At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Well, what did you write?

Do you know what a GCF is?

Yes, greatest common factor

I just don't know where to start..

Well, I would work it in reverse.

Do you think you could explain to me how i could factor out the variables?

3y^2 and 9x^2. Correct?

-6x^4y^5-15x^3+9^3?

Yes, please. Hahaha, sorry I'm not getting this right away.

\(5x^3y^2+3xy^3 \)
\(xy^2(5x^{3-1}y^{2-2}+3x^{1-1}y^{3-2}) \)
\(xy^2(5x^2+3y) \)

How do you know what to subtract the exponents by?

Oh, i see

So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)

Okay, I will be right back!

Would it be this? 3x^2y^2(-2x^2y^3-5x+3y)?

Exactly!

I get it now, do you think you could help with one more? I feel as if it is much simpler

Sure.

15x^4 + 9x^3 +8x

OK. So, you need to factor this, right?

Yes, and i think you would take out the x , correct? since its x^1?

Yep!

I need help with this next step

Which next step?

When you subtract from the exponents

So, it would now be 15x^4-^1 +9x^3-^1

Almost. The 8 still stays there... just the x goes away.
\(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)

Oh, haha i forgot about the 8

=)
And after the subtraction, that is:
\(x(15x^3 + 9x^2 +8)\)
So, anything else you can factor?

x^2

You already factored the x out of the 8, so you can't really factor x more.

Then what can i factor? The constant?

Well, are there any common factors in the constants?

I would say no, right?

I messed something up actually, the 9x^3 is supposed to be 9y^3.

Oops.... Hmmm.... well... that would make it different.

So it is?
\(15x^4 + 9y^3 +8x\)

correct

Hmmm... well, there is no GCF at all in that.

Since I made this equation I could keep the variable x

Sorry, I got disconnected

What would I multiply again?

So, the y then goes next to the other variables except the one in front?

And after I multiply it, it is no longer out front....

Oh okay, I see! Thanks for helping me =)