Question

Use the binomial theorem to expands the binomial. (d-4b)^3

Answer 1

Anybody? Help please.

Answer 2

"explains"? Do you mean "expand"?

The binomial theorem says
\[(x+y)^n=\sum_{k=0}^n\binom nkx^{n-k}y^k\]where \(\dbinom nk=\dfrac{n!}{k!(n-k)!}\). In your example, you can set \(x=d\), \(y=-4b\), and \(n=3\).

Then
\[\begin{align*}
(d-4b)^3&=\sum_{k=0}^3\binom 3kd^{3-k}(-4b)^k\\[1ex]
&=\binom 30d^3(-4b)^0+\binom 31d^2(-4b)^1\\
&\qquad+\binom 32d^1(-4b)^2+\binom 33d^0(-4b)^3\\[1ex]
&=\cdots
\end{align*}\]

Answer 3

@HolsterEmission Yes thats what I meant, Spellcheck must have changed it before I posted the question. I still don't get it? What's the answer?

Answer 4

I don't understand this at all

Answer 5

@sunnnystrong Do you happen to know anything about this?

Answer 6

Hmm.. not really sure. I'm pretty sure you just expand it? @holsteremission explained it pretty well.

Answer 7

I don't really know anything about this stuff, I just started it so what @holsteremission said didn't really make any sense to me

Answer 8

Hey Love!

It's a `bi`nomial because you have `two` terms inside of the brackets.

When you expand a binomial, the powers `count down on the first term`,
and `count up on the second term`.

\(\rm (x+y)^5=\text{__ }x^5y^0+\text{__ }x^4y^1+\text{__ }x^3y^2+\text{__ }x^2y^3+\text{__ }x^1y^4+\text{__ }x^0y^5\)

Each term in the expansion gets a coefficient.
If the formula for binomial coefficient is confusing,
you can instead look to Pascal's Triangle.

This is what the 5th row looks like

1 5 10 10 5 1

Those will be our coefficients,\[\rm (x+y)^5=1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5\]

Answer 9

I feel kinda of stupid, I just switched to a new school and this is what they are studying and i've never even seen this stuff before @sunnnystrong

Answer 10

Try to understand what I posted.
It's really just a pattern to follow.

Answer 11

@zepdrix Okay I think I understand the pattern

Answer 12

@Love_333 you're not stupid haha XD there are also online youtube videos that can help. or khanacademy (just use google)

Answer 13

Think of your binomial like this,\[\large\rm (d+-4b)^3\]Treat the subtraction as if it's part of the second term.

Answer 14

So we're going to have a bunch of terms added together after we expand,\[\large\rm =\text{__ }d^3(-4b)+\text{__ }d^2(-4b)+\text{__ }d^1(-4b)+\text{__ }d^0(-4b)\]The first term, d, is counting down from 3.

Answer 15

And the second term (-4b) is counting up from 0,\[\large\rm =\text{__ }d^3(-4b)^0+\text{__ }d^2(-4b)^1+\text{__ }d^1(-4b)^2+\text{__ }d^0(-4b)^3\]

Answer 16

And then if we look at Pascal's Triangle,
the third row (counting the top row as the 0th row) will give us our coefficients.

1 3 3 1

Answer 17

\[\large\rm =1d^3(-4b)^0+3d^2(-4b)^1+3d^1(-4b)^2+1d^0(-4b)^3\]

Answer 18

So we've got the weird pattern stuff out of the way.
Now it's just a matter of simplifying.

Answer 19

Start by replacing the 0 power terms with 1's.
(Anything to the zero power is 1).\[\large\rm =1d^3(1)+3d^2(-4b)^1+3d^1(-4b)^2+1(1)(-4b)^3\]

Then suppress any 1's that you see.

1 exponents don't mean anything.
And also 1 multiplication doesn't mean anything significant.\[\large\rm =d^3+3d^2(-4b)+3d(-4b)^2+(-4b)^3\]It's starting to look a little better, ya?

Answer 20

Yeah, I think so. So whenever you have an equation like this if there are ones do you always take them out like that?

Answer 21

Yes, one is a stupid number.
Always get rid of them if you can :)

Answer 22

Okay, thats good to know!

Answer 23

From here, we need to expand any powers of (-4b).

Recall that when you apply an exponent to a group,
you apply it to each thing in the group,
\(\large\rm (ab)^x=a^xb^x\)

Answer 24

So then,
\(\large\rm (-4b)^2=(-4)^2b^2\)

We'll have to do this with the last term as well.

Answer 25

Okay, so what's the next step after that?

Answer 26

Simplify the stuff involving the 4's.

(-4)^2 = ?
(-4)^3 = ?

Answer 27

16 and -64 ?

Answer 28

\[\large\rm =d^3+3d^2(-4b)+\color{orangered}{3d(-4b)^2+(-4b)^3}\]Ya, so apply the power expansion I was talking about before,\[\large\rm =d^3+3d^2(-4b)+\color{orangered}{3d(-4)^2b^2+(-4)^3b^3}\]and put in those values,\[\large\rm =d^3+3d^2(-4b)+\color{orangered}{3d(16)2b^2+-64b^3}\]

Answer 29

Recall that multiplication is `commutative`,
this means that we can multiply things in any order.

So instead of having 3d^2(-4b),
we would like to think of it like this,
-4*3d^2b
bringing all of the number stuff to the front.

Answer 30

You want to do that with all of them.

Answer 31

So our second term would look like this after multiplying the number stuff,\[\large\rm =d^3-12d^2b+3d(16)2b^2+-64b^3\]with a -12 in front.

Do that with the other terms as well.

Answer 32

Okay so.... \[d^3 = 12d^2b + 48d^2b - 64d^3\] ? Did I do that right?

Answer 33

I feel like I didnt XD

Answer 34

You accidentally put an equals sign in place of the subtraction sign.
Looks good besides that though!

Yayyy good job \c:/

Answer 35

I know I just saw that! Really??

Answer 36

Oh oh and also that third term should be db^2 not d^2b

Answer 37

Oh... and the last term should be b^3 not d^3
I guess I spoke too soon :D lol

Answer 38

Oh crap, I messed up with my letters I was too focused on the numbers sorry

Answer 39

XD

Answer 40

But it looks good as long as I correct those things?

Answer 41

Yes.

Answer 42

Okay great! Thank you sooo much

Answer 43

np c: