Question

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=?????

Answer 1

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`
*