Question

Expand the following using either the Binomial Theorem or Pascal’s Triangle.

Answer 1

Expand the following using either the Binomial Theorem or Pascal’s Triangle.

(2x + 4)^3

Can someone walk me through this?

Answer 2

(2x)^3+3.(2x)^2.4+3.(2x).4^2+4^3

Answer 3

=8x^3+3.4.4.x^2+3.2.16.x+64
=8x^3+46x^2+96x+64

Answer 4

is it clear

Answer 5

I'm not sure what the periods are in there for?

Answer 6

it is very simple like
(a+b)^3 = a^3+3a^2b+3ab^2+b^3

Answer 7

here a=2x,
b=4

Answer 8

hint:
here are the coefficient of the expansion:

Answer 9

coefficients*

Answer 10

Yes, I have (a + b)^3 = 1a^3b^0 + 3a^2b^1 + 3a^1b^2 + 1a^0b^3
Is that right?

Answer 11

yes! It is right!

Answer 12

Awesome, now can you tell me if I plugged everything in correctly?
(a + b)^3 = 1(2x)^3(4) + 3(2x)^2(4)^1 + 3(2x)^1(4)^2 + 1(2x)(4)^3

Answer 13

no, since the correct expression is:
\[\Large {\left( {2x} \right)^3} + 3{\left( {2x} \right)^2} \cdot 4 + 3 \cdot \left( {2x} \right) \cdot 4 + {4^3}\]

Answer 14

Oh, thank you! That is so much simpler. Are there any other steps I have to take after that?

Answer 15

oops.. I have made a typo, here is the right expression:
\[\Large {\left( {2x} \right)^3} + 3{\left( {2x} \right)^2} \cdot 4 + 3 \cdot \left( {2x} \right) \cdot {4^2} + {4^3}\]

Answer 16

@sorryimfrankie
in the last term 1.(2x)^0.4^3

Answer 17

after that expression, you have to compute the subsequent quantities, using the rules of algebra:
\((2x)^3=...?\)
\((2x)^2=...?\)
\(4^2=...?\)
and,
\(4^3=...?\)

Answer 18

(2x)^3 = (6x)
(2x)^2 = (4x)
4^2 = 16
4^3 = 64
Correct?

Answer 19

the last 2 expression are correct.
for the first two expressions, please note that:

\[\huge \begin{gathered}
{\left( {2x} \right)^3} = {2^3} \cdot {x^3} = ...? \hfill \\
{\left( {2x} \right)^2} = {2^2} \cdot {x^2} = ...? \hfill \\
\end{gathered} \]

Answer 20

expressions*

Answer 21

(2x)^3= 2x.2x.2x=8x^3
like this do remaining terms i already did above

Answer 22

I'm sorry dayakar, you're just confusing me

Answer 23

hint:
\[\huge {\left( {2x} \right)^3} = {2^3} \cdot {x^3} = 8 \cdot {x^3} = 8{x^3}\]

Answer 24

ok, whom you are following

Answer 25

Sorry Michele, I didn't see your last response. The 4th one would be 4x^2, right?

Answer 26

Sorry if I'm not getting all of this, my medication is making me kind of slow at things

Answer 27

yes! correct!

Answer 28

Oh awesome, thanks so much! And is that all I have to do?

Answer 29

@Michele_Laino
i'm sorry ,plz continue

Answer 30

next we can write this:
\[\Large \begin{gathered}
{\left( {2x} \right)^3} + 3{\left( {2x} \right)^2} \cdot 4 + 3 \cdot \left( {2x} \right) \cdot {4^2} + {4^3} = \hfill \\
\\
= 8{x^3} + 3 \cdot \left( {4{x^2}} \right) \cdot 4 + 3 \cdot \left( {2x} \right) \cdot 16 + 64 \hfill \\
\end{gathered} \]
now, what is:
\[\Large 3 \cdot \left( {4{x^2}} \right) \cdot 4=...?\]
and what is:
\[\Large 3 \cdot \left( {2x} \right) \cdot 16=...?\]

Answer 31

ok! @dayakar

Answer 32

hint:
we can write this:
\[\Large 3 \cdot \left( {4{x^2}} \right) \cdot 4 = 3 \cdot 4 \cdot 4 \cdot {x^2} = ...?\]

Answer 33

48x^2 and 96x?

Answer 34

correct!

Answer 35

so the final answer is:
\[\Large \begin{gathered}
{\left( {2x} \right)^3} + 3{\left( {2x} \right)^2} \cdot 4 + 3 \cdot \left( {2x} \right) \cdot {4^2} + {4^3} = \hfill \\
\hfill \\
= 8{x^3} + 3 \cdot \left( {4{x^2}} \right) \cdot 4 + 3 \cdot \left( {2x} \right) \cdot 16 + 64 = \hfill \\
\hfill \\
= 8{x^3} + 48{x^2} + 96x + 64 \hfill \\
\end{gathered} \]

Answer 36

Awesome! I guess that wasn't too hard after all. Thank you so much! :)

Answer 37

:)