Can someone help me factor this complex equation?

[ManyNames]

Homework Statement

A simple way to factor the left hand side of the given equation.

Homework Equations

E^2-E^2(\frac{\frac{F^2vt}{c}}{M})+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})=-p+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})

The Attempt at a Solution

I would, just that i am confused on eactly how to.


Could anyone help me simplify by factoring the left hand side of this equation in an easy and sensible manner?

E^2-E^2(\frac{\frac{F^2vt}{c}}{M})+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})=-p+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})

Thanks up-front.

 

[noumed]

\frac{\frac{F^2vt}{c}}{M}

That's one of your common terms.
Hint:
({\frac{\frac{a^2}{b}}{b}})^2 = \frac{\frac{a^4}{b^2}}{b^2}

 

[ManyNames]

That's one of your common terms.
Hint:
({\frac{\frac{a^2}{b}}{b}})^2 = \frac{\frac{a^4}{b^2}}{b^2}

Like

(E+\frac{\frac{F^2vt}{c}}{2}M)(E+\frac{\frac{F^2vt}{c}}{2}M)?

 

[noumed]

that won't work. if you multiply out something like:
(A+B) * (A+B)
you get
(A^2 + 2AB + B^2)

the expression you wanted to simplify is in the form of
(A^2 - A^2B + B^2)
if you had
A = E
and
B = \frac{\frac{F^2vt}{c}}{M}

looking at the whole equation, couldn't you have subtracted (\frac{\frac{F^4v^2t^2}{c^2}}{M^2})
from both sides and just end up with a simple
E^2-E^2(\frac{\frac{F^2vt}{c}}{M})=-p?

 

[ManyNames]

that won't work. if you multiply out something like:
(A+B) * (A+B)
you get
(A^2 + 2AB + B^2)

the expression you wanted to simplify is in the form of
(A^2 - A^2B + B^2)
if you had
A = E
and
B = \frac{\frac{F^2vt}{c}}{M}

looking at the whole equation, couldn't you have subtracted (\frac{\frac{F^4v^2t^2}{c^2}}{M^2})
from both sides and just end up with a simple
E^2-E^2(\frac{\frac{F^2vt}{c}}{M})=-p?

You see, i am quite familiar with simple factoring of equations. I am just getting very confused with all these symbols and how to use them when i do factorize. Could you not just show me please, and put me out of my misery. My heads sore.

As for the latter question friend, the added (\frac{\frac{F^4v^2t^2}{c^2}}{M^2}) I cannot subtract.

 

[noumed]

(A^2 - A^2B + B^2)
if you had
A = E
and
B = \frac{\frac{F^2vt}{c}}{M}

Why don't you just work with the equation above instead of all that mess? It's much easier to work with 2 variables, A and B rather than all that frazzle dazzle. The purpose of this forum is not to give out answers but provide hints. See if you can factor
(A^2 - A^2B + B^2)

Hint:
You can rewrite the equation above to:
-(A^2(B-1) - B^2)
That's a difference of two squares factorization.

 

[ManyNames]

Why don't you just work with the equation above instead of all that mess? It's much easier to work with 2 variables, A and B rather than all that frazzle dazzle. The purpose of this forum is not to give out answers but provide hints. See if you can factor
(A^2 - A^2B + B^2)

Hint:
You can rewrite the equation above to:
-(A^2(B-1) - B^2)
That's a difference of two squares factorization.

Agreed. That way is so much easier.

If that is all, thanks a bunch. It was really helpful.

 

[ManyNames]

(But i'll still try and solve this - I'm not off yet ;) ) Plus, I've never seen an equation of the form (A^2-A^2B+B)... i don't get out much. :)

 

[ManyNames]

I need to ask so i understand this more clearly, you want me to substitute the B's of (A^2+A^2B+B^2) for \frac{\frac{F^2vt}{c}}{M}, but what about \frac{\frac{F^4v^2t^2}{c^2}}{M^2}? Is this going to be A? What about E^2-E^2?

I am a terrible problem solver. There is quicker way through this, and trying to get through my thick skull as noble as it is, is not it.

 

[noumed]

I need to ask so i understand this more clearly, you want me to substitute the B's of (A^2+A^2B+B^2) for \frac{\frac{F^2vt}{c}}{M}, but what about \frac{\frac{F^4v^2t^2}{c^2}}{M^2}? Is this going to be A? What about E^2-E^2?

I am a terrible problem solver. There is quicker way through this, and trying to get through my thick skull as noble as it is, is not it.

If you let
B = \frac{\frac{F^2vt}{c}}{M}
then
B^2 = \frac{\frac{F^4v^2t^2}{c^2}}{M^2}

Right?

Edit:
and
A = E
so
A^2 = E^2

 

[ManyNames]

If you let
B = \frac{\frac{F^2vt}{c}}{M}
then
B^2 = \frac{\frac{F^4v^2t^2}{c^2}}{M^2}

Right?

Edit:
and
A = E
so
A^2 = E^2

Right...

 

[noumed]

So if you just keep in mind that A and B represents those things, once you've factored
(A^2 - A^2B + B^2)
You can plug in the original values of A and B into your answer to get what you wanted in the first place.

 

[ManyNames]

So why is there two A^2's in (A^2-A^2B+B^2) - just a typo?

 

[noumed]

Because
A = E
And there are two E^2's in the original expression.

 

[ManyNames]

Oh no, i get it now... my mistake

 

[ManyNames]

Oh please can you just show me. I am trying over at this end, but none of the answers i got i don't like. Just tell me please?

 

[ManyNames]

Is this what you are on about? (A^2-B)(A^2+B^2)... No?

 

[noumed]

It's similar to that, but there's a sleight of hands because of the -A^2B term inside. I gave you a pretty good hint -(A^2(B-1) - B^2)... You're almost there! =D

 

[ManyNames]

Patience please, but what is the (minus one) part? Why has it been plugged in there? I need to know these things, otherwise i am lost. I assume it has something to do with subtracting one of the E^2?

 

[noumed]

Nope the minus sign in there is just another sleight of hands. Multiply it out and you should verify that:
-(A^2(B-1) - B^2) = A^2 - A^2B + B^2

 

[ManyNames]

Well, i get A^22B^2=A^2-A^2B+B^2? This aint right is it?

 

[noumed]

Can't quite follow what you did there... What I did was basically rewrite the expression
A^2 - A^2B + B^2
because I do not know how to factor that expression at that state, but I do know how to factor
(B^2 - A^2(B-1)) or -(A^2(B-1) - B^2)

All three expressions are equivalent to each other.

 

[ManyNames]

Then i too admit i do not know how to factor the latter state, so please assist me in my teaching. Cheers,

 

[ManyNames]

And take me through it too...?

 

[noumed]

I suppose you've struggled far enough.
Let's look at:
-(A^2(B-1) - B^2)
We agree that the expression is equivalent to
(A^2 - A^2B + B^2)
Which is equivalent to your original complex expression that you wanted to factor.

Disregard the minus sign and the (B-1) on the first expression, and you're left with something you might be familiar with:
(A^2 - B^2)
Which is the difference of squares that I mentioned about.
We know how to factor that, right?
(A-B)(A+B)

Now, what to do with (B-1} that we left out? Since we're dealing with difference of squares, let's play with \sqrt{(B-1)} and stick that to A to get:
(A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)
If you put the minus sign back into the expression above to get
-(A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)
And multiply it out, you should get back to the original expression:
-(A^2(B-1) - B^2)

After you've verified that
-(A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)
is indeed the factor of the first expression of this post, then all you have to do is replace A and B with its original forms.

So there. You've basically factored that complex equation with a few sleight of hands coupled with trial and error.

 

[ManyNames]

Certainly NOT what i had in my mind lol

Thank you, i will memorize this process for future solving. Nice to meet you, and thanks for your patience.

 

[ManyNames]

That's brilliant. It works out fine.

 

[noumed]

No problem. The idea is not to memorize any cookbook process. It's to break down a complex problem into small parts that you are familiar with. That's why in most exams your professor or teacher might ask you to state all the known equations that are relevant to the problem before you attempt a solution. Keep that in mind, practice diligently, and you should do great.

 

[ManyNames]

Argh, I'm dead chuffed. I would never have been able to solve it on my own... well maybe, but i would have took time.