Question

I need to factor a four part polynomial. Z^3+3z^2-81z-243

Answer 1

Have you tried grouping?

Answer 2

\[Z ^{3}+3z ^{2}-81z-243\]

Answer 3

yes but I do not remember how to remove all the factors.

Answer 4

give me one moment and let me try it on my own and I will tell you what I come up with.

Answer 5

Good plan :)

Answer 6

\[z ^{2}(z+3)*-9z(-9+27)\]

Answer 7

but I don't know where to go from there.

Answer 8

well, the first step looks great, but how did you take a z out of the 243....?

Answer 9

whoops! it should -9(9z+27)

Answer 10

sorry!

Answer 11

is that as simplified as it can go though??

Answer 12

could it also be (z^2-9) (10z+30)?

Answer 13

Nope. You were on the right track before.

Lets look at -81z-243

What are the like terms here?

Answer 14

-81, -243

Answer 15

so how many times does a -81 go into both... ;)

Answer 16

oh. LOL alright!

Answer 17

now what do you have so far with grouping?

Answer 18

\[z ^{2}(z+3) -81(z+3)\]

Answer 19

see how z+3 is in both groups??

so now if I told you that the factor would be

\[(z^2-81)(z+3)\]

How do you think I got that?

Answer 20

so now I have \[(z ^{2}-81)(z+3)^{2}\]

Answer 21

wait don't you have to square the last set of parenthesis?

Answer 22

because it's a COMMON term. there's a more in depth explanation I can go into, but basically, just remember you do NOT square the terms left over.

Answer 23

ok.

Answer 24

How do I give you medals?

Answer 25

nevermind I figured it out, you should have one for the last answer too.

Answer 26

like that :P

So now to solve for z, you would set BOTH set = 0

so you have \[z^2-81=0\]
and
\[z+3=0\]

Solve for z, and you should have three terms. Let me know what you get.

Answer 27

no I don't think I have to solve it, it was the top term in a rational expression problem.

Answer 28

oh nice, haha okay!

Answer 29

\[z ^{3}+3z ^{2}-81z-243\div z ^{2}+3^{2}\] except that in place of the division sign think upper and lower, numerator/denominator.

Answer 30

I have to divide and simplify these irrational expressions for my college checkpoint, and that is fine, except that last week we did polynomials and I absolutely cannot retain those lessons for some reason.

Answer 31

well know that you know you have (z^2-81)(z+3) on top, what do you do with the bottom?

Answer 32

ummm I hadn't thought that far yet, this was the first exercise with a 4 part polynomial on the top, and the first one with a binomial on bottom. but I made it that way, it is actually a division exercise for two rational expressions. That was as far as I simplified it.

Answer 33

The original equation is \[(z ^{2}-81/z)\div(z-9/z+3)\]

Answer 34

Think what can you cancel out from the numerator and the denominator.

Answer 35

once again think of fraction instead of equation, so I flipped the z-9/z+3 around and multiplied the reciprocal to get the z^3+3z^2-81z-243/z^2+3z

Answer 36

I'm kinda stuck. I think I did something wrong when I multiplied the reciprocals.

Answer 37

Right, I got where you got it!

you're right ALL the way up to the
\[(z^2-81)(z+3) \over z^2+3z\]

Answer 38

now what can you do with the denominator?

Answer 39

right, I actually used FOIL on the top to make a polynomial because I thought the bottom binomial was already factored... hmm lemme look.

Answer 40

I have no clue. I don't remember how to factor binomials either.

Answer 41

wait. z(z+3)

Answer 42

YEAH!

so you have
\[(z^2-81)(z+3) \over z(z+3)\]

can you cancel anything out???

Answer 43

yep, z+3... how the heck are you making that bar so perfect? I don't know how to do that!

Answer 44

I have to go, I have work in the morning. I will be online again tomorrow night if you are on I would love some more help, otherwise I'm sure someone else will pick up where we left off. Thank you!

Answer 45

\over :)

and definitely! Have a great night.

Answer 46

just fyi:
\[(z-9)(z+3)(z+9)\]