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Algebra, Turingtest :D

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FOIL Inopeki :D (a+b)(c+d)
nice (x+3)(x-4)

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Other answers:

good, simplify
Good, (no need to write the 1 though) (3p+t)(2p-4t)
Oh right 6p^2+3p*-4t+t*2p+5t?
not 5t... everything else is right though
no... look closely
-4t^2 :P
Huh :O
You don't see it? (-4t)(t)=-4t^2 look at the last terms in the parentheses
Ohhh, now i see it
so you have what then, before simplifying?
-4t^2 :P
Oh right xd
can you simplify it?
6p^2+3p*-4t+t*2p-4t^2 6p^2+6p^2-4t^2-4t^2?
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) ???
-12pt? 2pt?
right, so then we have...?
yes :D and how does that simplify?
I dont know actually, maybe 6p^2-14pt+pt-4t^2?
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.
But dont i need to multiply them to get -10pt?
no, just add the coefficients factoring is one way to see it... factor out 'pt' from the middle two terms: \[6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2\]
Oh wait, i get it now, it would me multiplication if it was -10pt^2
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...
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
if you multiplied you get \[(2pt)(-12pt)=-24p^2t^2\]
Oh right
Since you would be multiplying both p and t with itself
Whatever, were getting off track :)
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)
hey I'm green! never been a mod for math before :)
Congratulations! You deserve it :)
thanks :) I can't delete things though, don't see the point oh well... (2s^2+p)(s-3p^3)
(2s^2+p)(s-3p^3) 2s^3+2s^3(-3p^3)+p(s)+3p^4
so far so good now try to simplify...
oh wait, small error
OH the -
2s^3+2s^3(-3p^3)+p(s)+3p^4 ^^ should be???
2s^3+2s^2(-3p^3)+p(s)-3p^4 ^^ plus that :)
good catch I missed the sign lol !!! yes excellent now can you simplify it
Thanks :D 2s^3+2s^2(-3p^3)+p(s)-3p^4 2s^3-6s^2p^3+sp-3p^4?
perfect last one: (a+b)(a-b)
a^2-ab+ab-b^2? a^2-2ab-b^2?
not quite, I don't think those terms in the middle add...
The simplification or the first one?
a^2-ab+ab-b^2 ^^^^^ what is -x+x ?
Oh right, 0
so (a+b)(a-b) is?
Some kind of coaching class going on here?
And that is a very important formula @pratu Inopkie wants to learn Algebra, a good cause in my book :)
Shall I help?
Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)
@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\[a^2-b^2\]
so a^2-b^2=(a+b)(a-b)?
Its an identity.
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
But isnt (x-2)(x+2) 0 because the 2s negate?
foil it out and tell me yourself
x^2+2x-2x+-2*2 Well pellet
right which simplified too what?
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)
isnt that f(1)=
yes, sorry, bad notation on my part f(1)=-3 f(0)-4 etc.
But how does that work?
which part?
Functions, maybe im not ready for that subject
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?
I just 'called' it a function, and made it graph-able...
Yeah, when x and y intersect its 0,0
f(x)=y+2x-3 So if we want to figure out y we need to plug in something for x
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.
hence it's better to think of graphs in terms of functions, represented in Cartesian coordinates by y
What are cartesian coordinates?
the regular ones you know x-axis and y-axis
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.
One question, what do we gain from functions, what are the numbers on the RHS?
which numbers on the RHS?
Like f(0)=0-4=-4<--- ^ |
well that's the simplified form of f(0): f(0)=-4 check it by plugging in x=0 into the function
But if we know that f(0)=-4 why do we do it?
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.
Then teach me! This is important!
Really, though of course actual physics problems are usually (though not always) a bit trickier than just plugging in a number but yes, functions are instrumental to calculus, and therefor physics As I said earlier calculus is about analyzing functions... I guess we should start to talk about functions in a very detailed way.
Yes please :D
the definition of a function for you will be: if you have something like y=f(x) (where f(x) is some arbitrary polynomial with x's in it) and for every x you put in you get exactly one number for y back, then y=f(x) is a function, and y is a function of x try to soak that in, I'll tell you more about it momentarily
So what you are saying is that if f(x)=y x=y?
Well not really
What you are saying is that if x is one and y is 2 and you add 1 to x you add one to y aswell?
no, just that we get only one value of y for each x for instance f(x)=y I defined that, whether or not it's a function however if for every x we put in we get only one y back, then f(x) is a function, and the variable y is a function of the variable x
scratch the *for instance :/
Gotta go eat dinner, brb
look at a graph like a circle (I know you haven't studied them much yet this way)|dw:1326043068666:dw|this is not a function because there is more than one y for each value of x... see you in a bit, I'm gonna have breakfast.
drawing a line from x to its corresponding y value you should know we need a vertical line|dw:1326043155512:dw|see how one x-coordinate corresponds to two different y values y1 and y2? But what about y=2x+5 ?|dw:1326043286670:dw|this graph has exactly one value of y for each value of x, so it is a function.
the idea above is what's called the 'vertical line test': if you can draw a vertical line at any point on the graph and intersect it twice, then it is not a function.
when you return look at this say we have a table of values for some y=f(x) and we want to know if f(x) is a function.... look at the table x |y=f(x) --------- -1 | 2 3 | 7 2 | 0 5 | 2 Is this a function? why or why nmot?
Its not. EXEPT the one that says 2 | 0<---------------< cause that means that there is only one value of x ^
The others seem to have two values of x
actually look at the x's each x has one y, so it is a function we have that f(5)=f(-1)=5 so two different x's give the same y, but that is okay the only thing against the rules would be to have something like this x | y ------ 1 | 3 2 | 5 1 | 4 because this means that f(1) has two values: f(1)=3 and 4 so this is not a function... look at the x to see which gives more than one f(x)... (I am using the terms f(x) and y here interchangeably) x | y ----- 2 | 0 3 | 1 5 | 6 3 | 6 1 | 0 function or not ant why?
That's right FFM does the curriculum meet your approval? I doubt it with your standards :P
x | y ----- 2 | 0 =f(2)=2? im sorry, i dont really understand 3 | 1 5 | 6 3 | 6 1 | 0
f(2)=0 that means when x=2, y=0, yes?
that's all there is to that so is there any f(x) on the list that gives more than one number for a single x?
So, x | y ----- 2 | 0 = f(2)=0 3 | 1 = f(3)=1 5 | 6 = f(5)=6 3 | 6 = f(3)=6 1 | 0 = f(1)=0 So, no.
actually what is f(3) ?
Oh! 1 and 6!
exactly, so not a function :)
f(3)=1+6 so its not a function!
YEah :D
And that means that f(3) could be graphed in a circle?
right not 1+6=7 of course, it equal the set\[f(3)=\left\{ 1,6 \right\}\] It doesn't mean it would be a circle but it would fail the vertical line test, watch...
Of course.
Y cant have to places
thanks FFM, praise form Caeser :) |dw:1326046327322:dw|now look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?
No way!
lol *Praise from Ceaser
no way to draw a vertical line and hit more than one point you say?
Not more than 3.
And there are 5 so..
more than 3? You only need to hit 2 points to show that this is NOT a function and here it is:|dw:1326046665544:dw|so again we show by this that f(3) has more than one value...
are the following functions?|dw:1326046773020:dw|(I'll fill in different graph above)
yes f(x)=y because we have defined it that way now:|dw:1326046848239:dw|function?
draw the vertical line that hits more than one point on the graph if that is true
No wait, there are no points
there are infinite points on a line.... a line intersects another line at a point|dw:1326047055394:dw|P is our intersect point can you draw a vertical line that hits the curve at more than one point?
Oh, so y is a vertical line?
Then this could be a function
|dw:1326047157974:dw| not function
But yours is
now you got the idea :D|dw:1326047282819:dw|function?
just one line is sufficient, but yes :)||dw:1326047413725:dw|function?
good :) x | y ----- 1 | 3 2 | 3 3 | 3 4 | 3 function?
For all of them
But if it was 1|3 1|4 It would not
good x | y ----- 0 | 3 0 | 2 0 | 1 0 | 0 I think you already proved you know this, but function?
0 has 4 values of y
nice job, exactly slightly different: f(x)=x+3 function?
I guess that depends on what you plug in for x?
Actually, no
It is a function
It doesnt show multiple values of y, just x+3
Yes! :D
perfect observation :) every x we put in gives us only one y. y=0 is y a function of x?
Only if we know that 0=x
not so, plug in various x into the function f(x)=0, we get the table x | y ---- 0 | 0 1 | 0 2 | 0 3 | 0 and so on... and I think you would agree that this chart constitutes a function.
tricky one, but now you know ;)
Yes, that chart constitutes a function. But if there were no chart?
They are all in agreement, the chart, the graph, everything. we don't need x=0 to get one value of y back; we always get one value of y. The same value of y, namely zero. do you know what the graph of y=0 looks like?
Thats 0,0
yes but I don't want the point (0,0) y=0 is a line: it's actually the x-axis the line x=0 is the y-axis think about that for a minute
Oh, right. |dw:1326048543616:dw|
Didnt really think of that
exactly, and does the graph of y=0 pass the vertical line test?
are you saying that you can draw a vertical line that intersects it twice? show...
OH you mean that? Never mind lol
so is the graph indicative of a function or not?
Yes, it is
perfect, what about the line x=0? (standard cartesian notation here , don't get hung up on the names of variables, we could easily have them to t and p or whatever)
yes, so is it a function?
be careful can we draw a vertical line that touches the graph at more than one point? what would the table of values look like?
No wait, this becomes f(x)=x right?
no, this would be f(x)|x=0 (read f of x such that x is zero) here f(x) takes on a range of values from negative to positive infinity: x | f(x) ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 etc. not a function, right?
I think i get it, its because there is no y?
but f(x) has endless possibilities
no this chart is the same as x | y ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 I told you that y and f(x) are interchangeable y is the name of the coordinate which we are using to represent f(x) so here they are the same the chart itself is of the same type as you saw before, we have one value of x corresponding to multiple (in this case infinite) values of y, so it is not a function.
Ohhh, but if you had y corresponding to multiple values of x it would be a function.
right :)
Ah, now i see!
let me try to clear a few things up... what is the dependent and what is the independent variable here?
The dependent variable is y because if its a function or not depends on y having multiple values corresponding to one value of x.
right, but I'm not sure about your reasoning, so what about f(y)=x=y^2+3y which is dependent and which is independent?
x is the dependent one i think
exactly, so here x is a function of y we could have therefor more than one y value for an x value in this case, but not more than one x value for each y, if it is to be a function. so really the rule is better stated this way: "a relation is a function if each value of the independent variable corresponds to exactly one value of the dependent variable" because we can change the names or swap positions of the variables
Because we can have more of y but not more of x
exactly, look at the new version of our rule above (in quotes)
Yeah, that sounds right.
so whenever we see f(t)=r for instance which is dependent and which ins independent?
We should move to another "thread" this one is lagging for me
I agree go ahead
r is the dependent and t is independent
exactly :)

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