I appreciate any help you can provide. I have gone back to college (42 y.o) and never had to learn algebra so I am incredibily lost and feel very stupid. I need to know if there is anything that can help me not get totally lost when trying to work polynomial problems.. I can't give you a specific problem because they are all causing issues for me. As soon as I see the formula I go math blank and it all looks like Greek to me. It does not make sense at all and I am in a panic. Is there any hints, help, talk that would relate to someone that was in grade school that would make any of this make sense to me? I really appreciate your help with this, I am scared to death that I may fail my course. Terri
Polynomials
- Thread starter thamrick
- Start date
N
njurkovi
Guest
You need to start with something. "I don't understand anything" isn't helpful.
Did you read the definition of a polynomial from your textbook?
Do you understand it?
If not, which part of it you don't understand?
If yes, what is your next stumbling block? etc...
Did you read the definition of a polynomial from your textbook?
Do you understand it?
If not, which part of it you don't understand?
If yes, what is your next stumbling block? etc...
In Algebra, a letter represents a number.
x + 2 = 5
Here, you can "see" that x = 3, right? Because 3 + 2 = 5
That's how it "starts"; not too scary, right?
Do you at least understand that?
You didn't tell us where you're at....except pretend you're scared :wink:
x + 2 = 5
Here, you can "see" that x = 3, right? Because 3 + 2 = 5
That's how it "starts"; not too scary, right?
Do you at least understand that?
You didn't tell us where you're at....except pretend you're scared :wink:
I understand that the fact that I am totally clueless probably doesn't give anyone a starting point. Okay, I did ask for elementary, and yes x + 3 = 5 is simple but to me that isn't really algebra. That is like putting a question mark in place of the x and is simple basic math. My problem is that I truly do not understand how they come up with x + 3 = 5 and x is suddenly 1. (not an actual example, but trying to better explain.) I am grabbing an example from one of my checkpoints to give you a better one.
12x^6-15x^5+21x^4-6x^3
1st real problem with this is how is the largest common factor to all terms of the polynomial possibly be 3x^3 ?
I have read all examples and definitions but to tell you what doesn't make sense would require me to copy and past the entire chapter. Again, thanks again! I am truly not an idiot; I have done all forms of business math (payroll, disbursement journals, time keeping etc..) and general math in life (checkbooks, cooking, measuring for carpet etc..)
12x^6-15x^5+21x^4-6x^3
1st real problem with this is how is the largest common factor to all terms of the polynomial possibly be 3x^3 ?
I have read all examples and definitions but to tell you what doesn't make sense would require me to copy and past the entire chapter. Again, thanks again! I am truly not an idiot; I have done all forms of business math (payroll, disbursement journals, time keeping etc..) and general math in life (checkbooks, cooking, measuring for carpet etc..)
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
thamrick said:
In each of the four terms, examine both prime factorization of the coefficient, as well as the number of factors of x:
12x^6 = 2 * 2 * 3 * (x*x*x*x*x*x)
15x^5 = 3 * 5 * (x*x*x*x*x)
21x^4 = 3 * 7 * (x*x*x*x)
6x^3 = 2 * 3 * (x*x*x)
Now ask, what factors are common to each term ?
2 is not a common factor because 2 is not a factor of 15x^5 or 21x^4.
5 is not a common factor because 5 is not a factor of 12x^6, 21x^4, or 6x^3.
7 is not a common factor because 7 is not a factor of 12x^6, 15x^5, or 6x^3.
The factor 3 is common to all four terms.
Three factors of x appear in each term, so x^3 is also a common factor.
Multiply the common factors together, to get the greatest factor common to each term.
GCF(12x^6 , 15x^5 , 21x^4 , 6x^3) = 3x^3
Dividing each term in the polynomial by this GCF gives us the remaining factor. That allows us to write the given polynomial factorized, like so:
12x^6 - 15x^5 + 21x^4 - 6x^3 = (3x^3)(4x^3 - 5x^2 + 7x - 2)
Cheers ~ Mark
Last edited:
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
thamrick said:
We want to examine the prime factorization of each coeffcient.
6 is not a prime number.
The prime factorization must contain only prime numbers.
12 = 2^2 * 3
I am honored, to be called Ted.
Last edited:
Really boils down to : what the highest value common to all terms?
Say you're asked to factor: 6x^5 + 15x^2
Here we can "see" that it is 3x^2:
3x^2 * 2x^3 = 6x^5
3x^2 * 5 = 15x^2
So factored form: 3x^2(2x^3 + 5)
Of course these are not all possible to "see"; that's when you go the way Ted, whoops Mark(!) showed you.
If you're wondering why 3x^2 * 2x^3 = 6x^5 not 6x^6 :
rule: a^x * a^y = a^(x+y)
Say you're asked to factor: 6x^5 + 15x^2
Here we can "see" that it is 3x^2:
3x^2 * 2x^3 = 6x^5
3x^2 * 5 = 15x^2
So factored form: 3x^2(2x^3 + 5)
Of course these are not all possible to "see"; that's when you go the way Ted, whoops Mark(!) showed you.
If you're wondering why 3x^2 * 2x^3 = 6x^5 not 6x^6 :
rule: a^x * a^y = a^(x+y)
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
It also helps to memorize the multiplication table.
When you consider the coefficients 12, 15, 21, and 6, as products in the multiplication table, then it's obvious that they're all divisible by 3 and that 3 is the only number that evenly divides into each of them (ignoring the trivial factor of 1).
One should be able to do this in their head, without resorting to writing out prime factorizations, when dealing with "small" Whole numbers. I mean, when dealing with numbers in the multiplication table.
If one does not know their multiplication table, manual factoring is very difficult.
I agree with Denis; writing out prime factorizations is mostly handy when dealing with "bigger" Whole numbers and factoring by hand.
Last edited:
Denis & Mark (sorry about that, looked once and saw Ted's name)
Thank you! I am going to try and get as much done when it comes to factors as I can then I will see if I can then figure out some of the rest. I think that might have been my biggest problem with polynomials (I hope). If not I will be sure to post another question. Thanks again! Terri
Thank you! I am going to try and get as much done when it comes to factors as I can then I will see if I can then figure out some of the rest. I think that might have been my biggest problem with polynomials (I hope). If not I will be sure to post another question.
Thanks again! Terri