Solve x^3 - 59x^2 - 420x + 10800 = 0.

Step-by-step solution

1. 1

   Try the rational root theorem first, and let it fail. Any rational root of a monic integer cubic must be an integer divisor of the constant term $10800=2^{4}\cdot 3^{3}\cdot 5^{2}$. Testing the plausible candidates: $f(10)=1700$, $f(12)=-1008$, $f(15)=-5400$, $f(20)=-13200$, $f(-15)=450$, $f(-20)=-12400$. None of them vanish, and the same holds for the remaining divisors, so there is no rational root and no factorisation over the integers - the cubic has to be solved numerically.

2. 2

   Locate the roots by sign changes. Those same evaluations of $f(x)=x^{3}-59x^{2}-420x+10800$ already bracket every root. Since $f(-20)=-12400<0$ and $f(-15)=450>0$, a root lies in $(-20,-15)$. Since $f(11)=372>0$ and $f(12)=-1008<0$, a second lies in $(11,12)$. And since $f(62)=-3708<0$ and $f(63)=216>0$, the third lies in $(62,63)$. Three sign changes for a cubic means all three roots are real and simple.

3. 3

   Refine each bracket with Newton's method. With $f'(x)=3x^{2}-118x-420$, iterating $x\mapsto x-\dfrac{f(x)}{f'(x)}$ from the midpoint of each bracket converges in a handful of steps to $x_{1}=-15.219737190,\qquad x_{2}=11.273122422,\qquad x_{3}=62.946614768.$ The derivative is comfortably away from zero at each root ($2070.85$, $-1368.98$ and $4039.13$), so Newton converges cleanly.

4. 4

   Verify with all three Vieta relations, not just the sum. For $x^{3}+ax^{2}+bx+c$ the roots satisfy $\sum r_{i}=-a$, $\sum_{i<j}r_{i}r_{j}=b$ and $r_{1}r_{2}r_{3}=-c$. Our values give $\sum r_{i}=59.000000,\qquad \sum_{i<j}r_{i}r_{j}=-420.000000,\qquad r_{1}r_{2}r_{3}=-10800.000000,$ matching $59$, $-420$ and $-10800$ exactly. Checking only the sum is not enough - a wrong pair of roots can still add up correctly while the product is badly off.

5. 5

   State the answer. $x\approx -15.219737,\qquad x\approx 11.273122,\qquad x\approx 62.946615.$ The huge spread between them is what the coefficients predicted: a leading behaviour of $x^{3}-59x^{2}$ pushes one root out past $60$, while the large constant $10800$ keeps the other two comparatively small.