Expand powers of binomial expressions

[3301]

Homework Statement

The problem comes in second term after 4(2z...)

Homework Equations

The Attempt at a Solution

so i got 4 (8z^3) 5k.
32 (z^3) * (20k). After that i left it like that but I supposed to get answer like this
160 z^3 * k

I can conjoin constants of z and k variables? Or am I missing something else

 

[SammyS]

Homework Statement

View attachment 95738
The problem comes in second term after 4(2z...)

Homework Equations

The Attempt at a Solution

so i got 4 (8z^3) 5k.
32 (z^3) * (20k). After that i left it like that but I supposed to get answer like this
160 z^3 * k

I can conjoin constants of z and k variables? Or am I missing something else

You are attempting to distribute multiplication over multiplication.

a×(b×c) ≠ (a×b)×(a×c)

Added in Edit:
To put it another way:

32 (z^3) * (20k) is just an extended multiplication. There is no adding going on.

32 (z^3) * (20k) is simply (32)⋅(z³)⋅(20)⋅(k)

 

[HallsofIvy]

Homework Statement

View attachment 95738
The problem comes in second term after 4(2z...)

Homework Equations

The Attempt at a Solution

so i got 4 (8z^3) 5k.
32 (z^3) * (20k).

4*8= 32 but where did the "20" come from? 4(8)(5)= 32(5)= 160.

After that i left it like that but I supposed to get answer like this
160 z^3 * k

I can conjoin constants of z and k variables? Or am I missing something else