These are my complete notes for Factoring Pt. 2 in Algebra II. I split Factoring into two parts, with the first part, found here, being more introductory and relating important concepts like the Quadratic Formula, Perfect Square Trinomials, and the X-method. The second part has somewhat higher level yet similar concepts such as Factoring Completely.
I color-coded my notes according to their meaning - All numbered notes (which I call rules) are red, and include examples and the basis for understanding a topic. Definitions are written in green, and other important information (such as large-scale drawings that are better visualized than explained) was written in blue. All of this information is preserved on this page, with logical flow and breaks. I use ascii line drawings sparingly - If I can convey information or a graph using an image online, I will do so.
All of the knowledge present in these notes are filtered through my personal explanations for them, the result of my attempts to understand and study them from my classes. In the unlikely event there are any egregious errors, contact me at jdlacabe@gmail.com.
Summary of Factoring Pt. 2 (Algebra II)
V. Factoring Pt. 2.
Rule 52. Whilst a difference of cubes, as described in Rule 23, is definitely factorable, a sum of squares (x² + 64) is not. (x² - 25) = (x + 5)(x - 5). (x² + 25) is not factorable. However, a sum of CUBES (x³ + 64) is very factorable, as is the difference of cubes. See Rule 53 for more.
Rule 53. Another law of factorability is the Sum of Cubes, as well as the associated Difference of Cubes. There is an entirely different system of factoring Quantities of Cubes than for Square factoring: A difference of cubes has the base of a² - b² and is factored to the equation of (a + b)(a - b). A different of cubes, on the other hand, has the base of a³ - b³ (being to the third power and all) and is factored to the equation of (a - b)(a² + ab + b²). A sum of cubes has the factor equation of (a + b)(a² - ab + b²), having the base of (a³ + b³). As evident, the first parentheses has the same sign as the equation it factors, with as sum of cubes (a³ + b³) having a first parentheses of (a + b) while the difference of cubes (a³ - b³) has the first parentheses of (a - b). The second sign in the second parenthesis, that in regard to ab, is opposite of the sign in the first parenthesis, and therefore opposite of the sign in the original equation:
Sum: (a³ + b³) = (a + b)(a² - ab + b²)
Difference: (a³ - b³) = (a - b)(a² + ab + b²)
To perform this factorization to plug in a & b, the base equation must be cube rooted to both variables: With 8x³ - 125 for example, ∛ applied to both sides would result in 2x - 5. No ±. Remember to plug in the absolute value of a and b into the equation, getting a result of (2x - 5)(4x² + 10x + 25) where a = 2x and b = 5. Switch the sign in the original equation to a sum of cubes and the answer will be (2x + 5)(4x² + 10x + 25).
Rule 54. On Factoring Completely, arguably the final key in the great constant of factoring. There are various rules, but most of the previously known concepts, such as GCF, still apply, but just on a larger scale with more steps. Complete Factoring incorporates the newly discovered Sum & Difference Cubes into trinomial Cbic equations. Generally, the first step in factoring a trinomial with a degree of three or greater (if another factoring option is not already obvious) is to GCF the entire equation. For example, in the equation x³ + 10x² + 24x, the GCF that can be removed is x, making the equation x(x² + 10x + 24). Now, the inner equation can be given the X method:
╲ 24 ╱
╲ ╱
0 ╲╱ 4
╱╲
╱ ╲
╱ 10 ╲
The completely factored equation is now x(x + 6)(x + 4), nothing more can be done. If exact answers as to what x is equal to are needed, than the outer x = 0, and x = -6, x = -4.
Factoring Completely is just a multistep, a more advanced form of factoring. However, one must always be checking if more can be done. For a complete simplification after a PST has been discovered, such as 6(x² - 4x + 4) factored into 6(x - 2)(x - 2), the equation can be condensed to 6(x - 2)².
An example of a furtherly elaborate equation is -16x⁵ - 250x². GCF: -2x²(8x³ + 125). Now, the internal equation is evidently in a sum of cubes (both the degree and the coefficient of a can be cube rooted). Cube rooting gives (2x)³ + (5)³, giving a = 2x and b = 5. The correct formula for a sum of cubes factorization is (a + b)(a² - ab + b²). All that's left to do is to plug in a and b, and Remember to put the GCF in front of the equation, kept for the entire problem. The answer would be -2x²(2x + 5)(4x² - 10x + 25). Equations with four variables can be dealt with the same way they always have: GCF-ing the first two and the last two separately, grouping. The paranthesized equation for each side should be the same: x²(x - 2) - 9(x - 2). As obvious, merge the sides and create (x² - 9)(x - 2). However, as Complete Factoring entails, there is more to be done: The first equation is a difference of squares. Therefore, the completely factored y is (x - 3)(x + 3)(x - 2), or x = 3, x = -3, x = 2. The "more that can be done" is almost always a difference of squares or sum/difference of cubes. In fact, in some cases, GCF is not immediately usable and the equation can just quickly be given using difference of squares or a sum/difference of cubes. Even if the equation looks quite complicated.
ALWAYS be checking if the coefficients or bases in the equation are Perfect Squares or Cubes, as this is a shortcut to the answer. Of course, the degree of a base (the exponent of x) being a perfect square or cube just means it is divisible by 2 or 3. An example of this shortcut is 625x⁸ - 256. On the surface, there is no easily identifiable GCF. Therefore, it must ascertained that the equation is a difference of squares and can have those properties applied. First, square root both sides (25x⁴)² - (16)². Using a standard formula, this equation is now (25x⁴ + 16)(25x⁴ - 16). Hilariously, the second equation is another difference of squares, and therefore more factoring can be done. The result/final answer is (25x⁴ + 16)(5x² + 4)(5x² - 4), nothing more can be done.
Rule 55. How can it be that X-Method is still applicable to non-quadratic equations? Well, it is possible to make an educated guess as to what the factors will be considering that exponents add when multiplied. In the equation 2x¹³ + 10x⁹ + 8x⁵, for example, after GCF it is 2x⁵(x⁸ + 5x⁴ + 4). Using the power of imagination, simply perform X-method normally and divde the degree by two and get a correct answer: 2x⁵(x⁴ + 4)(x⁴ + 1). This strategy will sometimes not work so use a mental trial and error to factor. The strategy generally works when the exponent of b is half of that of a.
Rule 56. As repeatedly stated in Rule 54, Complete Factoring sometimes allows for the values of x to be found. However, this should only be done when the question is specifically asking for it or is equal to 0. "Find the zeroes" provides different final answers than "Factor the Polynomial Completely", the latter stopping short of finding the values of x. As stated before, if an X is GCF-ed to the front of the equation, then it is automatically equal to zero.
Rule 57. Similar to a previous Rule of Thumb of simplifying a to zero when possible, Factoring out a negative in Complete Factoring whenever possible is a must, for not doing so could lead to an incorrect answer. It just makes things way easier in general.
Rule 58. Specialties of finding the values of x in situations somewhat specific to Complete Factoring: When after X-method the x base is still exponential (in a manner described in Rule 55) and the question wants an exact values for x, then give a ± √ with the same root as whatever degree x is for the answer. Having the value not being perfect is a prerequisite. Think of this like an exponential form of Rule 29. Another way to represent the values of x is with curly brackets, as in x + {2, 3, 0}. Also, sometimes solutions can be repeated. Just keep one.
Solutions are zeroes are x-intercepts.
Rule 59. The existence of the ± when square rooting is only made by the answer being X, such as x² = 4. It wants to catch all positive values of x. If the statement is simply √4 however, the answer is just 2. Continued at Rule 64.
Rational Zeros Theorem:
If F(x) is a polynomial function with integral coefficients and if p/q (a rational number in lowest terms) is a zero of F(x), then p is a factor of the constant term of F(x), and q is a factor of the leading coefficient of F(x).
*we can use the Factor and Remainder Theorem in tandem with this.
Take F(x) = 9x³ - 18x² + 11x - 2, for example. The constant term at the end of the function is 2 (always make the number positive, because the list of (p / q) itself will be ±), and so the factors for p are 1 and 2. The leading coefficient of the term with the highest degree is 9, and so q is equal to its factors: 1 and 9, and 3 and 3. The Rational Zeros Theorem says that if there is going to be a rational zero for this polynomial, it must be in the list of P's over Q's:
p = ± {1, 2}
q = ± {1, 3, 9}
(p / q) = ± {1/1, 1/3, 1/9, 2/1, 2/3, 2/9}
Unironically, the best move right now is to plug in each one of the damn p/q's until you find a zero. Once you have at least one, then you can use synthetic division to find the rest:
F(1) = 0
By synthetic division, F(x) / (x - 1) = 9x² - 9x + 2.
Thus, 9x² - 6x - 3x + 2
3x(3x - 2) - 1(3x -2)
(3x - 1)(3x - 2)
Thus, (x - 1)(3x - 1)(3x - 2) are the zeroes of the function.
When you have a particularly lengthy problem in which the degree is greater than 3, such as 2x⁴ - 9x³ + 4x² - 35x - 50, applying the Rational Zeros Theorem and then Synthetic Division will still leave you with unfactorable problem: 2x³ - 11x² + 15x - 50. Thus, you simply have to do entire process again on that lower polynomial. Repeat until you reach an easily factorable bottom. If you reach the bottom and have found a non-easily factorable answer, such as, in this case, f(dx) = 2x² - x + 10, use quadratic formula, obviously. If you end up with imaginary answers, then just leave your answer as an unsimplified form using the zeroes you found before using synthetic division: (x + 1)(x - 5)(2x² - x +10).
In order to find a polynomial from the roots, you just need to multiply the roots (in their x + or - form) and see what you get. For example, given the roots x = 3, x = -4 + 3i, and x = -4 - 3i, then one can multiply them together as so: (x - 3)(x + 4 - 3i)(x + 4 + 3i). This multiplies out to x³ - 5x² - 15x - 27.
The Fundamental Theorem of Algebra:
For every polynomial function with the degree of "n", there are exactly "n" zeros, real or imaginary, rational or irration.
For example, f(x) x²³ - 15x¹⁹ - 1 has 23 zeroes, because of the degree. This can be used in conjunction with another theorem:
The Conjugate Zeros Theorem:
States that if P(x) is a polynomial with real coefficients and if (a + bi) (where a and b are real and b ≠ 0) is a zero of P, then (a - bi) is also a zero of P.
Complex roots/solutions/zeros with non-zero imaginary components ALWAYS come in pairs. (a ± bi)
You can NEVER only have one zero with an imaginary number. The Conjugate Zeros Theorem will not allow it. See the diagram below for proof (and also for help for Descartes' Rule of Signs):
Descartes' Rule of Signs:
Gives us an idea of how many real zeros there could be in a polynomial, and whether they are positive or negative.
(1) The number of positive real zeros equals the number of variations in signs of the coefficients of P(x), or is less the number of variations by a positive even integer.
(2) The number of negative real zeros equals the number of variations in signs of the coefficients of P(-x), or is less the number of variations by a positive even integer.
Take this polynomial for example: P(x) = x⁴ - 2x³ - 2x² + 3x + 1. In this polynomial there are two sign changes, seen after the first term and third term, and so there are 2 or 0 positive real zeros, given 1) of Descartes' Rule of Signs.
When you plug in -x in order to find the negative real zeros, you get P(-x) = x⁴ + 2x³ - 2x² - 3x + 1. There are also 2 sign changes here, and so there are either 2 or 0 negative real zeros.
Thus, the polynomial can have any of the following:
2 positive real zeros and 2 negative real zeros
2 positive real zeros and 2 non-real zeros
2 negative real zeros and 2 non-real zeros
4 non-real zeros
This can occasionally be useful when used in conjunction with other theorems.
