Solve 1/(s + 2) + 1/(s + 4) + 1/(s + 9) - 1/(s + 6) = 0 for s.

Step-by-step solution

1. 1

   Record the excluded values first. The four denominators vanish at

   $s=-2,\ -4,\ -6,\ -9$

   None of these can be a solution, whatever the algebra produces. Noting them up front means any spurious root that appears later is caught immediately.

2. 2

   Clear the denominators by multiplying through. Multiplying every term by $(s+2)(s+4)(s+6)(s+9)$ turns the equation into a polynomial one. Collecting the four resulting cubics gives

   $2s^{3}+33s^{2}+180s+300=0$

   The quartic terms cancel because three fractions are added and one is subtracted, dropping the degree from 4 to 3.

3. 3

   Test for rational roots — and reject the tempting ones. By the rational root theorem the candidates are $\pm\frac{p}{q}$ with $p\mid 300$ and $q\mid 2$. Checking the plausible ones: $s=-5$ gives $2(-125)+33(25)-900+300=-25\neq 0$, $s=-4$ gives $-128+528-720+300=-20\neq 0$, and $s=-\tfrac{15}{2}$ gives $-\tfrac{3375}{4}+\tfrac{7425}{4}-1350+300\neq 0$. No rational root exists, so the cubic does not factor over the rationals.

4. 4

   Always expand a proposed factorisation. A guess such as $(s+5)(2s+15)(s+4)$ multiplies out to $2s^{3}+33s^{2}+175s+300$ — it matches the leading, quadratic and constant coefficients but the linear one is $175$, not $180$. Checking that single coefficient is what exposes the error.

5. 5

   Find the real root numerically. Since $f(-3)=3>0$ and $f(-3.1)=-0.452<0$, a root lies between them; Newton iteration converges to

   $s\approx-3.0864085$

   This is not one of the excluded values, so it is a genuine solution.

6. 6

   Account for the other two roots and verify. The remaining roots are the complex pair $s\approx-6.7068\pm 1.9024i$, so there is exactly one real solution. Substituting $s=-3.0864085$ back: $\tfrac{1}{-1.0864}+\tfrac{1}{0.9136}+\tfrac{1}{5.9136}-\tfrac{1}{2.9136}=-0.9205+1.0946+0.1691-0.3432=0.0000$ ✓.