Question

Will Give medal! Part C: If the two middle terms were switched so that the expression became 3x3y - 3x2y + 12xy - 12y, would the factored expression no longer be equivalent to your answer in part B? Explain your reasoning.

Answer 1

Part B:
By factoring out GCF from first two terms it will be: 3y(x(x2+4)−3x2−12)

Then By factoring the GCF from last two terms it will be:
3y(x(x2+4)−3(x2+4))

By factoring out the (x^2+4) from the terms inside parenthesis it comes out as : 3y(x2+4)(x−3)

Answer 2

@zepdrix

Answer 3

@Gabylovesyou @goformit100

Answer 4

@hedyeh99

Answer 5

Oops, extra 3 in the first term:
\[3x^{3}y-3x^{2}y+12xy - 12y\]

Answer 6

Is this correct?

Answer 7

yes

Answer 8

Well, I would have the first two terms grouped together and the 2nd two terms as a group and factor out their greatest common factor:
\[3x^{3}y - 3x^{2}y\]and
\[12xy-12y\]
Can you see what the greatest common factor would be in each of those two groups?

Answer 9

umm it would be 3 because 3*4=12

Answer 10

Well, i just mean within the two groups separately. So start with:
\[3x^{3}y-3x^{2}y\]What's the GCF of just those two terms?

Answer 11

3xy

Answer 12

There's an even higher power of x that can be removed.

Answer 13

Wait, It would be 3x^2y

Answer 14

Yep. So if I factor that out, I have:
\[3x^{2}y(x-1)\]Now how about the 2nd two terms:
\[12xy-12y\]What would the GCF be out of those 2 terms?

Answer 15

12y

Answer 16

Bingo. So if I rewrite the whole expression with those GCF's factored out, I have:
\[3x^{2}y(x-1) + 12y(x-1)\]Now do you see what I could do next from here?

Answer 17

Combine the (x-1)?

Answer 18

You could say combine in a sense. I have a group of (x-1)'s added to a 2nd group of (x-1)'s. But essentially what we're doing is factoring out an (x-1). So that gives us:
\[(3x^{2}y+12y)(x-1)\]Now we're almost done. That first group within parenthesis has something that can be factored out of it. So what could I take out?

Answer 19

Umm... The 12y? I suppose

Answer 20

12 is too high. The first term in that group only has a 3 in it :)

Answer 21

Okay, so say the y term maybe?

Answer 22

y works. But you can take out a 3. It's just 12 was too much to take out, 3 works fine as a number, though. So you take out 3y. Which gives:
\[3y(x^{2}+4)(x-1)\]That's as simplified as you'd be able to get it. So hopefully that;s enough info to answer your question and understand what I did :3

Answer 23

OKay, so From my answer to part B, 3y(x^2+4)(x−3) it would come out as that they werent equivilent

Answer 24

(x-3) sorry, i don't know why it made all those question marks

Answer 25

The way B was written up top was kind of a mess, but it seems that yes they would be different, lol.

Answer 26

lol okay thank you very much by the way :)

Answer 27

Np :)