Q.Solve the reciprocal equation $x^5-5x^4+9x^3-9x^2+5x-1=0$.

Step 1. The coefficients $1,-5,9,-9,5,-1$ satisfy $a_k=-a_{5-k}$ (e.g. $a_0=1=-a_5=-(-1)$; $a_1=-5=-a_4=-5$; $a_2=9=-a_3=-(-9)$), so this is a class-two (anti-symmetric) reciprocal equation of odd degree $5$. An odd-degree class-two reciprocal equation always has $x=1$ as a root.

Step 2. Confirm: $f(1)=1-5+9-9+5-1=0$. Yes, $x=1$ is a root.

Step 3. Depress by dividing by $(x-1)$ synthetically on $1,-5,9,-9,5,-1$ with $a=1$: $b_0=1$; $b_1=-5+1=-4$; $b_2=9+(-4)=5$; $b_3=-9+5=-4$; $b_4=5+(-4)=1$; remainder $=-1+1=0$. Quotient: $x^4-4x^3+5x^2-4x+1$.

Step 4. This quartic's coefficients $1,-4,5,-4,1$ are palindromic, so it is a class-one reciprocal equation of even degree $4$. Since $x=0$ is not a root, divide by $x^2$:

Step 5. With $t=x+\dfrac1x$ (so $x^2+1/x^2=t^2-2$): …

Odd-degree class-two reciprocal equation: depress by the guaranteed root $x=1$, then solve the remaining …

• Misclassifying the sign pattern (class one vs. class two), which leads to testing the wrong guaranteed root ($x=-1$ instead of $x=1$, or vice versa). …