Question

So for math (Algebra II) I'm using pascal's triangle to factor and this is my problem:
Expand (4x – 3y)^4 using Pascal’s Triangle

It just doesn't look right so far, so could someone double-check what i've done put to this moment?
The 4th row of the Pascal triangle is 1,4,6,4,1. So we multiply that to 4x with the powers in descending order and get:

4x^4+16x^3+24x^2+16x+1. Now we multiply that to y, going the opposite (ascending order) with the powers and get:
4x^4(-3y^0)+16x^3(-3y^1)+24x^2(-3y^2)+16x(-3y^3)+1(-3y^4)=

Answer 1

coefficient: 1 4 6 4 1
(4x – 3y)^4
\(= 1(4x)^4 (3y)^0+ 4(4x)^3 (3y)^1+6(4x)^2 (3y)^2+4(4x)^2 (3y)^1+1(4x)^0 (3y)^4\)
=.....

Answer 2

@Callisto, why isn't everything multiplied by -3y?

Answer 3

My bad.... all 3y => -3y

Answer 4

Oh okay no problem.

Answer 5

This is what I got afterwards-
4x^4 - 48x^3y - 72x^2y^2 - 48xy^3 - 3y^4
Is this correct @Callisto?

Answer 6

Nope....

Answer 7

For the first term \(1(4x)^4(-3y)^0 = 4^4 x^4 (1) = 256x^4\)

Answer 8

K then for the next term
768x^3y

Answer 9

-768x^3y*

Answer 10

Tyvm I get it now :) @Callisto

Answer 11

Wow~! You got the second term right! Congrats! And you're welcome! :)

Answer 12

Lol thx :) I was just confused at the end part and multiplying them all together :)

Answer 13

So for the final answer I got-
256x^4 - 768x^3y + 864x^2y^2 - 432xy^3 + 81y^4

Answer 14

Big applause to @Loujoelou :)

Answer 15

Thx :)

Answer 16

Welcome!~