Factor x^3 - 5x^2 + 3x + 2, or show that it does not factor over the rationals, and find its real roots.

Step-by-step solution

1. 1

   List the candidate rational roots. By the Rational Root Theorem, a rational root $\tfrac{p}{q}$ in lowest terms has $p$ dividing the constant $2$ and $q$ dividing the leading coefficient $1$. So $q=\pm1$ and the only candidates are

   $x=\pm1,\ \pm2$

2. 2

   Test every candidate.

   $f(1)=1-5+3+2=1,\qquad f(-1)=-1-5-3+2=-7$

   $f(2)=8-20+6+2=-4,\qquad f(-2)=-8-20-6+2=-32$

   None is zero, so there is no rational root and hence no linear factor with rational coefficients. A cubic factors over $\mathbb{Q}$ only if it has a rational root (any factorisation must include a linear piece), so this cubic is irreducible over $\mathbb{Q}$.

3. 3

   Do not confuse irreducible with rootless. A real cubic always has at least one real root, because it runs from $-\infty$ to $+\infty$ continuously. "Does not factor over the rationals" says nothing about real roots — it only says none of them is a fraction.

4. 4

   Count the real roots with sign changes. Evaluate at a few points: $f(-1)=-7$, $f(0)=+2$, $f(1)=+1$, $f(2)=-4$, $f(5)=+17$. The sign changes between $-1$ and $0$, between $1$ and $2$, and between $2$ and $5$ — three sign changes, so all three roots are real and irrational.

5. 5

   Locate them numerically. Bisecting on a fine grid over $[-10,10]$:

   $x_1\approx-0.3914,\qquad x_2\approx1.2271,\qquad x_3\approx4.1642$

   So the real factorisation is $(x+0.3914)(x-1.2271)(x-4.1642)$ to four decimals — perfectly valid over $\mathbb{R}$, just not over $\mathbb{Q}$.

6. 6

   Check with Vieta's formulas. The roots should sum to $-b/a=5$: $-0.3914+1.2271+4.1642=4.9999$ ✓, and their product should be $-d/a=-2$: $(-0.3914)(1.2271)(4.1642)=-2.0000$ ✓. Both agree to grid accuracy, confirming that three real roots have been found and none was missed.