<s>[INST] If $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$ if $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$
My try
Since this is a factor I tired finding A, B, C & D such that,
$(x^2+kx+1)(Ax^3+Bx^2+cx+D)$ =$px^5+qx^2+r$
if I can find k in terms of p, q & r I can try to get the given condition, But when I tried this equating coefficients, I ended up with quadratic expression for k? Is there a better way to approach this question? Please Help! thanks
 [/INST] Let $px^5+qx^2+r=(x^2+kx+1)(px^3+ax^2+bx+r)$
Let's compare the coefficients of $x$
$0=kr+b\iff b=-kr$
That of $x^4$
$0=a+kp\iff a=-kp$
That of $x^3$
$0=b+ka+p=-kr+k(-kp)+p$
which is on rearrangement, a quadratic equation in $k$
Compare the values of $a$ to find $k$
Coefficients of $x^2$
$q=r+kb+a=r+(-kr)(k-1)$
Replace the values of $a,b$ to form another quadratic equation in $k$
Solve the two quadratic equations for $k,k^2$
Use $k^2=(k)^2$ to eliminate $k$
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