I am trying to solve the problem no. 41(i) on page $33$, Higher Algebra by Barnard and Child.

The problem statement is:

If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$, prove that $$(a^2-c^2)(a^2-c^2+bc)=a^2b^2.$$

By synthetic division, I have found the remainder of $ax^5+bx^2+c$ upon division by $x^2+px+1$ is $$(ap^4-3ap^2-bp+a) x +(ap^3-2ap-b+c).$$

As $x^2+px+1$ is a factor of $ax^5+bx^2+c$, the memainder must be indentically zero. That is:\begin{align*} ap^4-3ap^2-bp+a&=0 \tag{1}\\ ap^3-2ap-b+c&=0 \tag{2} \end{align*} Then subtracting $(1)$ from $(2)\times p$ gives $$ap^2+cp-a=0.$$

I do not know how to eliminate $p$ from the equations. I do not even know if this approach will solve it. And I have not been able yet to think of other approach.

If it is easy, then please give me some hint. Otherwise a solution, please!

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    $\begingroup$ I need my morning coffee before I try this seriously, but may be you can use reciprocal polynomials? The (complex) zeros of $f(x)=x^2+px+1$ are reciprocals of each other. This implies that $f(x)$ must be a common factor of both $g(x)=ax^5+bx^2+c$ as well as its reciprocal $\tilde{g}(x)=cx^5+bx^3+a$. This is because $g(\alpha)=0$ if and only if $\tilde{g}(1/\alpha)=0$. $\endgroup$ May 27 '18 at 7:15
  • $\begingroup$ The idea of reciprocal polynomials is not introduced yet, maybe introduced in some later chapter. And I am afraid you are talking about problem no. $41(ii)$, which I have been able to solve. @JyrkiLahtonen $\endgroup$ May 27 '18 at 7:21
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    $\begingroup$ @JyrkiLahtonen 's hint makes it fall out if you look at the fact $f|cg-a\tilde{g}$. $\endgroup$ May 27 '18 at 7:55
  • $\begingroup$ Feel free to flesh that ut to an answer @ancientmathematician ! $\endgroup$ May 27 '18 at 7:57
  • $\begingroup$ Thanks, Jyrki Lahtonen and ancientmathematician! Jyrki Lahtonen, now I have got your hint. And solved it. It was a fairly appropriate hint; and only my lack of proficiency failed me to catch it. $\endgroup$ May 27 '18 at 8:07

Denote $f_1(t) = at^4 - 3at^2 - bt + a$, $f_2(t) = at^3 - 2at - (b - c)$, then $f_1(p) = f_2(p) = 0$. Therefore, defining\begin{align*} f_3(t) &= t f_2(t) - f_1(t) = at^2 + ct - a,\\ f_4(t) &= a(tf_3(t) - f_2(t)) - c f_3(t) = (a^2 - c^2)t + ab,\\ f_5(t) &= (a^2 - c^2)^2 f_3(t) - a (f_4(t))^2 + 2a^2b f_4(t) - c(a^2 - c^2) f_4(t)\\ &= a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)), \end{align*} then$$ f_1(p) = f_2(p) = f_3(p) = f_4(p) = f_5(p) = 0 \Longrightarrow a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)) = 0, $$ and $a ≠ 0$ implies$$ (a^2 - c^2)(a^2 - c^2 + bc) = a^2 b^2. $$

  • $\begingroup$ I have not verified every step. But I have got the idea of your solution. It appears to be more of art than of science! Nice effort! $\endgroup$ May 27 '18 at 8:12

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