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.

Solving a Quadratic with a Straightedge and Compass

In most high school algebra classes, students learn to solve a quadratic ax² + bx + c = 0 using a variety of different methods.  The most common methods taught are factoring, graphing, completing the square, or using the quadratic equation.  However, a method for solving a quadratic using a straightedge and compass is described near the end of the book Number: The Language of Science by Tobias Dantzig (first published in 1930).

To solve the quadratic equation ax² + bx + c = 0 using a straightedge and compass, let p = -^b/_a and q = ^c/_a.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(p, 0), Q(0, q), R(p, q), and S(p, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C, construct circle C with radius CU.  Circle C will then intersect the x-axis at the two solutions X₁ and X₂.

For example, let’s say you wanted to solve the quadratic x² – 6x + 5 = 0.  Then a = 1, b = -6, and c = 5, and p = -^b/_a = 6 and q = ^c/_a = 5.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(6, 0), Q(0, 5), R(6, 5), and S(6, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(3, 3), construct circle C with radius CU.  Circle C intersects the x-axis at x = 1 and x = 5, which are the two solutions to x² – 6x + 5 = 0. 

We can show that this method will always work by finding a Cartesian equation for circle C, setting y equal to zero (for the x-axis), and simplifying the answer to being the quadratic equation (x = ^{(–b ± √(b^2 – 4ac))}/_2a).  The center of circle C is the midpoint of U(0, 1) and R(p, q), which is (^p/₂, ^{(q + 1)}/₂), and the radius of circle C is half the distance between U(0, 1) and R(p, q), which is r = ½√(p² + (q – 1)²).  Therefore, the equation for circle C is (x – ^p/₂)² + (y – ^{(q + 1)}/₂)² = (½√(p² + (q – 1)²))² or when y = 0, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²).  Applying some algebra, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²) è (2x – p)² + (q + 1)² = p² + (q – 1)² è (2x – p)² = p² + (q – 1)² – (q + 1)² è (2x – p)² = p² + (q² – 2q + 1) – (q² + 2q + 1) è (2x – p)² = p² – 4q è 2x – p = ±√(p² – 4q) è 2x = p ± √(p² – 4q) è x = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

If the quadratic equation ax² + bx + c = 0 has two imaginary solutions, circle C will not intersect the x-axis.  However, the solution can still be found with a straightedge and compass by performing a few additional steps.  First, construct a perpendicular bisector of QR through C called CM.  Second, construct a tangent OT of circle C.  Third, construct circle O with radius OT.  Circle O will then intersect line CM at the two solutions Z₁ and Z₂, in which the x-coordinate is the real part and the y-coordinate is the imaginary part. 

For example, let’s say you wanted to solve the quadratic x² – 4x + 5 = 0.  Then a = 1, b = -4, and c = 5, and p = -^b/_a = 4 and q = ^c/_a = 5.  As before, use a compass and straightedge to plot out the coordinates U(0, 1), P(4, 0), Q(0, 5), R(4, 5), and S(4, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(2, 3), construct circle C with radius CU.  This time, however, circle C does not intersect the x-axis, so construct a perpendicular bisector of QR through C called CM, construct a tangent OT of circle C, and construct circle O with radius OT.  Circle O intersects the line CM at Z₁(2, 1) and Z₂(2, -1), and so the two solutions are x = 2 ± i.

We can also show that these additional steps will always work by combining the Cartesian equation for circle O and the equation of the vertical line CM and showing that its solution (x, ±y) can be entered into x’ = x ± yi to simplify into the quadratic equation (x’ = ^{(–b ± √(b^2 – 4ac))}/_2a).  As mentioned above, the center of circle C is (^p/₂, ^{(q + 1)}/₂), and so the vertical line CM can be represented by x = ^p/₂.  The length of segment CO is the distance between O(0, 0) and C(^p/₂, ^{(q + 1)}/₂), which is CO = ½√(p² + (q + 1)²).  Segment CT is a radius of circle C, which we also know from above is CT = ½√(p² + (q – 1)²).  Since ΔCTO is a right angle triangle, CT² + OT² = CT², or substituting, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))².  Applying some algebra, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))² è OT² = ¼(p² + (q + 1)²) – ¼(p² + (q – 1)²) è OT² = ¼(p² + q² + 2q + 1) – ¼(p² + q² – 2q + 1) è OT² = ¼(4q) è OT² = q è OT = √q, which means the radius of circle O is √q.  The Cartesian equation for circle O is then x² + y² = q.  Combining this with x = ^p/₂, (^p/₂)² + y² = q è y² = q – (^p/₂)² è y² = ¼(4q – p²) è y = ±½√(4q – p²).  Now if x = ^p/₂ and y = ±½√(4q – p²), x’ = x + yi = ^p/₂ ± ½√(4q – p²)i = ^p/₂ ± ½√(4q – p²)√(-1) = ^p/₂ ± ½√(p² – 4q) = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x’ = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

Therefore, there exists a method for solving a quadratic with a straightedge and compass.  The steps themselves are fairly simple, but the proof is not.  Even more difficult than the proof must have been the invention of the method itself, which must have been by some creative genius.

Factoring ax^2 + bx + c

Factoring the quadratic expression ax² + bx + c is a common component of most high school math curricula.  The simplest type of factoring problems is when a = 1, for example, x² + 8x + 15.  The algorithm for finding the factored form of x² + bx + c is to find two numbers, p and q, such that pq = c and p + q = b, and writing the solution as (x + p)(x + q).  So for x² + 8x + 15, the two numbers p and q such that pq = 15 and p + q = 8 are 3 and 5, so the solution is (x + 3)(x + 5).  With practice, factoring x² + bx + c is straightforward and can be done mentally.

 The harder factoring problems are when a > 1, for example, 8x² + 14x + 3.  To factor ax² + bx + c when a > 1, the textbook I teach out of (McDougal Littell) says to find factors of a and factors of c, and guess and check with those numbers until it multiplies out to the original question.  So for 8x² + 14x + 3, the factors of 8 are 1 & 8 and 2 & 4, and factors of 3 are 1 & 3.  The different possible combinations (and their solutions) for all these numbers are:

(x + 1)(8x + 3)	= 8x2 + 3x + 8x + 3	= 8x2 + 11x + 3
(x + 3)(8x + 1)	= 8x2 + x + 24x + 3	= 8x2 + 25x + 3
(2x + 1)(4x + 3)	= 8x2 + 6x + 4x + 3	= 8x2 + 10x + 3
(2x + 3)(4x + 1)	= 8x2 + 2x + 12x + 3	= 8x2 + 14x + 3

The only possibility that multiplied with the correct middle term of 14x is the last one, therefore, 8x² + 14x + 3 = (2x + 3)(4x + 1).

For the most part I like the McDougal Littell textbook, and to be fair this is the common algorithm for factoring ax² + bx + c when a > 1, but in my opinion this is not the best way.  It is long and cumbersome, and gets even worse for larger numbers with more factors.  It also relies on trial and error and a little bit of luck, all things good mathematicians should avoid when solving problems.  I can think of several better ways to factor ax² + bx + c when a > 1, and in fact I teach one of the ways to my own high school students.

Method 1 – Factoring b and ac

The first alternate algorithm for factoring ax² + bx + c when a > 1 is the way I learned when I was a student and the way I teach my own high school math students.  First, find two numbers p and q such that pq = ac and p + q = b, and then split bx into two terms px + qx.  Then factor out the first two terms (ax² + px) and then factor out the last two terms (qx + c).  If done correctly, there will be a common term (in parenthesis) which can be combined to give a factored solution to the problem.  For example, to factor 8x² + 14x + 3, first find two numbers p and q such that pq = 8∙3 = 24 and p + q = 14, which are 2 and 12, and then split 14x into 2x and 12x, giving 8x² + 2x + 12x + 3.  Then factor out the first two terms 8x² + 2x to 2x(4x + 1) and the last two terms 12x + 3 to 3(4x + 1), so that 8x² + 2x + 12x + 3 = 2x(4x + 1) + 3(4x + 1).  There is a common term (4x + 1) that can make 2x and 3 combine, so 8x² + 14x + 3 = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8x2 + 2x + 12x + 3	(2 and 12 add up to 14 and multiply to 8∙3)
	= 2x(4x + 1) + 3(4x + 1)	(factor the first two and the last two terms)
	= (2x + 3)(4x + 1)	(combine the like term of 4x + 1)

Reasoning: It is helpful to think of the factored answer as (ex + f)(gx + h), where e, f, g, and h are unknowns that need to be solved.  Multiplying (ex + f)(gx + h) gives egx² + (eh + fg)x + fh, so matching that with ax² + bx + c, a = eg, b = eh + fg, and c = fh.  That means finding two numbers that multiply to ac and add up to b is equivalent to finding eh and fg.  So ax² + bx + c = egx² + ehx + fgx + fh = ex(gx + h) + f(gx + h) = (ex + f)(gx + h).

Method 2 – Factoring with Fractions

The second alternate algorithm for factoring ax² + bx + c when a > 1 is to factor the a out first, making the middle term ^b/_a and the last term ^c/_a, then re-write the last term ^c/_a as ^ac/_a^2, which makes ax² + bx + c = a(x² + ^b/_ax + ^ac/_a^2).  To factor, you need to find two numbers p and q such that pq = ^ac/_a^2 and p + q = ^b/_a, but both p and q will have a denominator of a so really the only mental math to be done is to find the numerators, or in other words, two numbers that multiply to ac and add up to b.  Then re-write ax² + bx + c as a(x + p)(x + q), simplify the fractions, and distribute the factors of a in such a way to remove the fractions.  For example, to factor 8x² + 14x + 3, factor out the 8 first so it is 8(x² + ¹⁴/₈x + ³/₈), and then re-write the last term as ^ac/_a^2 which is ²⁴/₆₄.  Next find two numbers that multiply to 24 and add up to 14, which are 12 and 2, so p = ¹²/_a = ¹²/₈ and q = ²/_a = ²/₈, so 8x² + 14x + 3 = 8(x + ¹²/₈)(x + ²/₈) or simplifying, 8x² + 14x + 3 = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x2 + 14/8x + 3/8)	(factor out an 8)
	= 8(x2 + 14/8x + 24/64)	(change 3/8 to ac/a^2 = 24/64)
	= 8(x + 12/8)(x + 2/8)	(12 and 2 add up to 14 and multiply to 24)
	= 8(x + 3/2)(x + 1/4)	(12/8 simplified is 3/2, 2/8 simplified is 1/4)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: Factoring out the a term leaves a quadratic expression with a = 1, however with fractions.  Forcing the middle term to have a denominator of a and the last term to have a denominator of a² forces both p and q to have a denominator of a, which makes it easier to find p and q even if they are fractions.

Method 3 – Factoring by Completing the Square

The third way to factor ax² + bx + c when a > 1 is to complete the square.  First factor the a, making ax² + bx + c = a(x² + ^b/_ax + ^c/_a).  Then complete the square on x² + ^b/_ax by adding ^{b^2}/_4a^2 (and subtracting ^{b^2}/_4a^2 to balance the equation), which makes a(x² + ^b/_ax + ^{b^2}/_4a^2 – ^{b^2}/_4a^2 + ^c/_a) = a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)).  If ax² + bx + c is factorable, then ^{b^2}/_4a^2 – ^c/_a should be a square number, so replace it with t², and replace ^b/_2a with s, which makes a((x + s)² – t²).  Factoring this difference of squares makes a(x + s + t)(x + s – t).  Then the factors of the a in front can be distributed in a way to remove the fractions and get the factored solution.  For example, to factor 8x² + 14x + 3, factor out the 8 first so it is 8(x² + ⁷/₄x + ³/₈).  Then complete the square by adding (and subtracting) the term ⁴⁹/₆₄ to make 8(x² + ⁷/₄x + ⁴⁹/₆₄ – ⁴⁹/₆₄ + ³/₈) which is 8((x + ⁷/₈)² – ²⁵/₆₄).  Factoring the difference of squares makes 8(x + ⁷/₈ + ⁵/₈)(x + ⁷/₈ – ⁵/₈) or 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x2 + 7/4x + 3/8)	(factor out an 8)
	= 8(x2 + 7/4x + 49/64 – 49/64 + 3/8)	(complete the square with 49/64)
	= 8((x + 7/8)2 – 25/64)	(finish completing the square)
	= 8(x + 7/8 + 5/8)(x + 7/8 – 5/8)	(factor the difference of squares)
	= 8(x + 3/2)(x + 1/4)	(simplify the fractions)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: ^{b^2}/_4a^2 – ^c/_a is a square number because it is equal to ^{(b^2 – 4ac)}/_4a^2 and in terms of e, f, g, and h (defined in Method 1) it is equal to ^{((eh + fg)^2 – 4egfh)}/_4(eg)^2 = ^{(e^2h^2 + 2efgh + f^2g^2 – 4efgh)}/_4e^2g^2 = ^{(e^2h^2 – 2efgh + f^2g^2)}/_4e^2g^2 = ^{(eh – fg)^2}/_(2eg)^2.

Method 4 – Factoring with the Quadratic Equation

The fourth way to factor ax² + bx + c when a > 1 is to use the quadratic formula, x = ^{(–b ± √(b^2 – 4ac))}/_2a, to show that ax² + bx + c = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a).  Then the factors of the a in front can be distributed in a way to remove the fractions and get the factored solution.  For example, 8x² + 14x + 3 = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a) = 8(x – ^{(–14 – √(14^2 – 4∙8∙3))}/_2∙8)(x – ^{(–14 + √(14^2 – 4∙8∙3))}/_2∙8) = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x – (–b – √(b^2 – 4ac))/2a)	(formula)
	∙(x – (–b + √(b^2 – 4ac))/2a)
	= 8(x – (–14 – √(14^2 – 4∙8∙3))/2∙8)	(plug in a = 8, b = 14, c = 3)
	∙(x – (–14 + √(14^2 – 4∙8∙3))/2∙8)
	= 8(x + 3/2)(x + 1/4)	(solve)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: The quadratic equation can be proved by completing the square, and so can this method.  In Method 3 we already showed that ax² + bx + c = a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)).  Continuing on, ^{b^2}/_4a^2 – ^c/_a = ^{(b^2 – 4ac)}/_4a^2 = (^{√(b^2 – 4ac)}/_2a)² so a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)) = a((x + ^b/_2a)² – (^{√(b^2 – 4ac)}/_2a)²).  Factoring the difference of squares gives a(x + ^{(b + √(b^2 – 4ac))}/_2a)(x + ^{(b – √(b^2 – 4ac))}/_2a) = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a).

Method 5 – Factoring by Graphing

The fifth way to factor ax² + bx + c when a > 1 is to graph the equation and use the x-intercepts.  Unfortunately this method relies on a graphing calculator, but it is fairly quick if a graphing calculator is available.  In this method, graph the function y = ax² + bx + c and find p and q, the two x-intercepts (as fractions).  Then ax² + bx + c can be re-written as a(x – p)(x – q), and the factors of a can be distributed in a way to remove the fractions p and q and get the factored solution.  So for 8x² + 14x + 3, you would graph and find the two x-intercepts, which are x = -1.5 = -³/₂ and x = -0.25 = -¹/₄. 

This means 8x² + 14x + 3 = 8(x – -³/₂)(x – -¹/₄) = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x – -1.5)(x – -0.25)	(two x-intercepts)
	= 8(x + 3/2)(x + 1/4)	(re-write as fractions)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: Since the x-intercepts = ^{(–b ± √(b^2 – 4ac))}/_2a, this method works for the same reason using the quadratic equation works, which is described in Method 4.

Conclusion

As you can see, there are several different ways to factor ax² + bx + c where a > 1: factoring b and ac, factoring with fractions, factoring by completing the square, factoring using the quadratic equation, and factoring by graphing.  (There are probably other methods as well that I have not considered.)  I feel that all these alternate methods are superior to the method of guessing and checking, which does not seem very mathematical at all.