# Homework Help: Factoring a complex equation:

1. Nov 11, 2008

### ManyNames

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Nov 11, 2008

### noumed

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

3. Nov 11, 2008

### ManyNames

Like

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

4. Nov 11, 2008

### noumed

that wont 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
$$(A^2 - A^2B + B^2)$$
$$A = E$$
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
$$E^2-E^2(\frac{\frac{F^2vt}{c}}{M})=-p$$?

5. Nov 11, 2008

### ManyNames

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.

6. Nov 11, 2008

### noumed

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.

7. Nov 11, 2008

### ManyNames

Agreed. That way is so much easier.

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

8. Nov 11, 2008

### 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. :)

9. Nov 11, 2008

### 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.

10. Nov 11, 2008

### noumed

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
$$A^2 = E^2$$

11. Nov 11, 2008

### ManyNames

Right...

12. Nov 11, 2008

### 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.

13. Nov 11, 2008

### ManyNames

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

14. Nov 11, 2008

### noumed

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

15. Nov 11, 2008

### ManyNames

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

16. Nov 11, 2008

### 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?

17. Nov 11, 2008

### ManyNames

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

18. Nov 11, 2008

### 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

19. Nov 11, 2008

### 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 $$E^2$$?

20. Nov 11, 2008

### noumed

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

21. Nov 11, 2008

### ManyNames

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

22. Nov 11, 2008

### 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.

23. Nov 11, 2008

### ManyNames

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

24. Nov 11, 2008

### ManyNames

And take me through it too...?

25. Nov 11, 2008

### 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.

Last edited: Nov 11, 2008