# 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=?????

Jul 9, 2018

See below

#### Explanation:

From

$p \left({x}^{2}\right) + x \cdot q \left({x}^{3}\right) + {x}^{2} \cdot r \left({x}^{3}\right) = \left(1 + x + {x}^{2}\right) \cdot s \left(x\right)$

we get

$p \left(1\right) + 1 \cdot q \left(1\right) + {1}^{2} \cdot r \left(1\right) = \left(1 + 1 + {1}^{2}\right) \cdot s \left(1\right) \implies$

$p \left(1\right) + q \left(1\right) + r \left(1\right) = 3 s \left(1\right)$

Given $p \left(1\right) = k s \left(1\right)$ and $r \left(1\right) = k p \left(1\right) = {k}^{2} s \left(1\right)$, we get

$\left(k + {k}^{2}\right) s \left(1\right) + q \left(1\right) = 3 s \left(1\right) \implies$

${k}^{2} + k - 3 + \frac{q \left(1\right)}{s \left(1\right)} = 0$

This equation can be solved easily for $k$ in terms of $\frac{q \left(1\right)}{s \left(1\right)}$

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 $q \left(1\right) = k r \left(1\right)$, we would have had $\frac{q \left(1\right)}{s \left(1\right)} = {k}^{3}$, and the final equation would have become
${k}^{3} + {k}^{2} + k - 3 = 0 \implies$
${k}^{3} - {k}^{2} + 2 {k}^{2} - 2 k + 3 k - 3 = 0 \implies$
$\left(k - 1\right) \left({k}^{2} + 2 k + 3\right) = 0$
Now, since ${k}^{2} + 2 k + 3 = {\left(k + 1\right)}^{2} + 2 \ge 2$, it can not vanish for real $k$. So we must have $k = 1$