Solve 43681x^6 - 22932.5x^5 + 3931.3x^4 - 221.1x^3 + 1.945e-7 = 0.

Step-by-step solution

1. 1

   Note why the constant term cannot simply be dropped. Without it the equation factors as $x^3(43681x^3-22932.5x^2+3931.3x-221.1)=0$ with a triple root at $x=0$. The tiny constant $1.945\times10^{-7}$ perturbs that triple root into three small roots clustered near the origin, so it controls the entire small-$x$ behaviour and must be kept.

2. 2

   Find the three large roots from the cubic factor. For $x$ well away from zero the constant is negligible and the roots are those of

   $43681x^3-22932.5x^2+3931.3x-221.1=0$

   Sign changes: $f(0.14)<0$, $f(0.15)>0$, $f(0.16)<0$, $f(0.23)>0$. Bisecting each bracket on the full sextic gives

   $x=0.14670508,\qquad x=0.15345189,\qquad x=0.22484227$

   The first two are only $0.0067$ apart, so a solver that steps too coarsely will miss one of them.

3. 3

   Resolve the three small roots near the origin. For $|x|\ll 1$ the sextic is dominated by $-221.1x^3+1.945\times10^{-7}$, giving the estimate $x^3\approx 8.797\times10^{-10}$, i.e. $|x|\approx 9.6\times10^{-4}$. Newton refinement on the full polynomial gives the real value

   $x=0.000963677$

   and the other two are the complex cube-root partners $x\approx-0.00048175\pm0.00082509\,i$.

4. 4

   Confirm the count. A degree-six polynomial has exactly six roots with multiplicity. We have four real ($0.000963677$, $0.14670508$, $0.15345189$, $0.22484227$) and one conjugate pair — six in total, so nothing is missing.

5. 5

   Check each real root by residual. Substituting them back gives residuals of $2.6\times10^{-23}$, $1.2\times10^{-16}$, $5.2\times10^{-18}$ and $3.4\times10^{-16}$ — all zero to machine precision ✓. By contrast a commonly quoted set containing $0.0887304343$ and $0.2575361876$ produces residuals of $-0.0156$ and $+0.2814$, so those are not roots of this equation.