Algebra – Part 4: Factoring

Until the previous post, we looked at polynomials (I know – I only wrote about dividing polynomials, but there are of course other formats of polynomials that exist, adding, subtracting, and multiplying. The reason I skipped them all was that they are too easy, so I just thought I didn’t even feel any need to mention them here.)

Here, doctoring is the exact opposite of polynomials – we doctor out what’s typical between numbers and variables. And what’s interesting about the very basic concept of doctoring is that its notion is something you can take advantage of when you’re programming, especially OOP (object-oriented programming). The basic idea of OOP is to factor out what’s common among child classes, those factored-out elements are written in their parent class.

The concept here is the same, and algebra’s factoring is much easier than that of OPP programming. Anyways, let’s dive into the world of factoring!!

1st: Greatest common factor of trinomials

So, as I mentioned above, factoring is literally factoring out what’s common among trinomials/polynomials. Here, we have 2x + 4xy + 10bx, and what’s common among the three is 2x. So, why not factor out it? Then the answer is 2x(1 + 2y + 5b). Easy right?

2nd: Greatest common factor of polynomials

Here’s the polynomial version of factoring. We have 10x²y² – 5xy³ / 5. First, let’s factor out 5xy² as the greatest common factor and what remains is 2x – y. Then, simplify it by 5. What’s left is xy²(2x – y).

3rd: Greatest common factor of quadratic polynomials

This time, we have x² – 4x -21. In this case, take a look at the two numbers, -4 and -21, and ask yourself what two numbers that could be -4 by adding/subtracting and -21 by multiplying. The answer is -4 and 3.

• -7 + 3 = -4
• -7 x 3 = -21

So, the answer is (x – 7)(x + 3).

4th: Difference of squares

There are six different problems and their solutions. This is not that difficult, and probably you don’t need any lecture to understand the logic.

5th: Zero theorem

Zero theorem asks you to prove either A or B is 0. The first example is x² – x – 20 = 0, and you’ll get (x + 4) (x – 5) = 0 when you factor it. Then, apply the aforementioned AB theory: A x B = 0, so either A or B is 0. In this case, (x + 4) is A while (x – 5) is B.

Take a look at (x + 4):

x + 4 -> x = -4.

Plug -4 to x -> (-4 + 4) (x – 5)

-> 0 multiplies (x – 5)

-> 0
Nest, take a look at (x – 5):

x – 5 -> x = 5.

Plug 5 to x -> (x + 4) (5 – 5)

-> 0 multiplies (x + 4)

-> 0

So, it proofs either A or B is 0 and it satisfies the zero theorem.

Here’s another example:

6th: Complete the square

In this practice, follow the following steps:

1. Confirm a = 1.
2. Add (b/2)² to both sides.
3. Factor the perfect square.
4. Solve for x.

What’s important here is step 2: Add (b/2)² to both sides.

In the below sample (image 08), x² = A, 6x = B and 4 = C.

So, 6/2 -> 3 -> (3)² = 9.

Add 9 to both sides of the equation.

x² + 6x + 9 + 4 = 0 + 9

x² + 6x + 9 became factorable.

(x + 3)(x + 3) + 4 = 0 + 9

(x + 3)(x + 3) = 0 + 9 – 4

(x + 3)² = 5

Take square root from both sides:

x + 3 = +/-√5

x = -3 +/-√5

So the answer is:

x = -3 + √5

x = -3 – √5

Image 09 is the formula you need to follow. And this formula is something you might need in a wide range of algebraic methods.

in the following image 10, there are three methods you need to remember. Let’s take a look at (b/2)² – C > 0. If (b/2)² – C is bigger than 0, you end up getting two real numbers. In the first example (image 08), we ended up getting -3 and +/- √5.

But like the 2nd, if b/2)² – C = 0, you end up getting one real number, and it logically makes sense.

And this is an important concept you need to know. i (imaginary number) is for √-1.

So, hypothetically, if the answer you’ve got is x = -2 +/-√-3, remember the square root rule: √ab = √a√b.

x = -2 +/-√3(-1)

x = -2 +/-√3√-1

Replace √-1 with I

x = -2 +/-√3i

So, the answer is:

x = -2 + √3i

x = -2 -√3i

7th: Quadratic formula

The final installment in our journey in factoring is the quadratic formula. Basically, what you need to know is to follow the following formula:

x = -b +/- √b² – 4ac/2a.

And here is the very basic example with a, b, and c. And please forgive me for not giving you enough detailed explanation of it. Just you know, I’m not a mathematician, just a 30-something years old guy who started loving algebra, so just look at the following images and what I’ve done here. And I’m sure that it logically makes sense, so just take a look at it and apply the knowledge to your own problems. Remember the spirit of try-and-error!

And from here, let’s take a look at an example: x² + 2x -8 = 0. And also, apply the aforementioned formula, x = -b +/- √b² – 4ac/2a, to it.