Expand (4 + x - 2y)^4 using the binomial theorem

[JSmith]

Expand (4 + x - 2y)⁴ using the binomial theorem.

How do I do this?

 

[Unknown008]

You can put two things together, and it would be better not to group the variables together, say, let \(\displaystyle a = 4+x\)

\(\displaystyle (a - 2y)^4\)

Can you expand this one?

[Note that I could have used \(\displaystyle a = 4-2y\) instead, that would work too!]

 

[soroban]

Hello, JSmith!

We are expected to use the Binomial Theorem twice.

\(\displaystyle \text{Expand }(4 + x - 2y)^4\,\text{ using the binomial theorem.}\)

We know: .\(\displaystyle \begin{Bmatrix}(a\pm b)^2 \:=\:a^2 \pm 2ab + b^2 \\ (a \pm b)^3 \:=\:a^3 \pm 3a^2b + 3ab^2 \pm b^3 \\
(a\pm b)^4 \:=\: a^4 \pm 4a^3b + 6a^2b^2 \pm 4ab^3 + b^4 \end{Bmatrix}\)


We have: .\(\displaystyle \big[(x-2y) + 4)\big]^4\)

. . . . . . \(\displaystyle =\;(x-2y)^4 + 4(x-2y)^34 + 6(x-2y)^24^2 + 4(x-2y)4^3 + 4^4\)

. . . . . . \(\displaystyle =\;(x-2y)^4 + 16(x-2y)^3 + 96(x-2y)^2 + 256(x-2y) + 256\)


Now expand each of those binomials:

\(\displaystyle \big[(x-2y) +4\big]^4 \;=\;\begin{Bmatrix}x^4 - 4x^3(2y) + 6x^2(2y)^2 - 4x(2y)^3 + (2y)^4 \\ +16\left[x^3 - 3x^2(2y) + 3x(2y)^2 - (2y)^3\right] \\ +96(x^2-4xy+4y^2) \\ +256(x-2y) \\ +256 \end{Bmatrix}\)


And simplify . . .

 

[JSmith]

That is what I got for the expanded version... How do I simplify this?

 

[JeffM]

View attachment 2757

That is what I got for the expanded version... How do I simplify this?

Before you simplify, you need to check your work.

For example, your fifth term above should be \(\displaystyle 16y^4,\ not\ 16y.\)

Once you have everything multiplied out, you will notice that you have terms where the sum of the exponents on x and y is 4, terms where the sum of the exponents on x and y is 3, and so on down to a term where the exponents on x and y sum to zero. Any two terms where the sum of the exponents differ cannot be combined, and obviously those terms with different x exponents or different y exponents cannot be combined. There are five terms where the x and y exponents can sum to 4, four terms where the x and y exponents can sum to 3, down to one term where the x and y exponents can sum to zero. So your final answer should have 15 terms. Messy problem.

 

[JSmith]

You can put two things together, and it would be better not to group the variables together, say, let \(\displaystyle a = 4+x\)

\(\displaystyle (a - 2y)^4\)

Can you expand this one?

[Note that I could have used \(\displaystyle a = 4-2y\) instead, that would work too!]

using your method, I got this:


(there should not be an x before the (4+x)^4

does this need to be simplified? Or can i simply rearrange the terms

 

[JeffM]

using your method, I got this:
View attachment 2758

(there should not be an x before the (4+x)^4

does this need to be simplified? Or can i simply rearrange the terms

Unknown and soroban were proposing different versions of the same method.

\(\displaystyle a = 4 + x \implies \left(4 + x - 2y\right)^4 = \left(a - 2y\right)^4 = a^4 - 4a^3(2y) + 6a^2(2y)^2 - 4a(2y)^3 + (2y)^4 = a^4 - 8a^3y + 24a^2y^2 -32ay^3 + 16y^4.\)

You seemed to be OK up to there, but then made at least one error in expanding the powers of (4 + x).

 

[Unknown008]

using your method, I got this:
View attachment 2758

(there should not be an x before the (4+x)^4)

does this need to be simplified? Or can i simply rearrange the terms

\(\displaystyle = (4+x)^4 + 4(4+x)^3(-2y)^1 + 6(4+x)^2(-2y)^2 + 4(4+x)(-2y)^3 + (-2y)^4\)

This line is correct. However, you need to apply the expansion correctly! This becomes:

\(\displaystyle = (4+x)^4 - 8y(4+x)^3 + 24y^2(4+x)^2 - 32y^3(4+x) + 16y^4\)

Which simplifies to:

\(\displaystyle = [(4^4) + 4(4^3)(x^1) + 6(4^2)(x^2) + 4(4^1)(x^3) + (x^4)] - 8y[4^3 + 3(4^2)(x^1) + 3(4^1)(x^2) + x^3] + 24y^2[16+8x+x^2] - 32y^3(4+x) + 16y^4\)

Which you expand:

\(\displaystyle = [256 + 256x + 96x^2 + 16x^3 + x^4] - 8y[64 + 48x + 12x^2 + x^3] + 24y^2[16+8x+x^2] - 32y^3(4+x) + 16y^4\)

\(\displaystyle = 256 + 256x + 96x^2 + 16x^3 + x^4 - 512y - 384xy - 96x^2y - 8x^3y + 384y^2+192xy^2+24x^2y^2 - 128y^3 - 32xy^3 + 16y^4\)

And that's it. Since we proceeded with all the terms independent of \(\displaystyle y\), then in \(\displaystyle y\), then in \(\displaystyle y^2\) and last in \(\displaystyle y^3\) and \(\displaystyle y^4\), there aren't any terms you can simplify.

 

[JSmith]

Thank you all so much for your help... I see now! You guys are great!