Question

Algebra, Turingtest :D

Answer 1

FOIL Inopeki :D
(a+b)(c+d)

Answer 2

ac+ad+bc+bd

Answer 3

nice
(x+3)(x-4)

Answer 4

x^2-4x+3x-12

Answer 5

good, simplify

Answer 6

x^2-1x-12

Answer 7

Good, (no need to write the 1 though)
(3p+t)(2p-4t)

Answer 8

Oh right
6p^2+3p*-4t+t*2p+5t?

Answer 9

not 5t...
everything else is right though

Answer 10

-5t

Answer 11

no... look closely

Answer 12

-5t^2?

Answer 13

-4t^2
:P

Answer 14

Huh :O

Answer 15

You don't see it?
(-4t)(t)=-4t^2
look at the last terms in the parentheses

Answer 16

Ohhh, now i see it

Answer 17

so you have what then, before simplifying?

Answer 18

6p^2+3p*-4t+t*2p+4t^2?

Answer 19

-4t^2
:P

Answer 20

Oh right xd

Answer 21

can you simplify it?

Answer 22

6p^2+3p*-4t+t*2p-4t^2
6p^2+6p^2-4t^2-4t^2?

Answer 23

no... how did you the p^2 in the middle?
we had
6p^2+3p(-4t)+t(2p)-4t^2
parentheses look cleaner btw, you should use them here

what is
3p(-4t)
and
t(2p)
???

Answer 24

-12pt?
2pt?

Answer 25

right, so then we have...?

Answer 26

6p^2-12pt+2pt-4t^2

Answer 27

yes :D
and how does that simplify?

Answer 28

I dont know actually, maybe
6p^2-14pt+pt-4t^2?

Answer 29

No

Answer 30

they are like terms in the middle
-12pt+2pt=-10pt

this goes for any combination of variables

a^2bc+3a^2bc=4a^2bc

wr^4dt^3+3wr^2dt+6wr^4dt^3=7wr^4dt^3+3wr^2dt
notice in this last example we couldn't combine all the terms because one had different exponents. We can only combine when they are the exact same in exponents and number of variables.

Answer 31

But dont i need to multiply them to get -10pt?

Answer 32

no, just add the coefficients
factoring is one way to see it...

factor out 'pt' from the middle two terms:
\[6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2\]

Answer 33

Oh wait, i get it now, it would me multiplication if it was -10pt^2

Answer 34

not necessarily. You can't multiply things unless the equation tells you too, or you do a trick like multiply both sides
if you have

-10pt^2+7pt
you cannot simplify that...

Answer 35

Oh ok. But what i meant was that if it were to tell us to multiply it would look like this -10pt^2 and not this -10pt

Answer 36

if you multiplied you get
\[(2pt)(-12pt)=-24p^2t^2\]

Answer 37

Oh right

Answer 38

Since you would be multiplying both p and t with itself

Answer 39

right

Answer 40

Whatever, were getting off track :)

Answer 41

not really, you need to get this down too. It's all related, and it's good to know where you're at.
(2s^2+p)(s-3p^3)

Answer 42

hey I'm green! never been a mod for math before :)

Answer 43

Congratulations! You deserve it :)

Answer 44

thanks :) I can't delete things though, don't see the point
oh well...
(2s^2+p)(s-3p^3)

Answer 45

(2s^2+p)(s-3p^3)
2s^3+2s^3(-3p^3)+p(s)+3p^4

Answer 46

so far so good
now try to simplify...

Answer 47

oh wait, small error

Answer 48

OH the -

Answer 49

2s^3+2s^3(-3p^3)+p(s)+3p^4
^^
should be???

Answer 50

2s^3+2s^2(-3p^3)+p(s)-3p^4
^^ plus that :)

Answer 51

good catch I missed the sign lol !!!
yes excellent
now can you simplify it

Answer 52

Thanks :D
2s^3+2s^2(-3p^3)+p(s)-3p^4
2s^3-6s^2p^3+sp-3p^4?

Answer 53

perfect
last one:
(a+b)(a-b)

Answer 54

a^2-ab+ab-b^2?
a^2-2ab-b^2?

Answer 55

not quite, I don't think those terms in the middle add...

Answer 56

The simplification or the first one?

Answer 57

a^2-ab+ab-b^2
^^^^^
what is -x+x ?

Answer 58

Oh right, 0

Answer 59

so
(a+b)(a-b)
is?

Answer 60

a^2-b^2?

Answer 61

Some kind of coaching class going on here?

Answer 62

And that is a very important formula
@pratu Inopkie wants to learn Algebra, a good cause in my book :)

Answer 63

ok...

Answer 64

Shall I help?

Answer 65

Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)

Answer 66

@Inopeki
We just showed that\[(a+b)(a-b)=a^2-b^2\]That relation is called the 'Difference of Squares' formula.
It is used all the time, as you will see, as a quick way to factor anything of the form\[a^2-b^2\]

Answer 67

so a^2-b^2=(a+b)(a-b)?

Answer 68

Its an identity.

Answer 69

Yes, it's always true for that form.
This is handy as you will see because if you remember how factoring helped us solve other equations, it will do the same for us here:

For instance\[x^2-4=0\]we can factor this..\[(x-2)(x+2)=0\]which as I mentioned befor by the 'zero factor property' means that we need to solve\[x-2=0\]and\[x+2=0\]to find the zeros of our polynomial

Answer 70

But isnt (x-2)(x+2) 0 because the 2s negate?

Answer 71

foil it out and tell me yourself

Answer 72

x^2+2x-2x+-2*2
Well pellet

Answer 73

right which simplified too what?

Answer 74

x^2-4

Answer 75

so that's not always zero, is it?
if x=1 then f(x)=-3, not zero
what the expression f(x) is equal to here depends on what we put for x.

There are in fact only two values for x that will make f(x)=0
Which should make sense by what I said yesterday:

The order of the equation is 2, so the total multiplicity of its roots is 2
(fundamental theorem of algebra)

Answer 76

isnt that f(1)=

Answer 77

yes, sorry, bad notation on my part
f(1)=-3
f(0)-4
etc.

Answer 78

But how does that work?

Answer 79

which part?

Answer 80

Functions, maybe im not ready for that subject

Answer 81

No you are, perhaps I haven't been clear enough
whenever I write
x^2-4
If I want to know where this graph hits the x-axis it is useful to call this a function f(x)

here we have
f(x)=x^2-4

Think about a graph: When the graph is toughing the x-axis it is at y=0, correct?

Answer 82

I just 'called' it a function, and made it graph-able...

Answer 83

Yeah, when x and y intersect its 0,0

Answer 84

oooh

Answer 85

f(x)=y+2x-3
So if we want to figure out y we need to plug in something for x

Answer 86

well, wait now, if you have a y in the function then it is not f(x)

here's one thing that I bet is confusing you:
f(x)=y
when we write a line equation
y=mx+b
we are saying something about the graph:
that the y coordinate is proportional to the x coordinate in this way.

However this is rather simplistic. It is really more that we have a function
f(x)=mx+b
and we have chosen to graph it by letting the y-coordinate of our graph be f(x)

they are effectively the same here, but a function can be graphed in many different types of coordinates, so y would not be used unless we are in Cartesian coordinates as we are now.

Answer 87

hence it's better to think of graphs in terms of functions,
represented in Cartesian coordinates by y

Answer 88

What are cartesian coordinates?

Answer 89

the regular ones you know
x-axis and y-axis

Answer 90

Yeah

Answer 91

there are other coordinate systems like polar, spherical, and cylindrical coordinates
If we tried to graph our function in polar coordinates, y would have no meaning.

however the zeros of the function are independent of how we choose to graph it. The math will be the same.

Hence when we want to know the zeros of some polynomial we call it f(x), and that allows us to put numbers into it and see what happens
f(x)=x^2-4
f(0)=0-4=-4
f(1)=1-4=-3
f(2)=4-4=0
so we found a zero simply by calling this a function and putting numbers into it.

Answer 92

One question, what do we gain from functions, what are the numbers on the RHS?

Answer 93

which numbers on the RHS?

Answer 94

Like f(0)=0-4=-4<---
^
|

Answer 95

well that's the simplified form of f(0):
f(0)=-4
check it by plugging in x=0 into the function

Answer 96

But if we know that f(0)=-4 why do we do it?

Answer 97

lol, why /don't/ we want to do that?!
Almost all things in physics can be represented by function. If you want to know when the water pressure of a system will become too high, what trajectory to shoot your spaceship, or where a quantum particle will be at a certain time, you must analyse functions in this way.

Answer 98

Really?