AI Journey Foreword

Note: Dates are in DD/MM/YEAR format.

Any inline equations will be as following: $a^2 + b^2 = c^2$

And any equations in a block will be:

$$c = \sqrt{x} + S_0 N(d_1) - Ke^{-rT} N(d_2)$$

The format of this page is as one giant article that I dump all my thoughts and learnings into relating to [math, ai, biology, chemistry, physics]. If you’re new to my website I suggest exploring my articles or my career log as they reflect the history prior to this recording. If you remain, welcome aboard space pirate.

For some context, the purpose of this is to both act as:

1. A public accounting mechanism to keep me on track w/ the commitment
2. An inspiration to those wondering how I got to where I am, so you can either follow it or modify it to suit your needs

AI is such an incredibly difficult field to break into, especially coming from a position such as mine, no math background (even algebra), since it truly relies on the fundamental pillars of life: probabilities, rates of change + how they change, and tracking what effects these variables, why they do and how they do it.

In hopes to bring bright minds to take the leap of faith in the most exciting time of humanity to date, I step forth into a commitment to encourage those that lack the strength.

To creating new life, frens.

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2024

24/04 — Identity

Before starting, I would like to mention that I’ve had deep trouble gathering all my thoughts into articles that get published. I’ve written about neuroscience, evolution and math but have never finished them due to never polishing up the editing or to merely not having enough information to finalise the pieces. This, on the otherhand, will act as a continous feed of thoughts so there is no requirement to finishing a piece — therefore there will never be a hault on production.

I will never regret doing more of what’s important to me, I will only regret not doing it sooner. Someday is not a day in the week.

I also would like to establish my identity for future reference.

Who am I?

A pioneer in the realm of artificial intelligence, specifically mutation based computating. I stick up for what I believe in and will never be swayed unless theres a reasonable explaination to reassess.

What is my goal?

The goal is to learn as much as possible in the shortest amount of time to optimally accelerate. There are 2 things that we can leverage for the rest of our lives: [knowledge, skills], everything else is temporary. Health is to optimise them. Money is buy time that have them. Friendships will come naturally as long as you’re doing right and not immoral to them.

My journey is towards building the most advanced mutation based AI and apply it to [nanoscience, rocketry, military] to push the limits of humanity.

Why does this goal matter to me?

I truly believe the only way to see AGI is to have it be built from non-human minds. It needs to be able to take a life on it’s own. Follow evolution and speed it up. That can be done w/ compute + mutation algorithms.

When the model is running rampet, I want to control it and put it into:

1. Nanotech: to remove viruses, kill cancer, stop body deterioration and cure death — experienced too much of these and they effect me deeply; I want to live to see #2.
2. Rocketry: to explore the galaxy and see what’s really out there — very curious to seek the answers of the universe.
3. Military (optional): to protect myself against bad actors (civil war, world-war, assassinations) + for galactic conquest.

How will I reach said goal?

• Build a habit of doing the thing for 2hrs a day w/ undistracted focus. Habits remove motivation from the equation.
• Be f*cking disciplined, be consistent at learning math each day, then the goal will be reached.
  • Why 4hrs? Short periods of focused work is the only valuable form of work, e.g. long unfocused work.
  • Why math? Everything important requires math. It’s the lowest level you can go in anything. Observing biology to make algorithms is math, building ai is math, finance, inventing, simply understanding the world is math. You can’t do anything relevant in deep tech without math.
• Don’t work any jobs. Only do contracting if need money. Figure out how to get money without a job.
  1. Lets do the math on the time trade-off when working a job.
     • If you work 9hrs 5/7 days, 9*5 = 45hrs * 52 weeks = 2,340hrs…
     • Now lets do 12hrs conservatively for commute, calls, relaxing bc you work a slave job and have no metal bandwidth to do shit, 2340 + (3*5*52=780) = 3,120hrs.
     • Instead of wasting 3,120hrs for the year on a job that simply gives you money and not anything working towards who you want to be, is that really worth it objectively?
  2. Mathematical financial breakdown If you pay £2,000 * 12 months then thats £24,000 per year, where:
     • average rent in UK is £1,276 p/m
     • bills: total £230
       • phone £10
       • water avg: £40
       • internet avg: £40
       • gas + electricity avg: £140
     • health insurance of £26.15 p/m
     • 1 meal a day for £5 * 30days = £150 p/m
     • vitamins sold per quarter is around, get multivit maybe £100 / 3 = £33 (d3, fish-oil, b12, zinc, magnesium, c, etc)
     • gym membership conservatively £30 p/m
     • total essentials expenses exluding rent: £30 + £150 + £40 + £230 = £450; add whatever rent you have, e.g. £1,600 + £450 = £2,050.
     • then obviously you have one off purchases for bed, desk, lamp, books (use online ones though), etc.
     • is it really that hard to make £24,000? If you can’t do that then you have other problems to focus on over taking a leap of faith to change the world.
• Surround myself with people that are like-minded. Those you spend the most time with will either fuck you over mentally or enhance you.

What do I set out to accomplish each day

3 essentials for each day

• Workout: full workout and/or 100 burpees to take break
• Read: 2-4hrs of math w/ intense focus
• Writing: document learnings + progress, to solidify learning

Topics to attack in order

0. algebra: need to know how to manipulate variables to do anything beyond algebra
1. linear algebra
2. calculus
3. probability + statistics
4. number theory (modular arithmetic)
5. group theory (fields, rings, etc.)
6. mathematical modelling
7. ai
8. read “The Art of Electronics”
9. start robotics, only after 70% of AI stuff has been learned
10. biology: mutation, evolution, adaptation

The most optimal day would be to optimise:

• 8hr math (4x 2hr blocks; 1 after wake up, 1 before sleep)
• 2hr bio (2x 1hr blocks)
• 3hr thinking + curious study (not intense, e.g. nanotech)
• 3hr existing
  • cook + eat: 1hr
  • gym + shower: 1hr
  • relax: 1hr

My current situation

• Consistent at gym, building routine
• Working 12-14hrs p/d , even some all-nighters
• Enough runway for multi-year if quit, according to the financial specification above
• No time to self study; the time I have remaining is recovering from the work-gauntlet
• Planning future before anything drastic
• Always stressed bc pressure and health deteriorating as result

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27/04 - College Algebra Prerequisities, Part 1

“The laws of nature are written in the language of mathematics.” - Galileo

Intro

For the retards, like myself, lets start with a quotient: the number returned from division, e.g. 15 is the quotient in $\dfrac{45}{3} = 15$.

What if we wanted to find the multiplicative inverse, aka the reciprocal? By swapping the numerator and denominator — since we know the answer is 15 aka $\dfrac{15}{1}$, $\dfrac{1}{15}$

A product, on the otherhand, is the number or expression resulting from the multiplication of two or more numbers or expressions, e.g. $10x^2 + 15x = 5x(2x + 3)$ where $5x(2x + 3)$ is the equivalent expression that is a product! This example contained factoring a polynomial (more on these later), which means finding an equivalent expression that is a product.

Sets

A set is a collection of objects that can be determined by $\{\}$, e.g. the roster method looks like $\{1, 2, 3, ...\}$.

Then we can move onto the intersection of sets. These are elements found in both sets, e.g. in $A=\{1,2,3\}$ and $B=\{3,4,5\}$ we see 3 is the common real number. We can say this as $A \cap B = \{3\}$. Think of this as a bridge where the number crosses from one side to the other!

Next, there is the use of the union $A \cup B$. Can you guess what this is? It’s the set of elements that are members of set $A$ or $B$ e.g. $\{1,2,3\} \cup \{3,4,5\} = \{1,2,3,4,5\}$! Think of this as a cup where we want one of each number flavour. We don’t want to overdose on number flavours so we add only one of each.

Real Number ($\Reals$)

Real numbers ($\Reals$) are, what I consider as, the letters of English alphabet. If we don’t know what letters there are how can we form words (the analogy for expressions and formulas)?

Subsets

There are subsets, think of as architypes in a game, that have different properties — try noticing the pattern between them as you go down.

Make a name for the new anime homie that assassinates his trbie, n’wiri (to remember the order: N, W, I, R, I)!

1. Natural numbers: nums we use for counting $\{1,2,3,...\}$
2. Whole numbers: natural numbers w/ 0, bc it felt lonely $\{0,1,2,3,...\}$
3. Integers: whole numbers that also went backwards $\{-3,-2,-1,0,1,2,3,...\}$
4. Rational numbers: rational numbers that can be expressed as the quotant (divide result) of 2 integers $\{-17 = \dfrac{-17}{1}, -3, 0, 2, 3, \dfrac{-2}{3} = -0.6666\}$
5. Irational numbers: all numbers where the decimals are neither terminating or repeating, so not quotient of integers $\{ \sqrt{2} \approx 1.414214, \pi \approx 3.142, -\dfrac{\pi}{2} \approx -1.571\}$

Everything is a rational number, aside from irrational numbers (duh).

Real Number Properties

The properties of real numbers are what we can do with said letters. What are the rules of the game so we can perform? These are super important to know. Without these you cannot go on to modify expressions, move terms around and solve equations — mastering these helps with calculus hardcore (from my experience).

1. Commutative: Changing the order doesn’t affect the sum; think as going to the store back and forth — the destination should be the same.
   • $13 + 7 = 7 + 13$
   • $\sqrt{2} \cdot \sqrt{5} = \sqrt{5} \cdot \sqrt{2}$
2. Associative: Changing grouping doesn’t affect the product; think as if i give you a pistol and you gave me an ak we can still shoot each other (lmao).
   • $-2(3x) = (-2 \cdot 3)x = -6x$
3. Distributive: Multiplying the outside var by the inner parentheses member(s); we’d use this to factor the like terms to make it simpler.
   • $a(b+c) = a \cdot b + a \cdot c$
   • $-3(4x - 2y +6) = -3 \cdot 4x - (-3) \cdot 2y - 3 \cdot 6$
4. Inverse: The inverse gives the same identity; we use this to simplify negative expressions.
   • Additive Inverse: $a + (-a) = 0$ and $(-a) + a = 0$
   • Multiplicitive Inverse: $a \cdot \dfrac{1}{a} = 1, \not = 0$ and $\dfrac{1}{a} \cdot a = 1, \not = 0$
5. Identity: Removing either 0 or 1 bc it doesn’t change anything; think of getting rid of your ex.
   • Additive Identity: $a + 0 = a$
   • Multiplicitive Identity: $a * 1 = a$

Absolute Value

The absolute value $|a|$ really means how far away are we from 0? If -3 is the answer then $|a| = 3$, for 3 it’s the exact same!

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28/04 - College Algebra Prerequisities, Part 2

Exponents

Exponential notation looks like $b^n$, where $b$ is the base and $n$ is the exponent. For example, $b^3 \cdot b^2 = (b \cdot b \cdot b)(b \cdot b) = b^7$

There are a few rules we must understand, so we can modify expressions at will:

1. Product Rule: Adding exponents together!
   • $b^m \cdot b^n = b^{m+n}$
   • $3x^2(5x+2^3) = 15x^3 + 16x^2$
2. Quotant Rule: The reverse of product rule.
   • $\dfrac{b^m}{b^n} = b^{m-n}, b \not = 0$
   • $\dfrac{30y^9x^{12}}{5x^3y^7} = \dfrac{30}{5} \cdot \dfrac{x^{12}}{y^3} \cdot \dfrac{y^9}{x^3} = 6x^{12-3}y^{9-7} = 6x^9y^2$
3. Zero-Exponent Rule: If b is any real number other than 0 then, $b^0 = 1$
   • $1 = \dfrac{b^2}{b^2} = b^{2-2} = b^0$
4. Negative Exponent Rule: If b is any real number other than 0 and n is a natural number (non-zero), then $b^{-n} = \dfrac{1}{b^n}$
   • $7x^{-5}y^2 = \ 7 \cdot \dfrac{1}{x^5} \cdot y^2 = \ \dfrac{7y^2}{x^5}$
   • $\dfrac{1}{6^{-2}} = \dfrac{1}{\dfrac{1}{6^2}}$ which is $1 / \dfrac{1}{6^2} = \dfrac{1}{1} \cdot \dfrac{6^2}{1}$, the reciprocal of $\dfrac{1}{6^2} = \dfrac{6^2}{1}$, then $6^2 = 36$. The opposite would go in reverse.
5. Power Rule: $(b^m)^n = b^{mn}$
   • $(2^2)^3 = (2^2) \cdot (2^2) \cdot (2^2) = 2^{2+2+2}$
6. Products To Powers: $(ab)^n = a^nb^n$
   • $(3x)^2 = 3^2x^2$
7. Quotients To Powers: $(\dfrac{a}{b})^2 = \dfrac{a^2}{b^2}$

Radicals

Lets disect what a square root (sqrt) w/ the radical expression $\sqrt[n]{a} = b$

• $\sqrt{}$ is the radical
• $a$ is the radicand, e.g. $\sqrt{4} = 2$ as $2^2 = 4$, where 4 is the principal sqrt.
• $n$ is the radical index. If n is: odd then b, even then |b|.

Think of it as radi, [cal, and, index].

The principal nth root of a real number a $\sqrt[n]{a} = b$ means that $b^{n} = a$

Radical Factorisation

The perfect square is when a root’ed number returns an integer. If we have $\sqrt{36} = 6$, would be 6, bc multiplying itself (6^2), “perfectly” fits into 36.

If there is no perfect square, there will be factors of it. A factor is a number or expression that can be multiplied by another factor to get a product.

There is the greatest sqrt factor, referring to the largest perfect square that divides the number evenly, e.g. $\sqrt{50x} = \sqrt{{25 \cdot 2x}}$ since $25$ can be factored further we get $25 = 4 \cdot 5 \therefore \sqrt{{25 \cdot 2x}} = 4\cdot5\sqrt{2x} = 20\sqrt{2x}$

Another example is $\sqrt{500} = \sqrt{25 \cdot 20} = \sqrt{25} \sqrt{20} = 5 \sqrt{20}$. Since 20 contains a perfect square factor, 4, we need to simplify futher: $5 \sqrt{20} = 5 \sqrt{4 \cdot 5} = 5 \sqrt{4} \sqrt{5} = (5 \cdot 2) \sqrt{5} = 10\sqrt{5}$.

The aim is to get the lowest numbers to form the original number to make the expression less complex, e.g. for 36 we’d get $18 \cdot 2$ but we can go futher and do $9 \cdot 2$, then to get 36 we have the lowest factorisation w/ $3^2 \cdot 2^2 = 36$

Combining Radicals

Now with out new-found knowledge we can fight the mini boss battle: $4 \sqrt{50x} - 6\sqrt{32x}$

$$4 \sqrt{50x} - 6\sqrt{32x} = (4 \cdot x) \sqrt{5} \sqrt{2}$$

Radical Expression Rules

1. Product: The sqrt of a product is the product of sqrts, $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$
2. Quotient: If a (nominator) and b (the denominator) are nonnegative real numbers and b != 0, then: $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$. The sqrt of a quotient is the quotient of the sqrts.

Add + Sub Radicals

• Add: $4\sqrt{11} + 2\sqrt{11} = (4+2)\sqrt{11} = 6\sqrt{11}$
• Sub: $\sqrt{5x} - 7 \sqrt{5x} = 1\sqrt{5x} - 7 \sqrt{5x} = (1-7) \sqrt{5x} = -6\sqrt{5x}$

Rationalising Denominators

This concept is where things started to click for me. Essentially, any number divided by itself is 1, e.g. $\dfrac{\sqrt{3}}{\sqrt{3}} = 1$.

So, 2 approximations that look entirely different, e.g. $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$, can mean the same thing.

1. let’s break this down $\dfrac{1}{\sqrt{3}}$

$$\\ = \dfrac{1}{\sqrt{3}} \cdot 1$$ $$\\ = \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}$$

2. here’s the fascinating part: $$\\ = \dfrac{\sqrt{3}}{\sqrt{9}}$$
3. simplyfiying further $$\\ = \dfrac{\sqrt{3}}{3}$$
4. therefore $$\\ \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$$

Conjugates

Radical expressions that involve the sum and difference of the same two terms are called conjugates (joint together).

The general rule for multiplying conjugates is:

$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b$$

Why is this? We apply the Distributive property into:

$$(\sqrt{a} \cdot \sqrt{b}) (\sqrt{a} - \sqrt{b}) = (\sqrt{a} \cdot \sqrt{a}) + (\sqrt{a} \cdot (-\sqrt{b})) + (\sqrt{b} \cdot \sqrt{a}) + ((-\sqrt{b}) \cdot \sqrt{b})$$

Make this more readable:

$$= a - (\sqrt{ab})+(\sqrt{ab})-b$$

The like terms cancel out:

$$= - (\sqrt{ab})+(\sqrt{ab})$$

Which leaves us with:

$$= a -b$$

Lets look at another example, $\dfrac{h}{{\sqrt{x+h}} - \sqrt{x}}$ then the conjugate of the denominator is $\sqrt{x + h} + \sqrt{x}$ (the opposite).

As an example, $\dfrac{7}{5 + \sqrt{3}}$

$$\\ = \dfrac{7}{5 + \sqrt{3}} \cdot \dfrac{5 - \sqrt{3}}{5 - \sqrt{3}}$$

Rember how this is =1. Now we convert the two denominators to $(\sqrt{a})^2 - (\sqrt{b})^2$

$$\\ = \dfrac{7(5 -\sqrt{3})}{5^2 - (\sqrt{3})^2}$$

Simplify the denominator

$$\\ = \dfrac{7(5 -\sqrt{3})}{25 - 3}$$

Rational Exponents

Lets take a look at two expressions: $(7^{\frac{1}{2}})^2 = 7^{\frac{1}{2} \cdot 2} = 7^1 = 7$ and $(\sqrt{7})^2 = \sqrt{7} \cdot \sqrt{7} = \sqrt{49} = 7$

From this we see that: $7^{\frac{1}{2}}$ means $\sqrt{7}$

And so we can make the definition, as long as n >= 2:

$$a^\frac{1}{n} = \sqrt[n]{a}$$

And, as long as a != 0,

$$a^{-\frac{1}{n}} = \dfrac{1}{a^{\frac{1}{n}}} = \dfrac{1}{\sqrt[3]{a}}$$

But what about rationals where the numeration > 1?

$$a^\frac{2}{3} = (a^\frac{1}{3})^2 = (a^2)^\frac{1}{3}$$

Thus,

$$a^\frac{2}{3} = (\sqrt[3]{a})^2 = \sqrt[3]{a^2}$$

And we take this further with a new definition,

$$a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$$

And if $a^{-\frac{m}{n}}$ is a nonzero real number, then

$$a^{-\frac{m}{n}} = \dfrac{1}{a^\frac{m}{n}}$$

Notice how from the previous definition, $a^\frac{1}{n} = \sqrt[n]{a}$, n is the denominator and the only thing that changed was the numerator m. The reason why () come into play in the latter is because adding the exponent of 1 to something doesn’t change anything! So really, the latter is the general definition of rational exponents!

Lets do an example:

$$\dfrac{32^\frac{5}{3}}{16x^\frac{3}{4}} = (\dfrac{32}{16})(\dfrac{x^\frac{5}{3}}{x^\frac{3}{4}})$$ $$= 2x^{\frac{5}{3} - \frac{3}{4}}$$

Now, if you’re mathematically illiterate like myself, we need to simplify this mofo right here. It’s a bit confusing but I’ve done the hard work for the tribe:

We begin by finding the common denominator w/ 3 and 4: 12, essentially the lowest number w/ both numbers — like factoring.

So for ${\dfrac{5}{3} - \dfrac{3}{4}}$ we’re really just multiplying the numerator 5 and denominator 3 by the opposite expression’s denominator 4:

$$\dfrac{5}{3} = \dfrac{5\cdot4}{3\cdot3} = \dfrac{20}{12}$$

And do for the other expression with the former,

$$\dfrac{3}{4} = \dfrac{3\cdot3}{4\cdot3} = \dfrac{9}{12}$$

Then we get

$$= \dfrac{20}{12} - \dfrac{9}{12} = \dfrac{20 - 9}{12}= 2x^{\frac{12}{11}}$$

We finally got through the core sht! Only [polynomials, factoring, trig] to go (fck me it takes ages to write about this stuff, but damn is it sticking).

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02/05 - Polynomials!

Gm, welcome to learning about pole-pee-no-my-balls.

As always, lets learn the Vocabulary first, wtf are nomials:

• Binomial is a single term, $7x^3$
• Monomial: is 2 terms, $7x^3 + 2x^2$
• Trinomial: is 3 terms, $7x^3 + 2x^2 + 5$
• Polynomial: is anything beyond 3 terms, $7x^3 + 2x^2 + 5x + 69$

We describe polynomials w/ negatives w/ parentheses to be explicit (like a rap song)

$$7x^3 - 9x^2 + 13x - 6$$

As, $$7x^3 + (-9x^2) + 13x + (- 6)$$

I trip up a lot with the - bullshit so this is quite important to note: $-ax^n = (-ax^n)$.

The n represents the degree. We order polynomials in cronological order from left to right in terms of degrees. E.g. $7x^3$ will be ordered before $7x^2$. The degree of a polynomial is the greatest of the degress of all its terms.

Taking a look at an algebraic expression w/ a polynomial in x (expression of vars + coefficients combined using addition, subtraction and multiplication): $$a_{n}x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_1x + a_0$$

• The degree of the polynomial is $n$. This is important because it is the biggest rate of change.
• The leading coefficient (numerical or constant factor that multiplies a var) (the term w/ the highest exponent/power) is $a_n$
• The constant term is $a_0$

Polynomial Sub, Add, Mul

Subtracting polynomials by combining like terms (variables and their exponent powers are the same), e.g.

$$-9x^3 - 13x^3 = (-9 + 13)x^3 = 4x^3$$

We are using the the distributive property to factor out the common term, $x^3$

Adding polynomials: $5x^4 + 6x^4 = 11x^4$

However we cannot Simplify $5x^4 + 3x^2$ because there are no like terms!

Multiplying polynomials:

$$(5x^4)(6x^4)$$

Combining like terms,

$$= (5\cdot6)(x^4 \cdot x^4)$$ $$= 30x^{4+4} = 30x^8$$

But what would we do if there is no monomials? E.g,

$$(2x + 3) (x^2 + 4x + 5)$$

Well we would use the distributive property (our lord and saviour!):

Distributing 2x

$$2x(x^2 + 4x + 5) = 2x^3 + 8x^2 + 10x$$

Then distributing 3

$$3(x^2 + 4x + 5) = 3x^2 + 12x + 15$$

Then combining like terms

$$= 2x^3+ (3x^2 + 8x^2) + (10x + 12x) + 15$$

Simplifying

$$= 2x^3+ 11^2 + 22x + 15$$

Product Of Two Binomials

Once again, we use the distributive property to find the product of two binomials. For example,

$$(3x - 5) (4x + 5)$$ $$= 3x(4x) + 3x(5) + (-5(4x)) + (-5(5))$$ $$= 12x^2 + 15x + (-20x) + (-25)$$ $$= 12x^2 + -5x -25$$

Similar to multiplying conjugates, we find the Product of the Sum and Difference of Two Terms as follows: $$(A + B)(A - B) = A^2 - B^2$$ because the two terms cancel eachother out $$(B \cdot A) (A \cdot (- B)) = AB - AB = 0$$

Squaring & Cubing Binomials

$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$

How do we break this down? Lets expand the exponents:

$$(A + B)^3 = (A + B) (A + B) (A + B)$$

Lets distribute the first 2:

$$(A + B) (A + B) = A^2 + AB + AB + B^2 = A^2 + 2AB + B^2$$

Now we distribute the third:

$$= (A^2 + 2AB + B^2) (A + B)$$ $$= (A^3 + 2A^2B + 2AB^2 + B^3)$$

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03/05 - Factoring Polynomials

Factoring

Factoring a polynomial containing the sum of monomials means finding an expression that is a product:

$$10x^2 + 15x = 5x(2x+3)$$

The goal of factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors, except possibily for a monomial factor, is prime or irreducible, aka being factored completely.

The greatest common factor (GCF) is the largest positive integer that divides two or more integers without leaving a remainder.

$$18x^3 + 27^2$$

9 factors into 18 (9*2) and 27 (9*3),

$$= 9x^2(2x) + 9x^2(3)$$

What about for,

$$x^2 (x+3) + 5(x + 3)$$

We see that $(x + 3)$ is the common factor in both terms, so we can reverse the distributive property into:

$$(x + 3) (x^2 + 5)$$

Factoring By Grouping

Some polynomials only have 1 as the greatest common factor. But by grouping the terms together we can actually create the common factor.

For example, these terms have the common factor of 1,

$$x^3 + 4x^2 + 3x + 12$$

But if we group them together as the following we can find a new common factor.

First, the common factor is $x^2$ with:

$$(x^3 + 4x^2) = x^2(x+4)$$

The second, with the common factor of $3$, becomes

$$(3x + 12) = 3(x + 4)$$

Combining them together we get

$$= x^2(x + 4) + 3(x + 4)$$

Notice the common factor again? It’s $(x + 4)$. We use this as one term then the other would be the remaining vars, $(x^2 + 3)$. Therefore,

$$= (x + 4)(x^2 + 3)$$

Factoring Trinomials