Answer to Question #147639 in Calculus for liam donohue

Question

This question summarizes a few simple facts about polynomials. Recall that a polynomial p can be written in the form
p(x)=∑i=0nαixi
where we assume that αn≠0. The degree of a polynomial is the largest exponent present, and so the degree of p is n. Fill in the blanks in the following questions:

The degree of p′(x) is
.

The degree of ∫p(x)dx is
.

The degree of p2(x) (i.e., the square of p) is
.

The (n+1)-th derivative of p is

Answer 1

An n-th degree polynomial is given as

p(x)= $\sum_{i=0}^np_ix^i$

That means

p(x) = a₀ + a₁x + a₂x² + a₃x³+ •••••+a_nxⁿ, a_n ≠ 0.

p'(x) = a₁+ 2a₂x + 3a₃x²+ •••••+na_nx^n-1, a_n ≠ 0.

So degree of p'(x) is (n-1)

$\int p(x) dx = a_0x + \frac{a_1}{2}x^2+\frac{a_2}{3}x^3+••• +\frac {a_n}{n+1}x^{(n+1)}+ C$

So degree of $\int p(x) dx$ is (n+1)

p²(x) = (a₀ + a₁x + a₂x² + a₃x³+ •••••+a_nxⁿ)(a₀ + a₁x + a₂x² + a₃x³+ •••••+a_nxⁿ)

Here highest degree term is a_n*a_nx²ⁿ and a_n*a_n≠0 as a_n ≠ 0..

So degree of p²(x) is 2n

After derivative of each order the degree of a polynomial decreases by 1.

So after n th order derivative the degree of a n-th degree polynomial decreases by n . So degree after n-th order derivative becomes zero. So the polynomial becomes a constant after n-th order derivative.

So (n+1) th derivative of p is zero.