# What is a coefficient in a polynomial?

• Aug 20th 2009, 10:15 AM
What 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 AM
e^(i*pi)
Quote:

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

The coefficient is the term which the variable is multiplied by.

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

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

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

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

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 AM
stapel
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 AM
Gelfand
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 PM
e^(i*pi)
Quote: