Golden Deer Pub Community Forums

XD Sparkling!

How's the arm doing out of the cast? Are you going to have to have any therapy for it?

;__; I broke my arm so badly that I still don't have full use of my left wrist/hand/arm. Hope you don't have the same problem! If you're having stiffness or motor control I can recommend some exercises I remember...

* Ghostly Field Club [CD]

* Changeability of Strange Dream [CD]

Today was cheered up (not that it needed cheering up; it just became even more cheerful) by the arrival of another box of anime from RACS. I decided I needed more magical girl anime in my collection


Magical Girl Lyrical Nanoha Complete Box Set


Magical Girl Lyrical Nanoha A's Complete Box Set


Kamichu! Complete Box Set

OHBOYOHBOYOHBOY

I was at the thrift shop with my mother and spotted an assortment of the Durants' History of Civilization. I begged her to take me home super-quick so I could check which volumes I had.

I ended up buying volumes 3, 4 and 11. I'm only missing volume 6 now!

Very exciting. I've been collecting this series for a while now, and it's out-of-print.

Have a friend in Japan picking me up a copy of Umineko Chiru. WHEEEEEEEEEEEE.

[spoiler]And unfortunately I haven't managed to evade all spoilers yet T.T I know the 1st Twilight victims.[/spoiler]

;__; bought some books to help me study math.

They are hard to understand... I don't get it... factor polynomials what...?

I suddenly feel empathy for people who claim history makes no sense to them. It must feel similar to the utter gibberish written on these pages.

Oh!! I know that!! Algebra II material :3

Like: 3x-12
Factored is 3(x-4)

Kino, I can't guarentee that I know everything - it's been a long time and such, and I'm not a very good explainer, but if you ever need, I can ATTEMPT to try to teach you? >_> I really do suck at teaching so just keep that in mind XD but I would be more than happy to try to teach to you whatever you need.

Factoring polynomials isn't all that hard. It's rather basic application of arithmetic, assuming you don't need to use the quadratic equation (even then, that's not complex either besides memorization).

Ignore this if you don't have issue with factoring polynomials, I just typed it out of reaction I guess.
[spoiler]ax^2 + bx + c = 0

That is the generalized form. The goal is merely to express that as the following:

(ex + y) * (fx + z)

Where:
e * f = a
e*z + f*y = b
y * z = c

Edit: Something to keep in mind:

If "b" and "c" are both positive, then "y" and "z" are both positive.

If "b" is positive and "c" is negative, then "y" and "z" have different signs, with the greater of the two being positive.

If "b" is negative and "c" is negative, then "y" and "z" have different signs, with the great of the two being negative.

If "b" is negative and "c" is positive, then "y" and "z" are both negative.

It looks more complicated when you try to generalize it, personally, but it's still rather simple as long as you don't overthink it. Just follow the steps:

1) Identify the factors (i.e., numbers when multiplied together that yield the intended product) of the term "c"
2) Identify which of the factors, when added together, yield "b"
2a) If term "a" is non-one, don't forget to take that into account when solving for which factors yield "b".
3) Arrange the binomials.
4) Check your work >> Seriously.

And now, examples:

x^2 + 7x + 6 = 0

The coefficient behind the x ("a") is a one so no worries there; also, both the second term ("b") and last term ("c") are positive so both numbers we're looking for here will be positive as well. Therefore, all you'll need are two positive numbers, y and z, that will multiply together to yield 6 and add to yield 7:
1 * 6 = 6; 1 + 6 = 7
[s]3 * 2 = 6; 3 + 2 = 5[/s]

(x + 1) * (x + 6) = x*x + x*6 + 1*x + 1*6 = x^2 + 7x + 6

___

2x^2 + 9x + 4 = 0

The leading coefficient isn't one so we now have a situation where the first terms in both binomials may possibly have a coefficient. From looking at it, we know that we'll need one of the terms to have a 2 and the other to have a 1 (2 * 1 = 2). This changes the process slightly as we now need two numbers, y and z, that multiply to yield 4 and 2y + z OR y + 2z yield 9:

1 * 4 = 4:
2 * 1 + 4 = 6
1 + 2 * 4 = 9

[s]2 * 2 = 4:
2 * 2 + 2 = 6
2 + 2 * 2 = 6[/s]

We need to get the 2 to multiply with the 4 so, when arranging the solutions, the "2x" and the "+4" terms must be in DIFFERENT binomials:

(2x + 1) * (x + 4) = 2x*x + 2x*4 + 1*x + 1*4 = 2x^2 + 9x + 4

___

x^2 - 5x + 4 = 0

The second term is NEGATIVE and the LAST term if positive; that tells us the two numbers multiplied together are both negative. How? Well, negative * negative = positive and negative + negative = negative. Besides that, it works out the same as before:
-1 * -4 = 4; -1 + -4 = -5
[s]-2 * -2 = 4; -2 + -2 = -4[/s]

(x + [-1]) * (x + [-4]) = (x - 1) * (x - 4) = x*x + x*-4 + -1*x + -1*-4 = x^2 - 5x + 4

___

12x^2 + 77x - 27 = 0

a = 12
b = 77
c = -27

a = 1 * 12, 2 * 6, 3 * 4
b =
c = -1 * +27, +27 * -1, -3 * +9, +3 * -9

Solving for b is a bit harder as there's so many combinations. Trial and error is about all I can really say for this point, though you should be able to eliminate some options (since the c is negative, you know that the two terms you're looking for are different. This also means that b is the result of a subtraction operation so any of the combinations that cannot yield a value above 77 are not possible solutions ... however, if you're not comfortable with the work, I wouldn't bother trying to shortcut things really):

[s]1 * -1 + 12 * 27 = -1 + 324 = 323[/s]
[s]1 * 27 + 1 * -12 = 27 - 12 = 15[/s]

[s]2 * -1 + 6 * 27 = -2 + 162 = 160[/s]
[s]2 * 27 + 6 * -1 = 54 - 6 = 48[/s]

[s]3 * -1 + 4 * 27 = -3 + 108 = 105[/s]
3 * 27 + 4 * -1 = 81 - 4 = 77

Alright, so we know that we need to have 3 * 27 and 4 * -1 to be possible; that means the 3 and 27 terms are separate (likewise for the 4 and -1):

(3x - 1) * (4x + 27) = 3x*4x + 3x*27 + -1*4x + -1*27 = 12x^2 + 81x - 4x + -27 = 12x^2 + 77x - 27

__________[/spoiler]