What is the degree of the zero polynomial p(x)?

A polynomial that consists only of a non-zero constant, is called a constant polynomial and has degree 0. The polynomial p ( x) = 0 is called the zero polynomial. It has no terms and so there is no leading term. It is best not to define the degree of the zero polynomial. Some books write its degree as −1 or − ∞.

Is X5 – 2×3 + 8x + 3 a monic polynomial?

The term independent of is called the constant term. Thus x5 − 2×3 + 8x + 3 is a monic polynomial of degree 5 with constant term 3, while. x4 − x2 + 1 is a non-monic polynomials of degree 4 with leading coefficient and constant term 1.

How do you add polynomials with no leading term?

The polynomial p ( x) = 0 is called the zero polynomial. It has no terms and so there is no leading term. It is best not to define the degree of the zero polynomial. Some books write its degree as −1 or − ∞. To add or subtract two polynomials, we collect the like terms. Let p ( x) = 3 x4 − 2 x2 + x − 1, q ( x) = 7 x5 + 2 x2.

Which polynomial has remainder 13 after division by x – 1?

The polynomial p ( x) = x5 − 7 x3 + ax + 1 has remainder 13 after division by x − 1. Find the value of the coefficient a. Factoring quadratics is an important technique which we used to solve quadratic equations. In a similar way, we would like to be able to develop some techniques to factor polynomials.

Which polynomials contain only even powers of X?

The left side is an even polynomial, the right side an odd one, so P ( x) = f ( x). So polynomials that contain only even powers of x are exactly the even polynomials. Lemma Q ( x) = x 2 + 1, P ( Q ( x) = Q ( P ( x)) and P ( x) is even.

How do you name a polynomial with no leading term?

For small degree polynomials, we use the following names. A polynomial that consists only of a non-zero constant, is called a constant polynomial and has degree 0. The polynomial p(x) = 0 is called the zero polynomial. It has no terms and so there is no leading term.

How do you solve a functional equation with a polynomial?

Let R ( x) = x 2 + 1. Remark there clearly exists a polynomial S ( x) such that P ( x) = S ( x 2 + 1), just shift Q. Now remark that P ∘ R = R ∘ P. Hence S ∘ R ∘ R = R ∘ S ∘ R ⟹ S ∘ R = R ∘ S, so S is a solution to the functional equation.