math help

The upside down v is the best we can do with these keyboards to raise a number or variable to a power. That is, x^3 is x cubed. Go to your local book store, and you will probably find books to help you re-learn algebra, which I think is your best first step. Everything in analysis comes from algebra. I should say, though, that if you don't remember the arithmetic of fractions, you'd better catch up on that first.

This is a polynomial expression

normally it is given in the form

a(n)x^n+a(n-1)x^(n-1)+....+a(1)x+a(0)

the a(i)'s where i=0,.....,n refer to constants.
the x is the variable to various powers

fer instance x to the first power is x^1, x squared is x to the second power x^2=x*x

x cubed is x^3=x*x*x

x to the nth power is x^n=x*x*.....*x n times

so you can note that n is also a variable and that powers (n) have to be 0 or a positive integer (counting number).

There is a proved theorem (see note) that says that any polynomial can be expressed as the product of factors; that is

a(n)x^n+a(n-1)x^(n-1)+....+a(1)x+a(0)=[x+r(0)][x+r(1)]......[x+r(n)]

and this is the knut in beginning algebra. You may remember this as a concept of factoring.

a few more definitions:
polynomial power refers to the greatest n
fer instance

a(n)x^n+a(n-1)x^(n-1)+....+a(1)x+a(0)

is an nth degree polynomial because of the n in the (^n) in the expression---as a result I would expect to have n roots; e.g. r(i)'s

Now for particulars

2 + 9x + 7x^2 + 6x^3

I would first take a mirror image to get this in a conventional form

6x^3+7x^2 +9x+2

now the 3 in the ^3 tells me that this is a third degree polynomial and that this polynomial has three roots; that is

6x^3+7x^2 +9x+2=6[x-r(1)][x-r(2)][x-r(3)]

where r(1),r(2),and r(3) are some constants---didja see there are three of them and the power of this polynomial is three?

Finding these's roots then is the knut, and sometimes this process is tedious (not hard but time consuming and wrought with errors in arithmetic). Few instance the solution of a 2nd degree equation may require the dreaded quadratic formula and consider that the solution for cubics is four times as dreaded. My solution, Thats what a spreadsheet is for.

Any who's that's my partial answer, other than echo-ing the limited font of this forum.

Rap

Note---one of the things I like about math, particularly algebra is the certainty. So when a theorem is proved---that's it---add it to your grammar, it's always gonna be that way, and that synomyn is gonna be that way. If your answer is wrong, it very likely could be off by a bunch and won't pass the smell test.