  In general, if there are k regular polygons around one vertex with sides with sides n₁, n₂, … n_k; then ∑ ^k _p=1 ¹/_np(n_p – 2)180° = 360°.  After distributing, the equation becomes ∑ ^k _p=1180° – ∑ ^k _p=1^360°/_np = 360°, which is 180°k – ∑ ^k _p=1^360°/_np = 360°; after some rearranging, the equation becomes -∑ ^k _p=1^360°/_np = -180k + 360°; and after dividing by -360°, the equation simplifies to ∑ ^k _p=1¹/_np = ¹/₂(k – 2) or

¹/_n1 + ¹/_n2 + … +  ¹/_nk= ¹/₂(k – 2)

For k = 3, ¹/_n1 + ¹/_n2 + ¹/_n3 = ¹/₂ (as proved above); for k = 4,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 = 1

for k = 5,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 + ¹/_n5 = ³/₂

and for k = 6,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 + ¹/_n5 + ¹/_n6 = 2

Since n ≥ 3, there are no integer solutions when k ≥ 7 (geometrically, 7 or more equilateral triangles cannot fit around a single vertex). 

Once again using trial and error and the fact that n is an integer and n ≥ 3, the only possible solutions for k ≥ 4 are (3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 4, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), and (3, 3, 3, 3, 3, 3).  For some of these solutions, a new ordering produces a new unique pattern, and so (3, 4, 3, 12), (3, 6, 3, 6), (3, 4, 6, 4), and (3, 3, 4, 3, 4) can be added to the list. 

All Combinations of More Than 3 Polygons

Fitting Snugly around a Single Vertex

However, the requirement for semi-regular tessellations that there is the same order around each vertex eliminates (3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 4, 6), and (3, 4, 3, 12); leaving only the solutions (3, 4, 6, 4), (3, 6, 3, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), (3, 3, 4, 3, 4), and (3, 3, 3, 3, 3, 3).   

(3, 4, 6, 4)

(3, 6, 3, 6)

(3, 3, 3, 3, 6)

(3, 3, 3, 4, 4)

(3, 3, 4, 3, 4)

Since (6, 6, 6), (4, 4, 4, 4), and (3, 3, 3, 3, 3, 3) are regular tessellations, that leaves 8 possible semi-regular tessellation patterns: 1 triangle and 2 dodecagons (3, 12, 12); 1 square, 1 hexagon, and 1 dodecagon (4, 6, 12); 1 square and 2 octagons (4, 8, 8); 1 triangle, 2 squares, and 1 hexagon (3, 4, 6, 4); 2 triangles and 2 hexagons (3, 6, 3, 6); 4 triangles and 1 hexagon (3, 3, 3, 3, 6); and 3 triangles and 2 squares (3, 3, 3, 4, 4) and (3, 3, 4, 3, 4).  (See here for more details.)

The Eight Semi-Regular Tessellations

Quasi-Regular Tessellations

If the semi-regular tessellation requirement that there is the same order around each vertex is removed, then there are countless other variations of tile patterns that can be obtained called quasi-regular tessellations.  Many quasi-regular tessellations can be formed by modifying an existing semi-regular tessellation.  For example, since 1 regular hexagon can be formed from 6 equilateral triangles, the semi-regular tessellation of (6, 6, 6) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 6, 6) and others are (3, 3, 3, 3, 3, 3); and the semi-regular tessellation of (4, 6, 12) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 4, 12) and others are (3, 3, 3, 3, 3, 3). 

A regular hexagon formed by

6 equilateral triangles

A regular dodecagon formed by

1 regular hexagon, 6 squares,

and 6 equilateral triangles

In addition, 1 regular dodecagon can be formed from 1 regular hexagon, 6 squares, and 6 equilateral triangles, so the semi-regular tessellation of (3, 12, 12) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 3, 4, 4) and some are (3, 4, 6, 4).  There are countless other variations of quasi-regular tessellations, but all of them use the same vertex combinations found in regular tessellations and semi-regular tessellations.

Some (of Many) Quasi-Regular Tessellations

Conclusion

There are many ways to tile an area with shapes all consisting of segments of the same length.  Three of those ways are by regular tessellations, in which all the shapes are the same.  These include 3 hexagons (6, 6, 6), 4 squares (4, 4, 4, 4), and 6 triangles (3, 3, 3, 3, 3, 3).  Eight other ways are by semi-regular tessellations, in which some of the shapes are different, but all of the shapes follow the same order around each vertex used in the tessellation.  These include 1 triangle and 2 dodecagons (3, 12, 12); 1 square, 1 hexagon, and 1 dodecagon (4, 6, 12); 1 square and 2 octagons (4, 8, 8); 1 triangle, 2 squares, and 1 hexagon (3, 4, 6, 4); 2 triangles and 2 hexagons (3, 6, 3, 6); 4 triangles and 1 hexagon (3, 3, 3, 3, 6); and 3 triangles and 2 squares (3, 3, 3, 4, 4) and (3, 3, 4, 3, 4).  Finally, there are several more ways to tile an area with shapes all consisting of segments of the same length using quasi-regular tessellations, in which some of the shapes are different and in which some of the vertices have different orders of shapes around them.  

Hexagonal Honeycombs

When bees build honeycombs, they do so in a pattern of tiled hexagons.  They create the borders of the hexagon with wax, leaving hexagonal holes that they can use to store honey, pollen, or eggs.

But have you ever wondered why bees use hexagons?  Why not use a simpler shape, like squares, triangles, or circles?

The quick answer is that the hexagon is the single tileable polygon that has the least perimeter for a given area.  This is important because bees have to spend a lot of energy to make the wax for the borders of each honeycomb cell.  Less perimeter means less wax, and less wax means less spent energy.

Let’s examine the math behind the claim that the hexagon is the single tileable polygon that has the least perimeter for a given area.  First of all, we must show that there are only three regular polygons that can be tiled: the triangle, the square, and the hexagon.  Secondly, we must show that out of those three polygons, the hexagon is the one with the least perimeter for a fixed area.

Tileable Regular Polygons

First of all, we need to show that there are only three regular polygons that can be tiled.  In all regular polygons with n sides, the interior angle θ is θ = ^{(n – 2)180°}/_n.  So for a triangle (the polygon with the least amount of sides possible), n = 3 and θ = ^{(3 – 2)180°}/₃ = 60°; for a square, n = 4 and θ = ^{(4 – 2)180°}/₄ = 90°;  for a pentagon, n = 5 and θ = ^{(5 – 2)180°}/₅ = 108°; for a hexagon, n = 6 and θ = ^{(6 – 2)180°}/₆ = 120°; for a heptagon, n = 7 and θ = ^{(7 – 2)180°}/₇ = 128⁴/₇°; for a octagon, n = 8 and θ = ^{(8 – 2)180°}/₈ = 135°; and so on.  We can see that the values of the interior angles of a regular polygon are greater than or equal to 60° (for the triangle) but less than 180° (which represents a straight line, because otherwise it won’t close to make a polygon) and so 60° ≤ θ < 180°.

However, in order for a regular polygon to be tileable, its interior angle must also divide evenly into 360°.  The largest factor of 360° is 360° ÷ 1 = 360°, which is too big to be an interior angle of a regular polygon.  The second largest factor of 360° is 360° ÷ 2 = 180°, which is also too big.  However, the third largest factor of 360° is 360° ÷ 3 = 120°, which is the interior angle of a regular hexagon; the fourth largest factor is 360° is 360° ÷ 4 = 90°, which is the interior angle of a square; the fifth largest factor of 360° is 360° ÷ 5 = 72°, which is not the interior angle of any regular polygon; and the sixth largest factor of 360° is 360° ÷ 6 = 60°, which is the interior angle of an equilateral triangle.  After this, the factors of 360° are less than 60°, which are too small to be an interior angle of any regular polygon.  So the triangle, square, and hexagon are the only regular polygons that can be tiled.

Least Perimeter

Secondly, we must show that out of the three tileable regular polygons of the triangle, square, and hexagon, the hexagon is the one with the least perimeter for a fixed area.  Consider a regular polygon of n sides inscribed in a circle with a radius of r.   The polygon can be divided into n triangular slices, and each of those slices can be bisected at each central angle, making 2n symmetrical right triangles with an angle of ^π/_n and a hypotenuse of r.  The two legs of each right triangle are then r sin (^π/_n) and r cos (^π/_n), which makes the area of one of those right triangles ½r²sin(^π/_n)cos(^π/_n), and making the area of the whole regular polygon 2n times this or A = nr²sin(^π/_n)cos (^π/_n).  Also, since the opposite leg is r sin (^π/_n), 2 times this would give the length of one side of the whole regular polygon, and n times that would give the perimeter of the whole regular polygon or P = 2nr sin (^π/_n).

Solving the area equation for r gives us

and substituting this into the perimeter equation gives us

which simplifies to P = 2√A·√n·√tan(^π/_n).

So for a triangle, n = 3 and P = 2√A·√3·√tan(^π/₃) = 2√(3√3)√A ≈ 4.559√A.  For a square, n = 4 and P = 2√A·√4·√tan(^π/₄) = 4√A.  And for a hexagon, n = 6 and P = 2√A·√6·√tan(^π/₆) = 2√(2√3)√A ≈ 3.722√A.

Shape	Perimeter
triangle	4.559·√A
square	4.000·√A
hexagon	3.722·√A

Therefore, the hexagon has the least perimeter for a fixed area.

Wax Width

We have shown that the hexagon is the single tileable polygon that has the least perimeter for any given area.  But in a honeycomb grid, the perimeter of each cell is actually a wax border with its own width that is shared by adjacent cells, and this may affect our assertion that the hexagon is the most efficient shape.  So to be thorough, we need to consider the wax to honey ratio for each tiled shape.  We will consider the potential candidates of the triangle, square, hexagon, and circle.

Triangular Honeycombs

Let’s say there is a type of bee called the Triangle Bee that builds its honeycombs with equilateral triangles.  The Triangle Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the triangular pattern, the wax walls will meet in hexagon shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual triangular tile, as colored below.

The wax walls for one tile consist of 3 sixths of a hexagon and 3 rectangles, and so its area W would be W = 3·¹/₆·^3√3/₂w² + 3·½sw or W = ^3√3/₄w² + ³/₂sw.  Since the area of an equilateral triangle is A = ^√3/₄s², solving for s would give us s = ^{2√(3√3)√A}/₃, and substituting back into W would give us W = ^3√3/₄w² + ³/₂(^{2√(3√3)√A}/₃)w or W = ^3√3/₄w² + √(3√3)√A·w.  The wax to honey ratio for the Triangular Bee is then (^3√3/₄w² + √(3√3)√A·w) / A ≈ (1.299w² + 2.280√A·w) / A.

Square Honeycombs

Now let’s say there is a type of bee called the Square Bee that builds its honeycombs with squares.  The Square Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the square pattern, the wax walls will meet in square shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual square tile, as colored below.

The wax walls for one tile consist of 4 quarters of a square and 4 rectangles, and so its area W would be W = 4·¼·w² + 4·½sw or W = w² + 2sw.  Since the area of a square is A = s², solving for s would give us s = √A, and substituting back into W would give us W = w² + 2√A·w.  The wax to honey ratio for the Square Bee is then (w² + 2√A·w) / A.

Hexagonal Honeycombs

Now let’s say there is a type of bee called the Hexagonal Bee that builds its honeycombs with hexagons (like regular bees).  The Hexagonal Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the hexagonal pattern, the wax walls will meet in triangular shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual hexagonal tile, as colored below.

The wax walls for one tile consist of 6 thirds of an equilateral triangle and 6 rectangles, and so its area W would be W = 6·¹/₃·^√3/₄w² + 6·½sw or W = ^√3/₂w² + 3sw.  Since the area of a regular hexagon is A = ^3√3/₂s², solving for s would give us s = ^{√(2√3)√A}/₃, and substituting back into W would give us W = ^√3/₂w² + 3(^{√(2√3)√A}/₃)w or W = ^√3/₂w² + √(2√3)√A·w.  The wax to honey ratio for the Triangular Bee is then (^√3/₂w² + √(2√3)√A·w) / A ≈ (0.866w² + 1.861√A·w) / A.

Circular Honeycombs

Finally, let’s say there is a type of bee called the Circular Bee that builds its honeycombs with circles.  The Circular Bees make wax walls that have at least a width of w that will separate each circular honey cell with an area of A and a radius of r.  The wax to honey ratio for the whole honeycomb would then be the same as one individual hexagonal tile, as colored below.

The area of the wax walls for one tile would be equivalent to the whole hexagon minus the circle, and so its area would be W = 2√3(½w + r)² – A.  Since the area of a circle is A = πr², solving for r would give us r = ^√π√A/_π, and substituting back into W would give us W = 2√3(½w + ^√π√A/_π)² – A = 2√3(¼w² + ^√π√A/_πw + ^A/_π) – A = ^√3/₂w² + ^{2√3√π√A}/_πw + (^2√3/_π – 1)A.  The wax to honey ratio for the Circular Bee is then (^√3/₂w² + ^{2√3√π√A}/_πw + (^2√3/_π – 1)A) / A ≈ (0.866w² + 1.954√A·w + 0.103A) / A.

Summary

The wax to honey ratio with respect to wax width and area are summarized for each shape below:

Shape	Wax to Honey Ratio
triangle	(1.299w2 + 2.280√A·w + 0.000A) / A
square	(1.000w2 + 2.000√A·w + 0.000A) / A
circle	(0.866w2 + 1.954√A·w + 0.103A) / A
hexagon	(0.866w2 + 1.861√A·w + 0.000A) / A

Since both w and A must be positive, the shape with the smallest wax to honey ratio coefficients would have to have the least wax to honey ratio, which is the hexagon.  Therefore, the hexagon is the most efficient shape to use for building a honeycomb.

Real honeybees build honeycomb cells that are an average of 4.85 mm wide (http://www.bushfarms.com/beesnaturalcell.htm), which gives a honey area of 81.484 mm², and with wax walls that have an average thickness of 0.5 mm (http://keepingbee.org/bee-honeycomb/).  Using these values, we can use the formulas above to calculate the numerical wax to honey ratio for each shape:

Shape	Ratio
circle	21.357%
triangle	13.025%
square	11.385%
hexagon	10.575%

Which means that the average honeycomb (made up of hexagons) has a 10.575% wax to honey ratio.

Conclusion

There are generally two opposing trains of thought concerning the fact that all bees happen to use the most efficient shape to build their honeycombs.  Evolutionists attribute the bees’ efficiency to natural selection.  They would say that at some point there may have been bees that tried to make square or triangular honeycombs, but they were not as efficient as the bees that made hexagonal honeycombs and were eliminated by the rules of the survival of the fittest.  On the other hand, creationists attribute the bees’ efficiency to Intelligent Design.  They would say that the instinct to build hexagonal honeycombs was put there by a God who not only created but also upholds all the geometrical laws of the universe.

Unfortunately, there are a few problems with using the theory of natural selection to explain why bees make their honeycombs in a hexagonal pattern rather than some other pattern.  First of all, although we have mathematically proved that the hexagon is the most efficient shape for a honeycomb of a fixed area, the square is not that far behind, and the difference of efficiency would be too small for natural selection to take place.  In fact, given that a real honeycomb has hexagon widths that vary from 4.6 mm to 5.1 mm (http://www.bushfarms.com/beesnaturalcell.htm), which would make honey cell areas vary from 73.300 mm² and 90.101 mm², a high end honey area made up of squares is actually more efficient than a low end honey area made up of hexagons! (A square honeycomb with w = 0.5 mm and A = 90.101 mm² would have a wax to honey ratio of 10.812%, whereas a hexagonal honeycomb with w = 0.5 mm and A = 73.300 mm² would have a wax to honey ratio of 11.165%.  Note that this does not contradict our above assertion because the areas are not the same.)  According to evolutionary natural selection, a mutation of bees that make large square honey cells should eliminate bees that make small hexagonal honey cells, but this of course has not happened.  Which brings us to the second problem of using the theory of natural selection to explain why bees make hexagonal honeycombs: out of the millions of beekeepers worldwide, and over three thousand years of beekeeping, there is not one recorded instance of a mutated colony of bees trying to make a honeycomb with something other than a hexagonal pattern.  The theory of evolution relies on these mutations to take place, both now and in the past, but there is just no evidence for this in bees.

If the reason bees make their honeycombs in a hexagonal pattern rather than some other pattern is not because of natural selection, then it must be because of some innate instinct.  But where did that instinct come from?  The only plausible explanation is that it came from an Intelligent Designer.  Bees build their honeycombs in the efficient hexagonal pattern because of an instinct that was put there by the same God who created the geometrical laws of the universe.

Synthetic Division with Higher Powers

Most high school algebra textbooks cover two methods for dividing polynomials: long division and synthetic division.  The common teaching pattern in these textbooks is to show long division as the default method that works for all cases, and then to show synthetic division as a handy shortcut that only works when the divisor is a first degree polynomial in the form of x – c. 

However, synthetic division can be adjusted to be used for divisors of higher powers, which for some reason is not taught in high school algebra textbooks.  To show how this can be done, we must first examine how synthetic division of first degree divisors relates to long division, and use the same logic to apply the method to higher degree divisors.

First Degree Divisor

Let’s say we were divide x – 2 into 3x⁴ – 4x³ – 8x² + 11x + 1.  The long division method would be as follows:

	3x3	+ 2x2	– 4x	+ 3
x – 2	3x4	– 4x3	– 8x2	+ 11x	+ 1
	3x4	– 6x3
		2x3	– 8x2
		2x3	– 4x2
			– 4x2	+ 11x
			– 4x2	+ 8x
				3x	+ 1
				3x	– 6
					7

If we were to re-write the problem without the repetitive parts (the first terms subtracting to zero, the dropped down terms, and the x’s) we would have the following:

	3	2	–4	3
– 2	3	–4	–8	11	1
		–6
		2
			–4
			–4
				8
				3
					–6
					7

We can further re-write the problem in three lines to save space:

	3	2	–4	3
– 2	3	–4	–8	11	1
		–6	–4	8	–6
		2	–4	3	7

Once again we can eliminate repetition by combining the first and last rows:

– 2	3	–4	–8	11	1
		–6	–4	8	–6
	3	2	–4	3	7

Finally, we can multiply the top left number and every second row number by -1, giving us the following:

2	3	–4	–8	11	1
		6	4	–8	6
	3	2	–4	3	7

We should now recognize that each second row number can be obtained by multiplying the number in the top left by the number in the previous column’s last row, for example, 2 x 3 = 6, 2 x 2 = 4, 2 x -4 = -8, and 2 x 3 = 6.  Also, each number in the last row can be obtained by adding the first and second rows together, for example, 3 + 0 = 3, -4 + 6 = 2, -8 + 4 = -4, 11 + -8 = 3, and 1 + 6 = 7.  We have now transformed our long division problem into a synthetic division problem, a method in which you alternate between adding columns and multiplying by the c value in the divisor x – c.

For example, if we were to use synthetic division to divide x – 2 into x⁴ – x³ – 19x² + 49x – 30, we would start by writing the divisor number opposite 2 in the top left corner and the dividend coefficients 1, -1, -19, 49, and -30 along the top:

2	1	–1	–19	49	-30

Bring down the 1 in the first column:

2	1	–1	–19	49	-30
	1

Multiply 2 x 1 = 2 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2
	1

Add -1 + 2 = 1 in the second column:

2	1	–1	–19	49	-30
		2
	1	1

Multiply 2 x 1 = 2 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2
	1	1

Add -19 + 2 = -17 in the third column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15	0

Multiply 2 x -17 = -34 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2	–34
	1	1	–17

Add 49 + -34 = 15 in the fourth column:

2	1	–1	–19	49	-30
		2	2	–34
	1	1	–17	15

Multiply 2 x 15 = 30 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15

Add -30 + 30 = 0 in the last column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15	0

The numbers in the last row are the coefficients of the quotient with a degree that is one less than the dividend, which in this case is 1x³ + 1x² – 17x + 15 + ⁰/_{(x – 2)} or x³ + x² – 17x + 15.

Second Degree Divisor

Now we will use the same logic to transform a long division problem into a synthetic division problem when the divisor has a higher degree, for example,  x² – 2x + 3 into 3x⁴ – 4x³ – 8x² + 11x + 1.  The long division method would be as follows:

	3x2	+ 2x	– 13
x2 – 2x + 3	3x4	– 4x3	– 8x2	+ 11x	+ 1
	3x4	– 6x3	+ 9x2
		2x3	– 17x2	+ 11x
		2x3	– 4x2	+ 6x
			– 13x2	+ 5x	+ 1
			– 13x2	+ 26x	– 39
				– 21x	+ 40

If we were to re-write the problem without the repetitive parts (the first terms subtracting to zero, the dropped down terms, the intermediate subtraction steps, and the x’s) we would have the following:

	3	2	– 13
– 2,  3	3	– 4	– 8	11	1
		– 6	9
		2
			– 4	6
			– 13
				26	– 39
				– 21	40

We can further re-write the problem in four lines to save space:

	3	2	– 13
– 2,  3	3	– 4	– 8	11	1
			9	6	– 39
		– 6	– 4	26
		2	– 13	– 21	40

Once again we can eliminate repetition by combining the first and last rows:

– 2,  3	3	– 4	– 8	11	1
			9	6	– 39
		– 6	– 4	26
	3	2	– 13	– 21	40

Finally, we can multiply the top left number and every second and third row number by -1, giving us the following:

2,  – 3	3	– 4	– 8	11	1
			– 9	– 6	39
		6	4	– 26
	3	2	– 13	– 21	40

We should now recognize that each second and third row numbers can be obtained by multiplying the two numbers in the top left by the number in the previous column’s last row, for example, (2, -3) x 3 = (6, -9), (2, -3) x 2 = (4, -6), and (2, -3) x -13 = (-26, 39).  Also, each number in the last row can be obtained by adding the first, second, and third rows together, for example, 3 + 0 + 0 = 3, -4 + + 0 + 6 = 2, -8 + -9 + 4 = -13, 11 + -6 + -26 = -21, and 1 + 39 + 0 = 40.  We have now transformed our long division problem into a synthetic division problem, a method in which you alternate between adding columns and multiplying by the (b, c) value in the divisor x² – bx – c.

For example, if we were to use synthetic division to divide x² – 3x + 2 into x⁴ – x³ – 19x² + 49x – 30, we would start by writing the divisor number opposites 3, -2 in the top left corner and the dividend coefficients 1, -1, -19, 49, and -30 along the top:

3,  – 2	1	– 1	– 19	49	– 30

Bring down the 1 in the first column:

3,  – 2	1	– 1	– 19	49	– 30
	1

Multiply (3, -2) x 1 = (3, -2) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2
		3
	1

Add -1 + 3 = 2 in the second column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2
		3
	1	2

Multiply (3, -2) x 2 = (6, -4) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4
		3	6
	1	2

Add -19 + -2 + 6 = -15 in the third column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4
		3	6
	1	2	– 15

Multiply (3, -2) x -15 = (-45, 30) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15

Add 49 + -4 + -45 = 0 in the fourth column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15	0

Finally, add -30 + 30 = 0 in the last column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15	0	0

The numbers in the last row are the coefficients of the quotient with a degree that is two less than the dividend, which in this case is 1x² + 2x – 15 + ^{(0x + 0)}/_{(x2 – 3x + 2)} or x² + 2x – 15.

Conclusion

Therefore, despite the caution of many algebra textbooks, synthetic division can be adjusted and used for divisors of higher powers.  This article showed synthetic division only for divisors of degree 1 and 2, but the same pattern can be extended to divisors of higher degrees.