Mastering Math Identities: Clear and Comprehensive Guide for AS Level Students

[iNsChris]

Hey guys,

Having trouble gettign my head around these indentities!
So confusing so does anyone have a website which goes through the explanations systematically.

My textbooks touches on it but the explanation doesn't cover all the variations.
and the "letts AS Revisions guide" doesn't cover it at all which is rather annoying as i waitied for it and thought it would!

Please help,

I need a grade A in maths AS and its just holding me back!

 

[Tide]

Which identities would those be?

 

[iNsChris]

"olving the square" ones. And others such as
x^3 + ax^2 -3x + b =(the three line = Symbol) (x -c)(x - 4) (x+ 5)
i understand the way to solve some using the "values" method.

But the "coefficents" or "Comparing coefficents" method is confusing.

 

[Tide]

I don't know what you mean by "solving the square."

Actually, comparing coefficients is probably the simplest way for determining the values but, conceptually, it might be easier to try something like the following. Replace x with specific values and obtain equations for the unknowns! In this case, good ones are x = 4 and x = -5 which give simple equations for a and b that you can easily solve. A third value, given a and b, would be x = 0 in order to obtain c. Oftentimes, trying another method will give you insights and perhaps even an appreciation of the "standard" techniques.

One of the really nice things about math is that there are often many ways of obtaining solutions to problems though math courses sometimes focus on a single approach.

 

[Zurtex]

Identities are always true, that's the 1st thing you need to remember. It doesn't matter what value of x or theta or whatever they are always true.

Simple identity:

1 + 1 \equiv 2

That is always true, it does not depends on a specific value of x. So if:

A + 1 \equiv 2

Where A is some constant, then we know A = 1. Carrying on to something more complex:

x + x + 1 + 1 \equiv 2x + 2

Notice that x doesn't equal anything particular, this is just always true. So taking another example:

x + 2x + x + 1 + 76 \equiv Ax + B

We can easily work out that B = 77 and A = 4 for any value of x. So key thing is, always make as simple as possible and only compare the things in front of the same number of x's. So following on taking your example:

x^3 + ax^2 - 3x + b \equiv (x-c)(x-4)(x+5)

x^3 + ax^2 - 3x + b \equiv x^3 + 5x^2 - 4x^2 - cx^2 + 4cx - 5cx - 20x + 20c

x^3 + ax^2 - 3x + b \equiv x^3 + (1-c)x^2 + (-20-c)x + 20c

Now the 1st thing to notice is that we have an easy relationship look at just the bit in front of x. We have a -3 in front of the one on the left hand side and a -20-c, now as they are always equal no matter what value of x that must mean that: -3 = -20-c so we can say that:

3 = 20 + c

c = -17

Now we have the c = -17 let's look at the bit in front of the x^2 term. We have that 1 - c is the same as a, now it is easy to get from this that a = 18. Finally we have 20c = b as both of these are on their own without an x, so this gives us that b = -340.

Hope that helps.

 

[iNsChris]

That kind of helps and i also saw the book get (1 - a) from ax^2 etc. etc.

How did you get that? Are you simply expanding it so that you have a number to work with and because there is no coefficient for A it must be 1?

(Note: I am talking about the +(1 - c)x^2 )

Also, Expanding three brackets at a time got me bit confused. I normally use the "FOIL" technique so i would have done:

First ones (x X x X x) = x^3
Outsides: x X 5 = 5x
Inside . . . . Then realize its all messed up!

How should i do those () () () ? As i say, I usually only do (x + x) (x - x) for example.

Once again there are to many unexplained steps :(

I don't see how you got from

https://www.physicsforums.com/latex_images/33/339434-5.png
to
https://www.physicsforums.com/latex_images/33/339434-6.png

I see you kept the x^3, 20c then everything else seems to have somehow changed :S

I need a website explaining all the possible techniques, I found Values (i.e. substituing x = 1, x = 2 etc) easy. But then it doesn't seem to work for all of them, or does it?

This CoEfficent technique seems to take huge steps and i don't see how they can go from one thing to the other.


would it be possible to go from:

x^3+5x^2-4x^2-cx^2+4x-5cx-20x+20c

to group similar items and get:

so i would then get:x^3+x^2-cx^2-cx-20x+20c

Then I am all stuck! and don't know what to do next.

I seem to be treating them like equations and thinking "well i have x^3 on both sides so i can cancel that out". But i know i can't.

Hence i really need a deep explanation .

Look forward to response,
INsChris

 

[Zurtex]

As my Gran always used to tell me when I was 1st learning algebra (she was a maths teacher), collect like terms!

If you have x + x^2 + x this is the same as: x^2 + 2x. Another example cx + 2x^3 - 3x is the same as: 2x^3 + (c - 3)x.

This is all I have done, for example looking purely at the x^2 term, I had 5x^2 - 4x^2 -cx^2 is the same as: x^2 - cx^2 taking x^2 as a common factor: (1 - c)x^2.

As for dealing with the product of 3 linear equations. Well if your confident enough with quadratic equations you can take this:

(ax+b)(cx+d)(ex+f)

And rearrange it like so:

ax(cx+d)(ex+f)+b(cx+d)(ex+f)

Now just expand the (cx+d)(ex+f) then finally expand whatever is left that needs expanding and collect like terms.

Hope that helps

 

[iNsChris]

Thank zurtex, that's made me realize i REALLY REALLY need to touch up on my expanding/Factoring alot!

Got three basic rules now:
Collect like terms
Factorise where possible
Look at it again and follow the above steps :P

Ok I'm off to relearn factorising and expanding as i have clearly forgotten a lot of it!

 

[iNsChris]

Definatly need to practice... this has puzzled me!

(x^2 - a/x^2) (x^2 - a/x^2)

I know the answer is:

http://img91.exs.cx/img91/8392/x2.jpg

Problem is when i tried it...

i got it all wrong!

i normally use the high school "FOIL" technique...

But i can't use it on that!

I get things like x^4 X a/x^2

gets messy :)

 

[Zurtex]

x^4 * \frac{a}{x^2} = x^4 * \frac{1}{x^2} * a = \frac{x^4}{x^2} * a = ax^2

Anyway, looking at this: (x^2 - a/x^2) (x^2 - a/x^2). Take it from this approch:

(y - z)^2 = (y - z)(y - z) = y^2 - 2yz + z^2

Now let y = x^2 and z = a/x^2

 

[iNsChris]

yeh i did it like that but it got sooooo messy and screwed up!

I mean what does: x^2 multiplyed a/x^2 =?

ax^2/x^4?
or
ax/x^4
Really confused lol! Need to learn this expanding stuff with / involved!

never done them much before and when i did, i got them all wrong

Please post your method fully explained if you have time.

 

[matt grime]

it equals a. (ie neither of your guesses, sorry)

a/x^2 is the same as ax^{-2} and whenever you multiply x^{p} by x^{q} the answer is x^{p+q} for all p and q.

so here you have p=2 and q=-2, so 2+-2 = 0 and x^0=1.

 

[iNsChris]

my teacher says its:

(x^2 +a/x^2)(x^2 +a/x^2) + b

= x^4 -2a +a^2/x^4 +b

 

[matt grime]

excepting the minus sign error(s), yes. (which is what foil would give you too. but don't use foil: it is unnecessary).

 

[iNsChris]

but i need to know how to get there lol :)

anyone got good website with expanations on / In expanding!

I have never seen Divisions in () ()'s

So need to get my head around them

thanks

 

[matt grime]

the division is a red herring!

As some one has pointed out (u+v)^2 = u^2+2uv+v^2 is true what ever u and v are. after you've done that you simplify the terms.

and if we need to rehash it:

(u+v)(u+v) = u(u+v) + v(u+v) = u^2+uv+vu+v^2, rearranging gives us the answer we want.

 

[iNsChris]

excepting the minus sign error(s), yes. (which is what foil would give you too. but don't use foil: it is unnecessary).

which minus sign errors?

when i try this i end up getting:

x^4 -2a/x^8 -a^2/x^4

Must have gone wrong with the +/- somewhere because my tutor has 2a over nothing.

But i have 2a/x^8.

So when simplifying i assume i should of had -a/-x^4 and -a/+x^4 therefore cancelling each other out.

I seem to have forgotten the basics on expanding and factorising :(

 

[matt grime]

you seem to have forgotten the basics on exponents:

you have x^2 times a/x^2 which simplifies to a, not a/x^8.

your post three back contains a minus sign error. read it again. notice you hav(x^2+a/x^2) in the brackets, not minus as you originally wrote.

 

[iNsChris]

yeh i definatly have. Know any websites which have good explanation of this stuff?

I need to revise it before continuing and I am really far behind in my studys! at this rate ill of read everything once and will have 0 Revision time :(

Do you have a webpage or any ebooks with this explained?

Can't find my old GCSE revision stuff!

and the sites i have been on do not have any explanations of this stuff (specially with Division involved).

 

[iNsChris]

if x^2 X -a/x^2 = a

then how did my tutor get

-a/x^2 X a/x^2 = -a^2/x^4 ?

Wouldnt it be the same rule meaning -a/x^2 X -a/x^2 = +a^2

I got +a^2/x^4

i guess it wouldn't as its both DIVIDED BY x^2 rather than
x^2 X a/x^2

Hmm, clearly need to revise this but can't find anywhere to do it!

 

[matt grime]

(1/x^2)*(1/x^2) = 1/x^4

that's how powers work.

remember your rules:

\frac{1}{x^n} = x^{-n}

x^p x^q = x^{p+q}

(x^r)^t = x^{rt}

 

[iNsChris]

K thanks,

I have another problem lol!

This is the full thing along with where i have got to so far:


Find Factors For:

f(x) = x^3 -3x^2 + 4

soon that was down to:

x^3 -3x^2 + 4 = (x -2) (x^2 + bx + c)

then simplyfied/factories down to:

x^3 -3x^2 +4 = x^3 + (b-2)x^2 + (x -2)c -2bx

So i can find b by:

-3 = b-2
b = -3+2
b= -1

Right?

However i think i need to factorise more to find C or X ?

 

[matt grime]

you can't 'find' x, it is the variable.

I think you ought to check your mutliplying out again and group together the coeffecients of powers of x.

you found the coeff of x^2, so find the coeff of x and the constant term

 

[iNsChris]

x^3 +bx^2 -2x^2 -2bx +xc -2c

got this down to:

x^3 +(b -2)x^2 + (c -2b)x -2c


so i get:

2c = +4
-3 = b-2

Agree?

If I'm correct my confidance will go up lol!

 

[matt grime]

I think you can have that confidence boost but for the minus signs that you keep dropping!

-2c=4