P(x^2)+x*q(x^3)+x^2*r(x^3)=(1+x+x^2)*s(x), p(1)=ks(1) and r(1)=kp(1). Then k=?????

1 Answer
Ananda Dasgupta
Jul 9, 2018

See below

Explanation:

From

#p(x^2)+x*q(x^3)+x^2*r(x^3)=(1+x+x^2)*s(x)#

we get

#p(1)+1*q(1)+1^2*r(1) = (1+1+1^2)*s(1) implies#

#p(1)+q(1)+r(1) = 3s(1)#

Given # p(1)=ks(1)# and #r(1)=kp(1)=k^2s(1)#, we get

#(k+k^2)s(1) +q(1) = 3s(1) implies#

#k^2+k-3+{q(1)}/{s(1)} = 0#

This equation can be solved easily for #k# in terms of #{q(1)}/{s(1)}#

However, I can't help feeling that there was one more relation in the problem which got missed out somehow. For, example, if we had one more relation like #q(1) = kr(1)#, we would have had #{q(1)}/{s(1)} = k^3#, and the final equation would have become
#k^3+k^2+k-3 = 0 implies#
#k^3-k^2+2k^2-2k+3k-3=0implies#
#(k-1)(k^2+2k+3)=0#
Now, since #k^2+2k+3=(k+1)^2+2 ge 2#, it can not vanish for real #k#. So we must have #k=1#

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