Solving (2c - 3d)^5: Help Needed!

[Bucs44]

I'm really confused when it comes to this stuff. I'm stuck on this problem:

(2c - 3d)^5

I have no idea where to begin. Please help!

 

[arildno]

Set a=2c, b=-3d.
Hence,
$$(2c-3d)^{5}=(a+b)^{5}$$
Does that help you?

 

[dontdisturbmycircles]

There is a deep connection between Pascal's Triangle and the binomial theorem. I suggest you look up the terms of pascal's triangle. Secondly, the general expansion of the binomial theorem goes like this:

$$_{n}C_{0}(a)^{n}(b)^{0}+_{n}C_{1}(a)^{n-1}(b)^{1}+...+_{n}C_{n}(a)^{0}(b)^{n}$$

where as you can see, if the a is raised to the power of k, b is raised to the power of n-k ALWAYS, so your exponents should add up to n. n is the power of the binomial, 5 in this case. Your binomial expansion should have n+1 terms in it (in this case 6).

And follow arildno's suggestion to set a=2c ($$(2x)^{2}=4x^{2}$$ by the way, not $$2x^{2}$$) and b as -3d. and expand it.

 

[Bucs44]

I'm still confused on that.

Breaking down the possibilities is where I get confused:

(a + b)^5 = aaaaa + aaaab + aaabb + aabbb + abbbb + bbbbb

I'm lost!

 

[matt grime]

You know how to do long multiplication, right? Well, this is long multiplication. So long multiply*. Failing that just use the formula.* if you want to multiple 47 by 31 in your head, how do you do it? Don't br confused and think that this is hard - it is arithmetic you were taught as a small child, just with letters. Oh, and the answer is 40*30 + 40*1 +7*30+ 7*1, remember? (This was possibly retaught unnecessarily as FOIL, if that helps).

And if you want to check your logic, what is 2^5? And what is (1+1)^5?

 

[dontdisturbmycircles]

I am guessing that you know (or should know) the binomial theorem! Since after all this thread is titled "binomial theorem".

Read this, http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut54_bi_theor.htm . I would show you all of it but I think that will suffice. Please post any concerns you have after you read that, I am sure that someone will be glad to help you Bucs44 (I will if I am still here)

Your expansion is an okay start to the understanding (although not very good yet) you are forgetting a very important part.. the $$_{n}C_{0}$$ and $$_{n}C_{1}$$ etc.

Please note when you read that page, $$(^{n}_{1})$$ represents $$_{n}C_{1}$$.. It's just different notation.

 

[Bucs44]

Ok - Here's where I am at this point:

5!/5!0! (32X^5) + 5!/4!1! (16x^4)(-3d) + 5!/3!2! (8x^3)(9d^2) + 5!/2!3! (4x^2)(-27d^3) + 5!/1!4! (2x)(81d^4) + 5!/0!5! (-243d^5)

Here is where I'm stuck - I have 32^5 + 5*4*3*2/4!1! ?? Is this right?

 

[dontdisturbmycircles]

Note that

$$\frac{5!}{5!0!}$$ = $$_{5}C_{0}$$

I just want to point this out because most calculators have the ability to evaluate such expressions. On TI calculators it is ussually under MATH->Probability or Statistics.

Okay, but that does not mean you shouldn't understand what is going on here, so let me explain it.

Lets look at the first term, which is correct if we are now calling the variable c x, (look at your first post).

$$\frac{5!}{5!0!}=1$$, because you cancel the 5!'s and 0! = 1.

So the first term in the expansion is just $$32x^{5}$$

For the next term you are correct once again. Now you have to simplify
$$\frac{5!}{4!1!}$$
Now since
$$5!=5x4x3x2x1$$
and
$$4!=4x3x2x1$$
we have
$$\frac{5x4x3x2x1}{4x3x2x1x1}$$
cancelling out gives us
$$\frac{5}{1}=5$$
so our second term is
$$5(16x^{4})(-3d)=-240x^{4}d$$

For the third term you are correct once again, now you should get the pattern. We look at what all the factorial notation really means and simplify it
$$\frac{5!}{3!2!}=\frac{5x4x3x2x1}{3x2x1x2x1}$$
and now we cancel out and get
$$\frac{5x4}{2x1}=10$$
so for the third term we have
$$10(8x^{3})(9d^{2})$$
which equates to
$$720x^{3}d^{2}$$

Can you finish it from here?

Note that the first three terms in the expansion are $$32x^{5}-240x^{4}d+720x^{3}d^{2}$$ and you should have 3 more terms after that.

 

[HallsofIvy]

Ok - Here's where I am at this point:

5!/5!0! (32X^5) + 5!/4!1! (16x^4)(-3d) + 5!/3!2! (8x^3)(9d^2) + 5!/2!3! (4x^2)(-27d^3) + 5!/1!4! (2x)(81d^4) + 5!/0!5! (-243d^5)

Here is where I'm stuck - I have 32^5 + 5*4*3*2/4!1! ?? Is this right?

? Where did you get that from? The only "32" I see in you first expression is (32X^5) what happened to the X?

5!/5!0! is equal to 1 so the first term is 32x^5 (what happened to the c and d of your original problem?). The NEXT term is then 5!/4!1!(16c^4(-3d) (I've replace x by c which is what I think you meant) 5!/4!1!= 5 so this is 5(16c^4)(-3d)= -240c^4d.

One easy way to calculat things like 5!/4!1! or 5!/3!2! is to factor as much as possible: 5*4*3*2*1/4*3*2*1 and everything in the denominator cancels: this is equal to 5. 5!/3!2!= 5*4*3*2*1/(3*2*1)(2*1) and we can cancel the 3*2*1 in both numerator and denominator leaving 5*4/2= 10.

Of course, I would probably quickly write down Pascal's triangle as suggested earlier.

 

[Bucs44]

Dontdisturbmycircles,

I can't thank you enough for your help - It makes sense now! Here's the final answer: 32x^5 - 240x^4d + 720x^3d^2 - 1080x^2d^3 + 810xd^4 - 243d^5

 

[dontdisturbmycircles]

Correct.

No problem, glad to help.