#I need a lesson over simplifying polynomials
172 messages · Page 1 of 1 (latest)
Ik, js
I can't be here doing calculus 1 stuff and struggling at simplifying derivatives
Not like I'm discouraged
Polynomials are what happens when we multiply together an unknown with itself (a few times) and add/subtract numbers
Daddy_314
Can I do it myself
Daddy_314
Aw ok
Daddy_314
The maximum power of x is 2 in your polynomial
But in mine it's 3
This is called the degree of the polynomial
Which is the maximum number of times the unknown was multiplied by itself
So your polynomial is of degree 2
Ok
So that's why ax²+bx+c is called a 2nd degree polynomial and equaling it to 0 makes it an equation?
Exactly
Sometimes we call them quadratics
But thats like a cat called Tom because we know it well
The neighbors can just call it cat
If we have x²*x² does that simplify to x⁴?
Yes
So it's just adding up the exponents
Bamtryfoil (noisificated)
Well
What you wrote is true
But anytime you see powers that are not integers
The object is not a polynomial
Taking notes
So $f(x) = \sqrt{x} x$ is not a polynomial in x
Daddy_314
Right
But if I take $u = \sqrt{x}$ for example
Then $f(x) = x \times \sqrt{x} = u^2 \times u = u^3$ is a polynomial in u
Daddy_314
So the rules are clear, powers of the variable have to be integers
Now the reason you may be traumatized by polynomials specifically
Is because one thing we try to do to them is factor them
That's the thing
And i have to do that for taking limits because some take us to false indeterminations
Yes
This typically happens when both polynomials
In the numerator and denominator
Share a common factor
And I know the l'hopital rule but I need to know factorization for 1st degree bachelor
So I can do the limits
Yeah dont use that
For polynomials
This is a strong tool
No need for that
Everytime
L'hopitals just makes very long derivatives when we have to do product and division rules
Factorization just should work better
Depends
If its only polynomials then
Either of them works
But if you struggle
Then you need to train
On factorization properly
Which means you need to do exercises and struggle
Until you make some progress
Daddy_314
Yes
Ohh
Preferably of simpler terms
Isn't 10*10 100 tho
I could always write $20 = 20 \times 1$ but this would be useless
Daddy_314
I see
Is there a way to factor any integer?
Without having to do it just on the spot?
Because then what about factoring big numbers like 89076
Yes !
Good question
This is a big deal in maths
It's called
The fundamental theorem of arithmetic
Any number can be written as a product of smaller, elementary numbers
And these elementary numbers cannot be decomposed into products
They're called prime numbers
And the reason we like to factor polynomials, is that it allows us to find its roots
A root of a polynomial is a number that makes it equal to zero
For example of we take
$P(x) = x^2-9$
If x = 3,
P(3) = 9-9 = 0
Daddy_314
Therefore 3 is a root
To know this, We could factor P into a product of smaller easier polynomials
We could prove that
$p(x) = (x+3)(x-3)$
Daddy_314
Its normal
I didnt give you details
Its just the concept for now
The main idea
This needs to be done later on
I have to go for now
Alright
I'm out of school
Mult tables?
Like multiples?
0 = 0
2= 2,4,6,8,10,12,14,16,18,20...
6= 6,12,18,24,30,36,42,48,54,60...
@loud totem hey! if you ask your question i could probably help as well! its a bit difficult for me to read through all of that lol
Please list all the 0 operations
0 * 0 = 0
0 * 1 = 0
0 * 2 = 0
Until 10
The follow-up question to that one:
What do you notice
Multiplication is reiterated sum?
I actually don't know
Yes
Is it possible to have two numbers (I call them A and B)
Such that
$A \times B = 0$
Daddy_314
And neither A nor B is zero ?
Same question with A , and B if
$A \times B = 12$
Can you deduce the values of A and B ?
I say no
Daddy_314
There are multiple values for B and a but I can say that A could be six and B 2
Or 3 And 4 for example
Or A = 0.5 and B = ...
What about this?
B = 24
So now that we saw this
This tells you how special 0 is
If a product of two numbers is 0, one of these numbers at least must be zero
This is not the case for any other number (like 12)
This fact is the basis on which everything around polynomials is built
Why ?
If for example $x \times (x+2) = 0$
Then we can be dead sure that $x = 0$ or that $x+2=0$
Daddy_314
Yup
I got that
Ok cool
Just learned ruffini's method of factoring 3rd degree + polynomials
idk man just get better