Question

Anyone know the easiest way to factor polynomials e.g. (x^2 + x^3)^3?

Answer 1

know the rules maam

Answer 2

I'll take a stab at it

Answer 3

\((x^2 + x^3)^3 = (x^2 + x^3)(x^2 + x^3)(x^2 + x^3)\)

\(=x^2(1 + x)x^2(1 + x)x^2(1+x)\)
\(=(x^2)^3(1+x)^3\)
\(=x^6(1+x)^3\)

That last expression represents the factorization as a product of two factors.

Answer 4

its true you could just simply expand which is more accurate from where you have x^6+3^7+3x^8+x^9 were you can easily see that x^6 is a factor already and so you could divide the polynomial by x^6 to get other factors.

Answer 5

this method is just the use of simple expansion but hero's method is direct factorisation which is easier and faster

Answer 6

Thank you Hero & nelotsak for your help!

Answer 7

You're welcome

Answer 8

I'm really struggling with Calculus right now which I last took in 1998 and have only used a very little bit since then. I am struggling with remembering/understanding the use of derivatives in Calculus which was part of that question I asked. The whole question is actually: Calculate y' if \[y = (x^2 + x^3 )^4\]

Answer 9

Yes, it would be easier to find the derivative of \(y = (x^2 + x^3)^4\)

Answer 10

So I'm gonna try to follow your method you showed me to get the final answer of: \[y \prime = 4x ^{7}(x+1)^{3}(3x+2)\]

Answer 11

Good luck with it. You already have the answer so...

Answer 12

I understand that the derivative of \[x ^{2}+x ^{3} = 3x ^{2}+2x\] but am confused to as to whether or not to factor the 3x & 2x out first or to the forth power. Thank you, again!

Answer 13

Well if you factored out the x you would have \(x(3x + 2)\) which gets you closer to what you want.

Answer 14

Cool. Thank you!

Answer 15

So do you see how to get to the derivative now?

Answer 16

Well when I first started to answer it I get \[4(3x^2 + 2x)^3\] but what I'm wondering is if I'm missing \[(x^2 + x^3)^3 and then finish factoring and solving \it as you showed me earlier\]

Answer 17

That was supposed to say: and then finish factoring and solving as you showed me earlier.

Answer 18

I messed up on one of the steps. I put an equal sign instead of a plus.

Answer 19

I hate that because it is not easy to re-write

Answer 20

Okay, I see it. I was confused how they were set equal to each other

Answer 21

Okay, good. Long as you understand it.

Answer 22

I do. I broke it down a little different and I'm not sure if it is correct:

\[= 4 ( 3x^2 + 2x) (x^2 + x^3) ^3\]
\[= 4x (3x + 2) (x^2)^3 (x +1)^3\]
\[= 4x(x^6)(3x+2)(x+1)^3 = 4x^7(3x+2)(x+1)^3\]

Answer 23

The difference is I started from the very beginning.

Answer 24

I messed up twice. I should never have had a plus symbol in certain spots.

Answer 25

I'm just going to delete what I wrote.

Answer 26

No need. I understood what you you meant.

Answer 27

So how would you do it differently if you need to solve for y' but are unable to factor y completely out of the equation e.g.

\[xy^4 + x^2y = x + 3y\]

Answer 28

You have to do implicit differentiation.

Answer 29

Also, Hero, do you know any general rules of thumb of when to use the sum, scalar multiple, product, quotient, chain, inverse or implicit rule?

Answer 30

Where is your textbook?

Answer 31

All those rules should be in it.

Answer 32

Right in front of me confusing the heck out of me. According to the math placement exam I was ready for Calculus 1 and understood it for a second but now I am LOST!

Answer 33

Which textbook do you have?

Answer 34

I'm looking at them right now. I'm using Calculus 7th ed. by James Stewart, McMaster University & University of Toronto

Answer 35

I see. Trust me, that's an easy book compared to the book I had to use when I was taking Calculus.

Answer 36

Good to hear...I think ;-)

Answer 37

If you want help with calculus, you should try asking some of the other 99 users like wio, king george, satellite73. Ask wio for help.

Answer 38

Okay, thank you. I have been meaning to look into getting a tutor to try to dumb it down for me so that I can start to understand again. I will take your advice. Thank you you again for your help and your time, Hero!

Answer 39

You're welcome.

Answer 40

Btw, what are 99 users in comparison to me, a "15 user"?

Answer 41

A 99 user has quite a bit of experience on OS. Has helped a lot of users, answered (or asked) many questions and has a long tenure on OS. Most 99+ users have been here for at least a year. A few have been here for just six months. I came here before they started doing smart score, so when they converted to it, I was already 99 user.

Answer 42

Cool, so I just need to lookup Wio and message him/her to see if they might possibly have a bit of time to help teach me something? :-D

Answer 43

Maybe. You have to catch the person at a good time.

Answer 44

If you show that you are willing to participate in the process and that you have a clue what's going on, then you're more likely to get helped.

Answer 45

Thank you. I'll try not to appear as lost but unfortunately that is sort of where I am.

Answer 46

USe the binomial theorem