Factor Theorem and Polynomial Roots

The Factor Theorem

If $f(r) = 0$ (plugging in $r$ gives zero), then $(x - r)$ is a factor of $f(x)$. Conversely, if $(x - r)$ is a factor, then $r$ is a root (zero).

Step-by-step: Factor $x^3 - 6x^2 + 11x - 6$

Step 1 — Test candidates. Try $x = 1$: $f(1) = 1 - 6 + 11 - 6 = 0$ ✓ → $(x - 1)$ is a factor.

Step 2 — Divide by $(x - 1)$ using synthetic division: quotient = $x^2 - 5x + 6$.

Step 3 — Factor the quotient: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Complete factorization: $(x - 1)(x - 2)(x - 3)$. Roots: $x = 1, 2, 3$.

Finding the first root

Use the Rational Root Theorem: candidates are $\pm$(factors of constant) / (factors of leading coefficient). For $x^3 - 6x^2 + 11x - 6$: try $\pm 1, \pm 2, \pm 3, \pm 6$.

Try it in the visualization

Adjust the roots. The cubic graph shows x-intercepts at each root. The factored form updates. Test $f(r)$ at each candidate to verify. For $x^3 - 6x^2 + 11x - 6$: $f(1) = 0$, $f(2) = 0$, $f(3) = 0$, so the factored form is $(x-1)(x-2)(x-3)$.