Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with x^3 for polynomials f (x) and g(x).

necessaryh 2021-09-05 Answered

Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with x3 for polynomials f (x) and g(x). Find the smallest possible degree of f (x)?
d1992dx1992(p(x)x3x)=f(x)g(x)

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Expert Answer

Neelam Wainwright
Answered 2021-09-06 Author has 102 answers

Given p(x) be a nonzero polynomial of degree less than 1992 having no non constant factor in common with x3x
Let d1992dx1992(p(x)x3x)=f(x)g(x), where f(x) and g(x) are polynomials.
To find the smallest possible degree of f(x):
BY division algorithm we know that,
If f(x) and g(x) are any two polynomials with g(x)0, then f(x)=g(x).q(x)+r(x), where r(x)=0 or deg r(x) where q(x) is the quotient and r(x) is the remainder.
Hence p(x) can be written as
p(x)=(x3x)q(x)+r(x)
Here, we can see that the degree of r(x) is 2 and the degree of q(x)is less than 1989
Then,
d1992dx1992(p(x)x3x)=d1992dx1992(r(x)x3x)
To find r(x)x3x
Using partial fractions, we have
r(x)x3x=Ax1+Bx+Cx+1
Hence,
d1992dx1992(r(x)x3x)=1992!(A(x1)1993+Bx1993+C(x+1)1993
=1992!(Ax1993(x+1)1993+B(x1)1993(x+1)1993+C(x1)1993x1993(x1)1993x1993(x+1)1993)

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