Factor or find the roots of x³ - 5x² - 7x + 41.

Step-by-step solution

1. 1

   Test for rational roots. By the rational root theorem, with leading coefficient $1$ any rational root must divide the constant term $41$ — and $41$ is prime, so the only candidates are $\pm1$ and $\pm41$:

   $f(1)=1-5-7+41=30,\qquad f(-1)=-1-5+7+41=42$

   and $f(\pm 41)$ is dominated by $41^3$, nowhere near zero. So there is no rational root, and the cubic does not factor over $\mathbb{Q}$.

2. 2

   Do not stop there — count the sign changes. Irrational roots still exist, and a cubic always has at least one. Tabulating $f$:

   $f(-3)=-10,\quad f(0)=41,\quad f(3)=2,\quad f(4)=-3,\quad f(5)=6$

   That is three sign changes: on $(-3,0)$, on $(3,4)$ and on $(4,5)$. So this cubic has three distinct real roots, not one — the common claim that it has a single real root near $-2.721$ is wrong twice over, since $f(-2.721)=2.88$ as well.

3. 3

   Confirm three roots with the discriminant of the derivative. $f'(x)=3x^2-10x-7$ has roots $x=\frac{10\pm\sqrt{184}}{6}$, i.e. a local maximum near $x=-0.60$ and a local minimum near $x=3.93$. Since $f(-0.60)\approx 43.2>0$ and $f(3.93)\approx -3.0<0$, the curve rises above the axis, dips below it, and rises again — exactly the configuration for three real crossings.

4. 4

   Locate each root by bisection. Narrowing each bracket to machine precision:

   $x_1=-2.7875545,\qquad x_2=3.2205249,\qquad x_3=4.5670296$

5. 5

   Verify with Vieta's formulas. For $x^3-5x^2-7x+41$ the three roots must satisfy $\sum x_i=5$, $\sum_{i<j}x_ix_j=-7$ and $x_1x_2x_3=-41$. Numerically the values give $5.000$, $-7.000$ and $-41.000$ ✓, so all three roots are correct and none is missing.

6. 6

   State the conclusion. The polynomial is irreducible over the rationals, yet it factors over the reals as

   $f(x)\approx (x+2.7875545)(x-3.2205249)(x-4.5670296)$