Question

Explain, in complete sentences, how you would expand (3x + 7y)^4 using Pascal’s Triangle.

I got 3x^4+84x^3y+126x^2y^2+84xy^3+7y^4. Can anyone verify to see if this is correct? :) Thank you very much.

Answer 1

It is nt correct.

Answer 2

Oh okay, can you show me what I did wrong?

Answer 3

what is the 5th row of pascal's triangle?

Answer 4

1,4,6,4,1

Answer 5

yes, and those will be your coefficients
now what are your terms?

Answer 6

x^4 + x^3y + x^2y^2 + xy^3 + y^4

Answer 7

no, you have ignored the coefficients in the terms themselves
they must remain...

Answer 8

1x^4+4x^3y+6x^2y^2+4xy^3+1y^4 ?

Answer 9

the terms are
3x and 7y
so you have
1 4 6 4 1
1(3x)^4(7y)^0+4(3x)^3(7y)^1+6(3x)^2(7y)^2+4(3x)^1(7y)^3+1(3x)^0(7y)^4

Answer 10

When I combine everything I still get what I still got from up above. o.O
1st term- 3x^4
2nd term- 84x^3y
3rd term- 126x^2y^2
4th term- 84xy^3
5th term= 7y^4

Am I combining some of these terms wrong?

Answer 11

did you try to include the coefficients from pascals triangle yet or no?
because what you have is not quite right any way I can slice it, but it's closer

Answer 12

I included the 1,4,6,4,1 into the multiplying.

Answer 13

let's just look at the terms first without the coefficients then to see where you went wrong

Answer 14

k

Answer 15

terms:\[\begin{array}91^{st}= (3x)^4(7y)^0=3^4x^4=81x^4\\2^{nd}=(3x)^3(7y)^1=3^3x^3\cdot7^1y^1=567x^3y^1\\3^{rd}=(3x)^2(7y)^2\\4^{th}=(3x)^1(7y)^3\\5^{th}=(3x)^0(7y)^4\end{array}\]you failed to distribute the exponents properly...
I will continue but this should be enough to see your mistake
we then put in the coefficients from Pascal's triangle at the end and add it all up

Answer 16

Oh I forgot to use the exponent on the numbers too. So it should actually be-
81x^4+567x^3y^1+441x^2y^2+1029x^1y^3+2401 correct? :)

Answer 17

\[\begin{array}91^{st}= (3x)^4(7y)^0=3^4x^4=81x^4\\2^{nd}=(3x)^3(7y)^1=3^3x^3\cdot7^1y^1=567x^3y^1\\3^{rd}=(3x)^2(7y)^2=3^2x^2\cdot7^2y^2=441x^2y^2\\4^{th}=(3x)^1(7y)^3=3x\cdot7^3y^3=1029xy^3\\5^{th}=(3x)^0(7y)^4=7^4y^4=2401y^4\end{array}\]yeah but you forgot the y^4 at the end ;)

Answer 18

and I wouldn't write them all summed up like that yet until we use the right coefficients from the triangle

Answer 19

oh yeah careless mistake there :) Wait we still have to multiply this with 1,4,6,4,1 right?

Answer 20

yes, by the 5th row of the triangle we have the coefficients of the terms as\[1(1^{st})+4(2^{nd})+6(3^{rd})+4(4^{th})+1(5^{th})\]

Answer 21

K so the final problem would be 81x^4+2268x^3y+2646x^2y^2+4116xy^3+2401 right? :)

Answer 22

the second coefficient is wrong, but that is due to a typo I made above I can see
find it and fix it! that should be your effort in this :)

Answer 23

besides that everything else is correct though

Answer 24

It should be 756x^3y right?

Answer 25

yes :)

Answer 26

\[(3 x+7 y)^4=81 x^4+756 x^3 y+2646 x^2 y^2+4116 x y^3+2401 y^4 \]

Answer 27

http://www.wolframalpha.com/input/?i=expand%20(3x%20%2B%207y)%5E4%20

Answer 28

Thx a ton @TuringTest for helping me with this problem :)

Answer 29

I know wolframalpha but I wanted to not use anything like that on my test so yeah :) Thx!

Answer 30

Yeah I just gave it for you to check your work. You should never rely on it; it is wrong more often than you would probably think at first.

Answer 31

...and you're very welcome!

Answer 32

Oh okay thx :)