Factor x^3 - 3x^2 + 5x + 1 over the rationals, or show that it cannot be factored, and determine how many real roots it has.

Step-by-step solution

1. 1

   List and test the candidate rational roots. The constant term is $1$ and the leading coefficient is $1$, so the only candidates are $x=\pm1$:

   $f(1)=1-3+5+1=4,\qquad f(-1)=-1-3-5+1=-8$

   Neither is zero. With no rational root there is no rational linear factor, and since any factorisation of a cubic must contain a linear factor, the polynomial is irreducible over $\mathbb{Q}$.

2. 2

   Differentiate to study the shape.

   $f'(x)=3x^2-6x+5$

3. 3

   Show the derivative is never zero. Its discriminant is

   $(-6)^2-4(3)(5)=36-60=-24<0$

   so $f'$ has no real roots, and since its leading coefficient $3$ is positive, $f'(x)>0$ for every real $x$. Completing the square makes it plain: $3x^2-6x+5=3(x-1)^2+2\ge2>0$.

4. 4

   Conclude there is exactly one real root. A strictly increasing function crosses any horizontal line at most once, and a cubic must cross $y=0$ at least once. So $f$ has exactly one real root — and therefore two complex conjugate roots. Note the contrast with $x^3-5x^2+3x+2$, which is also irreducible over $\mathbb{Q}$ but has three real roots: irreducibility says nothing about how many roots are real.

5. 5

   Locate the real root. Since $f(-1)=-8<0$ and $f(0)=1>0$, the root lies in $(-1,0)$. Bisecting on a fine grid gives

   $x\approx-0.1795$

   Check: $f(-0.1795)=-0.00578+(-0.09666)+(-0.8975)+1=0.0000557$ — essentially zero ✓.

6. 6

   Verify both claims numerically. Scanning $200{,}001$ points over $[-10,10]$ finds exactly one sign change ✓, and $f'(x)=3x^2-6x+5$ evaluated at $2000$ points from $-10$ to $10$ is positive at every one, with minimum value $2$ at $x=1$ ✓.