Q.Solve the equation $6x^4-5x^3-38x^2-5x+6=0$ if it is known that $\dfrac13$ is a solution.

Step 1. Confirm $\tfrac13$ is a root. $6\left(\tfrac1{81}\right)-5\left(\tfrac1{27}\right)-38\left(\tfrac19\right)-5\left(\tfrac13\right)+6=\tfrac6{81}-\tfrac{15}{81}-\tfrac{342}{81}-\tfrac{135}{81}+\tfrac{486}{81}=\tfrac0{81}=0$ ✓.

Step 2. Use the Type I reciprocal structure to get a second root for free. Coefficients $6,-5,-38,-5,6$ satisfy $a_r=a_{4-r}$ (palindromic, Type I), so if $\tfrac13$ is a root, so is its reciprocal $3$. Check: $6(81)-5(27)-38(9)-5(3)+6=486-135-342-15+6=0$ ✓.

Step 3. Divide by $(x-\tfrac13)(x-3)$, i.e. by $3x^2-10x+3$. $6x^4-5x^3-38x^2-5x+6\div(3x^2-10x+3)$: quotient $2x^2+5x+2$. …

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