Solve the quartic equation x⁴ − 10x³ + 18x² − 40x − 18 = 0, giving the real roots to seven decimal places.

Step-by-step solution

1. 1

   Rule out rational roots first. By the rational root theorem any rational root divides $-18$, so the candidates are $\pm1, \pm2, \pm3, \pm6, \pm9, \pm18$. Testing every one gives a non-zero value — for instance

   $f(9) = 6561 - 7290 + 1458 - 360 - 18 = 351 \ne 0, \qquad f(-1) = 1 + 10 + 18 + 40 - 18 = 51 \ne 0$

   so there is no linear factor over $\mathbb{Q}$. In particular $x = 9$ is not a root, a tempting guess because $9 \mid 18$.

2. 2

   Rule out an integer split into two quadratics. Writing $f(x) = (x^2+ax+b)(x^2+cx+d)$ forces

   $a+c=-10, \quad ac+b+d=18, \quad ad+bc=-40, \quad bd=-18$

   An exhaustive sweep over every integer divisor pair $(b,d)$ of $-18$ and every integer $a$ finds no solution. Beware the near-miss $b=9, d=-2$: it satisfies the first two equations with $ac = 11$, $a,c = -5 \pm \sqrt{14}$, but then $ad+bc = -35 - 11\sqrt{14} \approx -76.2 \ne -40$. All four conditions must be checked, not three.

3. 3

   Locate the real roots by a sign table. Evaluate $f$ on a grid:

   $f(-1) = 51, \quad f(0) = -18, \quad f(8) = -210, \quad f(9) = 351$

   The sign flips on $(-1, 0)$ and on $(8, 9)$, so by the intermediate value theorem there is a real root in each. Since $f$ is negative across the whole span $[0, 8]$, there are exactly two real roots.

4. 4

   Refine with Newton's method. With $f'(x) = 4x^3 - 30x^2 + 36x - 40$, iterate $x_{n+1} = x_n - f(x_n)/f'(x_n)$ from the bracket midpoints:

   $x \approx 8.4610649, \qquad x \approx -0.3736462$

   At 40-digit working precision these satisfy $|f(x)| < 10^{-36}$, so the digits shown are correct.

5. 5

   Account for the remaining two roots. Dividing out the real pair leaves a quadratic with negative discriminant, giving the complex conjugates

   $x \approx 0.9562906 \pm 2.1861170\,i$

   So the real quadratic factorisation is $\left(x^2 - 8.0874187x - 3.1614449\right)\left(x^2 - 1.9125813x + 5.6935992\right)$, with no rational coefficients anywhere.

6. 6

   Confirm with Vieta's formulas. The four roots must sum to $10$ and multiply to $-18$:

   $8.4610649 - 0.3736462 + 2(0.9562906) = 10.0000000 \ \checkmark$

   $(8.4610649)(-0.3736462)\left(0.9562906^2 + 2.1861170^2\right) = -18.0000 \ \checkmark$

   Both identities hold, which cross-checks the numerics independently of the Newton iteration.