i am in year ... dude
When the limit of x approaches 0 the degree on n is greater than 0.
The polynomial P(x)=(x-3)(x-0)(x+3)(x-1) is of the fourth degree.
If x^2 is second degree, and x (which is x^1) is first degree, then a constant would be zeroth degree, I think since x^0 = 1 for any non-zero x.
The degree of a polynomial is the highest degree of its terms. The degree of a term is the sum of the exponents of the variables that appear in it.
Yes, a constant function has degree 0; i.e., f(x)= a where a does not equal 0 is a mathematical function of degree 0. Lowell F. Lynde, Jr. (Univ. of Arkansas at Monticello)
If you mean: x-3-4 = 0 then x = 7 I mean x + ln (x-3) - 4 =0
If you mean: 4(x-2) = 0 then 4x-8 = 0 and x = 2
Yes, f(x) = 2 is a polynomial of degree 0 (because there are no x terms).
Do you mean x = x^2 ? If so, you can write this as a quadratic x^2 - x = 0 x(x - 1) = 0 The roots are x = 0 and x = 1.
I assume you mean a polynomial of degree 4. In general, a polynomial of degree "n" can be separated into "n" linear factors. As a result, a polynomial of degree "n" has exactly "n" solutions - unless two or more of the factors are repeated; in which case the corresponding solution is said to be a multiple solution. As an example: (x - 2)(x - 5) = 0 has two solutions, namely 2 and 5; while (x - 3)(x - 3)(x + 5) = 0 is of degree three, but has only two solutions, since the solution "3" is repeated. An equation such as x2 + 1 = 0 cannot be factored in the real numbers, so if you insist that the solutions be real, there are zero solutions. However, the polynomial can be factored in the complex numbers; in this case: (x + i)(x - i) = 0, resulting in the two complex solutions, -i and +i.
Good question! The zero polynomial "0" could result from any of the following: (0), (0)x, (0)x2, (0)x3, etc. Since you don't know which it came from, you can't say what the degree is.
The smallest is 0: the polynomial p(x) = 3, for example.