If I have a polynomial 3X+2Y+1, what are the coefficients? Is the constant, which is 1, considered a coefficient?

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- Aug 20th 2009, 10:15 AMblaskodeWhat is a coefficient in a polynomial?
If I have a polynomial 3X+2Y+1, what are the coefficients? Is the constant, which is 1, considered a coefficient?

- Aug 20th 2009, 10:36 AMe^(i*pi)
The coefficient is the term which the variable is multiplied by.

In $\displaystyle 3x + 2y + 1$ the coefficient of x is 3, the coefficient of y is 2 and 1 is just a constant.

For non linear powers such as $\displaystyle 3x^2-2x+1 = 0$ then the coefficient of $\displaystyle x^2$ is 3, the coefficient of $\displaystyle x$ is -2 and 1, although a constant can be thought of as the coefficient of $\displaystyle x^0 \, \: \, x \neq 0$

Coefficients don't have to be just numbers though, they can be constants. For example: $\displaystyle (3k-2)x^2 + \frac{1}{k-1}x - 2$ (where k is a positive constant)

Coefficient of $\displaystyle x^2 = 3k-2, of x = \frac{1}{k-1}, constant = -2$

Coefficients are used mainly for working out cubics by inspection and using the factor/remainder theorem. In chemistry they are widely used to balance equations - Aug 20th 2009, 10:37 AMstapel
To some extent, whether or not constants are considered their own coefficients will depend upon the particular textbook's

**definition**. In general, the "numerical coefficient" is just whatever number is multiplied on the variable portion of the term. But "1" could be considered the "numerical coefficient" of the term "1", too. (Wink) - Aug 20th 2009, 10:44 AMblaskodeGelfand
I was working out of I. M. Gelfand's Algebra and encountered the following problem:

"Imagine that the polynomial (1+x-y)^3 is converted to the standard form. What is the sum of all the coefficients?"

Now, if "-1" is considered as a coefficient, this is an easy problem: (1+1-1)^3 = 1, but if "-1" is not considered a coefficient, I don't know how to do the problem without expanding the polynomial to standard form. Expanding this polynomial to standard form isn't difficult, but what if the polynomial were (1+2x)^200 instead? - Aug 20th 2009, 12:23 PMe^(i*pi)
- Aug 20th 2009, 12:37 PMblaskodeGelfand
I'll scour the book and see if Gelfand provides an answer. I am glad to hear that there isn't a standard definition--kind of like how the defition of "whole number" isn't standard--some books consider "0" a whole number; others do not.