   Chapter 8.5, Problem 53E ### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

#### Solutions

Chapter
Section ### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Let p(x) be a nonzero polynomial of degree less titan1992 having no nonconstant factor in common with x 3 − x Let d 1992 d x 1992 ( p ( x ) x 3 − x ) = f ( x ) g ( x ) for polynomials f(x) and g(x), Find the smallest possible degree of f(x).

To determine

To calculate: The smallest possible degree of f(x). Let p(x) be a nonzero polynomial of degree less p(x) than having no non-constant factor in common with x3x. Let d1992dx1992(p(x)x3x)=f(x)g(x) for polynomials f(x) and g(x).

Explanation

Given:

polynomials f(x) and g(x)

Formula used:

Dividend = Divisor×Quotient+Remainder

Calculation:

The division algorithm is:

Dividend = Divisor×Quotient+Remainder

Thus, we can write p(x) as:

p(x)=(x3x)q(x)+r(x)

Here, the degree of r(x) is 2 and that of q(x) is less than 1989.

Then,

d1992dx1992(p(x)x3x)=d1992dx1992(r(x)x3x)

Now, using partial fractions we can write r(x) as:

r(x)x3x=Ax1+Bx+Cx+1

Thus,

d1992dx1992(r(x)x3x)=1992!(A(x1)1993+Bx1993+C(x+1)1993)

Now, simplify the right hand side as:

d1992dx1992(r(x)x3x)=1992!(A(x<

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