Number of real roots, counting multiplicities, of the cubic $p_t(x) = (1+t^2)x^3 - 3t^3 x + t^4$

Solution

Step 1 — It is always a cubic. Since $1 + t^2 \neq 0$ for every real $t$, the polynomial $p_t$ is a genuine cubic. A real cubic has either $1$ or $3$ real roots counting multiplicity (non-real roots occur in conjugate pairs), so it is enough to track the sign of the discriminant.

Step 2 — Compute the discriminant. For $ax^3 + bx^2 + cx + d$,

With $a = 1+t^2,\ b = 0,\ c = -3t^3,\ d = t^4$, only two terms survive:

Factoring out $27\,t^8(1+t^2)$ gives

Step 3 — Analyze the sign. The factor $27\,t^8(1+t^2) \ge 0$, vanishing only at $t = 0$. Hence $\operatorname{sign}(\Delta) = -\operatorname{sign}(t^2 - 4t + 1)$, where $t^2 - 4t + 1$ has roots $t = 2 \pm \sqrt{3}$.

• $\Delta > 0$ — three distinct real roots — when $2-\sqrt{3} < t < 2+\sqrt{3}$.
• $\Delta < 0$ — one real, two complex — when $t < 2-\sqrt{3}$ or $t > 2+\sqrt{3}$ (with $t \neq 0$).
• $\Delta = 0$ — a repeated real root (all real) — at $t = 2 \pm \sqrt{3}$, and also at $t = 0$.

Step 4 — The special value $t = 0$. The factor $t^8$ forces $\Delta = 0$ at $t = 0$, even though $0 < 2 - \sqrt{3}$. Directly,

a triple root at $x = 0$ — so $3$ real roots counting multiplicity. This isolated value, where the count jumps from $1$ to $3$, is the subtle point of the problem.

Conclusion.

Quick checks.

• $t = 1$: $\ p_1 = 2x^3 - 3x + 1 = (x-1)(2x^2 + 2x - 1)$ — three real roots. ✓
• $t = -1$: $\ p_{-1} = 2x^3 + 3x + 1$, with $p' = 6x^2 + 3 > 0$ — strictly increasing, one real root. ✓
• $t = 0$: $\ p_0 = x^3$ — triple root, $3$ counting multiplicity. ✓