The Imaginary Part of a Parabola

Parabolas are commonly taught in most high school algebra classes. All parabolas are curves defined by a quadratic equation and are symmetrical, open in one direction, and have a vertex.  Generally, a parabola defined by a quadratic equation in standard form y = ax² + bx + c has a line of symmetry at x = ^-b/_2a, opens upward if a is positive (but downward if a is negative), and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a).  For example, a parabola defined by the quadratic equation y = x² – 6x + 13 have variables a = 1, b = -6, and c = 13, which means it has a line of symmetry at x = ^-b/_2a = ^--6/_2·1 = 3, opens upward because a is positive, and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a) = (^--6/_2·1, ^{-(-6)^2 + 4·1·13}/_4·1) = (3, 4). 

Some other points that are on the parabolic curve can be found by using the given equation.  For example, in the equation y = x² – 6x + 13, when x = 0, y = 0² – 6·0 + 13 = 13, so (0, 13) is on the curve, and when x = 1, y = 1² – 6·1 + 13 = 8, so (1, 8) is on the curve.  Similar calculations can be made to show that (2, 5), (3, 4), (4, 5), (5, 8), (6, 13), (and so on) are also on this parabolic curve.

However, we can also find imaginary points on any parabolic curve as well.    In the equation y = x² – 6x + 13, when y = 0, then 0 = x² – 6x + 13, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13, yields x = ^{–-6 ± √((-6)^2 – 4·1·13)}/_2·1 = ^{6 ± √(36 – 52)}/₂ = ^{6 ± √-16}/₂ = ^{6 ± 4i}/₂ = 3 ± 2i, so (3 + 2i, 0) and (3 – 2i, 0) should also be on the curve.  Similarly, when y = 3, then 3 = x² – 6x + 13 or 0 = x² – 6x + 10, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13 – 3 = 10, yields x = ^{–-6 ± √((-6)^2 – 4·1·(13 – 3))}/_2·1 = ^{6 ± √(36 – 40)}/₂ = ^{6 ± √-4}/₂ = ^{6 ± 2i}/₂ = 3 ± i, so (3 + i, 3) and (3 – i, 3) should also be on the curve.  Similar calculations can be made to show that (3 + 3i, -5), (3 – 3i, -5), (3 + 4i, -12), (3 – 4i, -12), (and so on) are also on the parabolic curve.

But how can you graph these imaginary points on a coordinate graph?  In an Argand graph, the imaginary part is graphed on a separate axes.  But since both the x-axis and y-axis are already being used, we will have to add a third dimensional z-axis to represent the imaginary part.  This results in the following graph for y = x² – 6x + 13:

It appears that the imaginary part is the same size as the real part, but reflected horizontally at the vertex and then rotated 90° into the third dimension.

Depicting three-dimensional graphs on a two-dimensional medium is difficult to do, even with the help of technology.  To make this easier to draw, let’s rotate the imaginary part back into two dimensions by letting the x-axis serve a double purpose of defining x-values and imaginary values.  This means that in our y = x² – 6x + 13 example, the coordinate (3 – i, 3) would transform to (3 – 1, 3) = (2, 3); the coordinate (3 + i, 3) would transform to (3 + 1, 3) = (4, 3); the coordinate (3 – 2i, 0) would transform to (3 – 2, 0) = (2, 0); the coordinate (3 + 2i, 0) would transform to (3 + 2, 0) = (5, 0); and so on; resulting in the following graph:

This transformation graphically changed the imaginary part to a real part.  We can denote this algebraically by multiplying square root part of the quadratic formula by i.  In other words, the imaginary part of any quadratic equation y = ax² + bx + c (or 0 = ax² + bx + c – y) can be represented by new x-values such that x = ^{–b ± i√(b^2 – 4a(c – y))}/_2a.  Solving for y results in another quadratic equation:

x = –b ± i√(b^2 – 4a(c – y))/2a	(x-values of transformation)
2ax = –b ± i√(b2 – 4a(c – y))	(multiply by 2a)
2ax + b = ±i√(b2 – 4a(c – y))	(add b)
(2ax + b)2 = -1(b2 – 4a(c – y))	(square both sides)
4a2x2 + 4abx + b2 = -b2 + 4ac – 4ay	(distribute)
4ay + 4a2x2 + 4abx + b2 = -b2 + 4ac	(add 4ay)
4ay + 4a2x2 + 4abx = -b2 + 4ac	(subtract b2)
4ay + 4a2x2 + 4abx = 4ac – b2	(rearrange b2)
4ay + 4a2x2 = -4abx + 4ac – 2b2	(subtract 4abx)
4ay = -4a2x2 – 4abx + 4ac – 2b2	(subtract 4a2x2)
y = -ax2 – bx + c – b^2/2a	(divide by 4a)

This proves that the imaginary part of a parabola is another parabola.  (A similar proof can be used to show that the imaginary part of a hyperbola is an ellipse, and that the imaginary part of an ellipse is a hyperbola.)  It also gives us a fast way to forcibly graph the imaginary part of a parabola on a graphing calculator (or some other technology) that would not normally do so.  For example, to graph the full graph of y = x² – 6x + 13, where a = 1, b = -6, and c = 13, we should also graph the imaginary part at the same time using the imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5.  Graphing y = x² – 6x + 13 (in bold) and y = -x² + 6x – 5 gives the full graph, both real and imaginary, of the parabola:

A great application for graphing the imaginary part of the parabola is to use it as a visual aid for finding the number and types of solutions to a quadratic equation, a common objective in most high school algebra classes.  To find the number and types of solutions to a quadratic equation, students are traditionally taught to calculate the discriminant, which is the part under the square root of the quadratic formula, namely b² – 4ac.  If the discriminant is negative, the square root will result in an imaginary number, and the quadratic formula will yield two imaginary solutions and no real solutions.  If the discriminant is equal to zero, the square root will also be equal to zero, and the quadratic formula will yield one real solution and no imaginary solutions.  Finally, if the discriminant is positive, the square root will also be positive, and the quadratic formula will yield two real solutions and no imaginary solutions.

Discriminant	Solutions
-	0 real, 2 imaginary
0	1 real, 0 imaginary
+	2 real, 0 imaginary

This approach is rather abstract, but the concept can now be reinforced by graphing both real and imaginary parts of the parabola, and examining which part intersects with the x-axis.  For example, let’s say we were asked to find the number and types of solutions to x² – 6x + 13 = 0.  The discriminant is b² – 4ac = (-6)² – 4·1·13 = -16, which is negative, so it will have 0 real solutions and 2 imaginary solutions.  Graphing y = x² – 6x + 13 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5) shows that the real part of the parabola does not cross the x-axis but the imaginary part of the parabola crosses the x-axis twice, visually reinforcing the result that there are 0 real solutions and 2 imaginary solutions. 

Using a different example, let’s say we were asked to find the number and types of solutions to x² – 6x + 9 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·9 = 0, so it will have 1 real solution and 0 imaginary solutions.  Graphing y = x² – 6x + 9 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 9 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 9) shows that the vertex crosses the x-axis exactly once.  Recall that the vertex is real, not imaginary, visually reinforcing the result that there is 1 real solution and 0 imaginary solutions. 

Lastly, let’s say we were asked to find the number and types of solutions to x² – 6x + 5 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·5 = 16, which is positive, so it will have 2 real solutions and 0 imaginary solutions.  Graphing y = x² – 6x + 5 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 5 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 13) shows that the real part of the parabola crosses the x-axis twice but the imaginary part of the parabola does not cross the x-axis, visually reinforcing the result that there are 2 real solutions and 0 imaginary solutions. 

x2 – 6x + 13 = 0 0 real solutions 2 imaginary solutions	x2 – 6x + 9 = 0 1 real solution 0 imaginary solutions	x2 – 6x + 5 = 0 2 real solutions 1 imaginary solution

In summary, a quadratic equation is actually comprised of two curves – one real and one imaginary.  The real curve is the traditional parabolic curve.  The imaginary curve is the same size and shares the same vertex as the real curve, but is reflected horizontally and rotated 90° into the third dimension.  Rotating the imaginary curve back into two dimensions helps make it easier to graph both real and imaginary parts of the quadratic equation, and also allows us to quickly determine the number and type of solutions by examining where it intersects with the x-axis.

The Garden Border Problem

The following problem is in the McDougal Littell Algebra 2 textbook and is a typical word problem for a section on solving quadratic equations: 

You have just planted a rectangular flower bed of red roses in a park near your home.  You want to plant a border of yellow roses around the flower bed as shown.  Since you bought the same number of red and yellow roses, the areas of the border and inner flower bed will be equal.  What should the width x of the border be? 

To solve this problem, you must first write an area equation.  The length of the whole garden is 12 feet plus the unknown widths of the left and right borders, which can be expressed as 2x + 12.  The width of the whole garden is 8 feet plus the unknown widths of the top and bottom borders, which can be expressed as 2x + 8.  The area of the whole garden is the area of the red rose garden (which is 8 feet times 12 feet or 96 feet squared) plus the area of the yellow rose garden (which is the same as the red rose garden or also 96 feet squared), which added together is 192 feet squared.  Since the area is length times width, the equation to solve is (2x + 12)(2x + 8) = 192.

The next step is to solve this area equation.  Multiplying (2x + 12)(2x + 8) gives us 4x² + 40x + 96, so 4x² + 40x + 96 = 192, and subtracting 192 to the left side gives us 4x² + 40x – 96 = 0, and dividing everything by the common factor 4 gives us x² + 10x – 24 = 0.  At this point, there are several methods for solving this quadratic (such as factoring, completing the square, quadratic equation, and graphing) but we will use the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 10, and c = -24.  Therefore, x = ^{-10 ± √(10^2 – 4·1·-24)} / _2·1 = ^{-10 ± √(100 + 96)} / ₂ = ^{-10 ± √196} / ₂ = ^{-10 ± 14} / ₂, which means x = -12 or x = 2.  Since x represents a geometrical dimension, it cannot be negative, and therefore the border width x must be 2 feet long.

You will notice that this answer conveniently comes out as an integer, and not as a decimal.  But what would happen if the problem started out with different dimensions for the inner garden?  Would the border width still be an integer?  Let us examine the same problem but with a starting inner garden of 8 feet by 8 feet, as pictured below:

  

This time both the length and the width of the whole garden can be expressed as 2x + 8, and the area of the whole garden is 2 times 8 feet by 8 feet, or 128 feet squared, giving us the equation (2x + 8)(2x + 8) = 128.  Multiplying (2x + 8)(2x + 8) gives us 4x² + 32x + 64 = 128, subtracting 128 to the left side gives us 4x² + 32x – 64 = 0, and dividing everything by the common factor 4 gives us x² + 8x – 16 = 0.  Using the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 8, and c = -16 gives us x = ^{-8 ± √(8^2 – 4·1·-16)} / _2·1 = ^{-8 ± √(64 + 64)} / ₂ = ^{-8 ± √128} / ₂ = ^{-8 ± 8√2} / ₂ = -4 ± 4√2.  Since x cannot be negative, the border width must be -4 + 4√2 feet long, which is not an integer answer.

Can we come up with different dimensions for the inner garden such that the border width solution comes out as an integer?  We already know one solution set is (8, 12, 2) from the original problem, and using the properties of proportions and dividing each number by two we can also include (4, 6, 1).  In fact, using the same argument we can include all solution sets in the form of (4k, 6k, k) where k is any positive integer.  To simplify things, we will say that (4, 6, 1) is a “garden border triple” that includes all solutions sets in the form (4k, 6k, k), so the garden border triple (4, 6, 1) includes (4, 6, 1), (8, 12, 2), (12, 18, 3), and so on (just like the Pythagorean triple (3, 4, 5) includes all solution sets in the form (3k, 4k, 5k)).

Are there other garden border triples other than (4, 6, 1)?  Just as there are different Pythagorean triple solutions to the formula a² + b² = c² ((3, 4, 5), (5, 12, 13), etc.), there are also different garden border triples.  And just as there are different Pythagorean triple generators (see here), there are different garden border triple generators.  To make one, we must generalize the garden border problem by calling the length of the inner garden b and the width of the inner garden h, as pictured below:

The length of the whole garden can then be expressed as 2x + b, the width of the whole garden can be expressed as 2x + h, and the area of the whole garden can be expressed as 2bh, giving us the equation (2x + b)(2x + h) = 2bh.  This time, however, we are going to solve this equation for b.  Multiplying (2x + b)(2x + h) gives us 4x² + 2bx + 2hx + bh = 2bh, subtracting bh on both sides gives us 4x² + 2bx + 2hx = bh, subtracting 2bx on both sides gives us 4x² + 2hx = bh – 2bx, factoring 2x from the left side and b from the right side gives us 2x(2x + h) = b(h – 2x), and dividing both sides by h – 2x gives us b = ^{2x(h + 2x)}/_{h – 2x}.

We can now use the formula b = ^{2x(h + 2x)}/_{h – 2x} to generate garden border triples.  If we let x = 1, then b = ^{2(h + 2)}/_{h – 2}.  Then if h = 3, b = ^{2(3 + 2)}/_{3 – 2} = 10, then the garden border triple is (3, 10, 1).  If h = 4, b = ^{2(4 + 2)}/_{4 – 2} = 6, then the garden border triple is (4, 6, 1) (which is a repeat of a triple we already knew).  If h = 5, b = ^{2(5 + 2)}/_{5 – 2} = ¹⁴/₃, then the garden border triple is (5, ¹⁴/₃, 1), and to eliminate the fraction we can multiply each number by 3 to get (15, 14, 3).  Continuing on in this fashion, we also arrive at (6, 4, 1) (a repeat), (7, ¹⁸/₅, 1) ≡ (35, 18, 5), (8, ¹⁰/₃, 1) ≡ (24, 10, 3), (9, ²²/₇, 1) ≡ (63, 22, 7), (10, 3, 1) (a repeat) and so on.  If we let x = 2, then b = ^{4(h + 4)}/_{h – 4}, and the resulting garden border triples are (5, 36, 2), (6, 20, 2) ≡ (3, 10, 1) (a repeat), (7, ⁴⁴/₃, 2) ≡ (21, 44, 6), (8, 12, 2) ≡ (4, 6, 1) (a repeat), (9, ⁵²/₅, 2) ≡ (45, 52, 10), (10, ²⁸/₃, 2) ≡ (15, 14, 3) (a repeat), and so on.  If we let x = 3, then b = ^{6(h + 6)}/_{h – 6}, and the resulting garden border triples are (7, 78, 3), (8, 42, 3), (9, 30, 3) ≡ (3, 10, 1) (a repeat), (10, 24, 3), and so on. 

The garden border problem is a common word problem given to students to practice solving quadratic equations.  Most solutions come out as a decimal answer, but there are a few scenarios in which the width, length, and border width are all integers, which we called garden border triples.  Generalizing the problem in terms of b and h and solving for b gave us a garden border triple generator b = ^{2x(h + 2x)}/_{h – 2x}.  The unique garden border triples we generated in this article were (3, 10, 1), (4, 6, 1), (15, 14, 3), (35, 18, 5), (24, 10, 3), (63, 22, 7), (5, 36, 2), (21, 44, 6), (45, 52, 10), (7, 78, 3), (8, 42, 3), and (10, 24, 3); but there are many, many more.

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.