Question

(Algebra 2 Honors)
Expand the following using the Binomial Theorem and Pascal’s triangle.
1. (x + 2)^6
2. (x – 4)^4
3. (2x + 3)^5
4. (2x – 3y)^4
5. In the expansion of (3a + 4b)^8, which of the following are possible variable terms? Explain your reasoning.
a^2b^3; a^5b^3; ab^8; b^8; a^4b^4; a^8; ab^7; a^6b^5

Answer 1

What do you know about the binomial theorem?

Answer 2

Like, nothing. .-.

Answer 3

ouch

Answer 4

\[(a +b)^{n} = a^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right)a^{n-1} b + \left(\begin{matrix}n \\ 2\end{matrix}\right)a^{n-2} b^{2} + ...+b^{n} \]

Answer 5

hopefully you have learned combinations

Answer 6

Holy poop on a shoe. o_o;;; Not yet.

Answer 7

lol

http://en.wikipedia.org/wiki/Combinations

Answer 8

I can't go to Wikipedia. I have Parental Controls on my computer (My mom sucks sometimes. -_-) and Wiki isn't one of the sites I'm allowed on. I can only get on OS and my school site(I do virtual school)

Answer 9

lol smart mom , its like you really are at school

Combinations
\[\left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{n!}{k! (n-k)!}\]

Answer 10

Hopefully you have learned factorials...

Answer 11

This is all like jibberish to me, guys.. ;-;

Answer 12

@Ranga Can you please help me, my math angel??

Answer 13

You simply need to plug into the binomial expansion formula given above.
You will have to know how to evaluate combinations. Only then you will be able to use the binomial formula.

Answer 14

Well poo.. I'm in trouble then. I have no idea what any of that is, or how to. My lessons are confusing, and this one doesn't explain well. It's like it's trying to be confusing and vague because it's the Honors assignment. .-.

Answer 15

Do you have access to math textbooks or books from the library that would explain these concepts? Because a textbook should be the primary source of learning and websites like these could serve to clarify some difficult points. But learning binomial theorem or calculus from scratch through websites like these is hard because you have to type every answer and typing mathematical equations is too time consuming and explaining concepts by typing is not the efficient way to learn things.

Answer 16

I don't have any math text books or anything. All have is this site, and my online lessons.

Answer 17

Borrow a good math book from a local library. Or go through your online lessons.

Answer 18

I don't have time to go to a library and look through books.. So I guess I'll just go back and try reading the part of the lesson that covers this set of questions.. They're introducing a bunch of new stuff in one lesson, and it's all for one huge assignment in five parts that all have to be submitted together.. And each task is harder than the last, according to the teacher.

Answer 19

I can provide the answers but that won't help you to learn and it would be unfair to students who don't have easy access to answers to get good grades. The concept is fairly straightforward and easy to learn. Once you understand the concept these problems are interesting to do and you may even enjoy doing homework instead of treating it as an unwanted chore that you are being forced to do. There are plenty of excellent online resources as well and you can ask your parents to allow you access to some popular and legitimate websites such as khanacademy, mathisfun, etc.

Answer 20

Oh holy crap, it's just teaching a more complicated way to factor, basically. OMG.

Answer 21

I think I can actually do this one on my own.. Would you mind checking my work for me, though, just to be safe, when I'm done??

Answer 22

I can do that.

Answer 23

Thank you! Heck with doing it the lesson's way, I'll do it my way and probably still get the right answer. LOL The lesson is basically a more complicated factoring system.

Answer 24

whichever method works for you...

Answer 25

Okay, @ranga, here's numbers 1 through 4.. How'd I do?

(x + 2)^6
(x + 2)(x + 2)(x + 2)(x + 2)(x + 2)(x + 2)
x^6 + 64x^6 + 64

(x – 4)^4
(x – 4)(x – 4)(x – 4)(x – 4)
x^4 – 256x^4 – 256

(2x + 3)^5
(2x + 3)(2x + 3)(2x + 3)(2x + 3)(2x + 3)
32x^5 + 243

(2x – 3y)^4
(2x – 3y)(2x – 3y)(2x – 3y)(2x – 3y)
16x^4 - 81y^4 + 1,296xy^4

Answer 26

@dape Since my math angel is offline, could you check these answers for me, please?

Answer 27

Your problem said to use the binomial theorem and Pascal's triangle.

Answer 28

First we will use the binomial theorem which gives us the exponents.

Answer 29

Are you with me so far?

Answer 30

Yup!

Answer 31

Let's forget about the coefficients for now.
(a+b)^6=
\[a^6b^0+a^5b^1+a^4b^2+a^3b^3+a^2b^4+a^1b^5+a^0b^6\]

Answer 32

Notice that the a exponents decrease and the a exponents increase and that the sum of the exponents is always 6.

Answer 33

Notice also that in your problem all "a's" should be replaced with x and all "b's" should be replace with 2.

Answer 34

Are you still with me?

Answer 35

Yeah, I'm with ya.

Answer 36

\[(a + b)^{6} = \sum_{k = 0}^{6}\left(\begin{matrix}6 \\ k\end{matrix}\right)a ^{6 - k}b ^{k} \quad where \quad \left(\begin{matrix}6 \\ k\end{matrix}\right) = \frac{ 6! }{ (6-k)! k! }\]

Answer 37

So now we should do the coefficients which come from Pascal's triangle:

Answer 38

You can expand the above summation by putting in k = 0,1,2,3,4,5,6 and adding up the individual terms.
\[(a + b)^{6} = \sum_{k = 0}^{6}\left(\begin{matrix}6 \\ k\end{matrix}\right)a ^{6 - k}b ^{k} = \left(\begin{matrix}6 \\ 0\end{matrix}\right)a ^{6 - 0}b ^{0} + \left(\begin{matrix}6 \\ 1\end{matrix}\right)a ^{6 - 1}b ^{1} + \left(\begin{matrix}6 \\ 2\end{matrix}\right)a ^{6 - 2}b ^{2} + \]

Answer 39

\[\left(\begin{matrix}6 \\ 3\end{matrix}\right)a ^{6 - 3}b ^{3} + \left(\begin{matrix}6 \\ 4\end{matrix}\right)a ^{6 - 4}b ^{4} + \left(\begin{matrix}6 \\ 5\end{matrix}\right)a ^{6 - 5}b ^{5} + \left(\begin{matrix}6 \\ 6\end{matrix}\right)a ^{6 - 6}b ^{6}\]