Question

Factor completely:

2z(4x-y)-z(4x-y)

Answer 1

Can you do this one: 2wr - wr

Answer 2

2w^2r^2

Answer 3

??

Answer 4

I'm having troubles

Answer 5

No good.

2wr - wr

Common w

w(2r - r)

Common r

wr(2 - 1) = wr(1) = wr

How did you get squares?

Answer 6

but I thought it would be (2z-z)(4x-y)

Answer 7

a bit confusing.

Answer 8

Look for common factors

2z(4x-y)-z(4x-y)

Common z

z[2(4x-y)-(4x-y)]

Common (4x-y)

z(4x-y)[2 - 1] = z(4x-y)[1] = z(4x-y)

Yours is fine.

(2z-z)(4x-y)

You just need to find the z and you'll nearly be done.

Answer 9

thank you for explaining :)