Factor Theorem: Finding Polynomial Roots

The Factor Theorem

If $f(c) = 0$ (plugging in $c$ gives zero), then $(x - c)$ is a factor of $f(x)$. This connects roots (zeros) and factors directly.

Step-by-step: Factor $f(x) = x^3 - 6x^2 + 11x - 6$

Step 1 — Verify $x = 3$ is a root. Compute $f(3)$:

$f(3) = 3^3 - 6(3^2) + 11(3) - 6 = 27 - 54 + 33 - 6 = 0 \checkmark$

Since $f(3) = 0$, the Factor Theorem says $(x - 3)$ is a factor.

Step 2 — Divide $f(x)$ by $(x - 3)$ using synthetic division:

$3 \;|\; 1 \quad -6 \quad 11 \quad -6$

Carry-multiply-add: $1, \; -3, \; 2, \; 0$. Quotient: $x^2 - 3x + 2$, remainder: $0$.

Step 3 — Factor the quotient. $x^2 - 3x + 2$: find two numbers that multiply to $2$ and add to $-3$: $-1$ and $-2$.

$x^2 - 3x + 2 = (x - 1)(x - 2)$

Step 4 — Complete factorization:

$x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$

Roots: $x = 1, 2, 3$.

Check: $(x-1)(x-2)(x-3)$ expanded: $(x^2-3x+2)(x-3) = x^3 - 6x^2 + 11x - 6$ ✓

Finding the first root

If no root is given, try the Rational Root Theorem: candidates are $\pm$ (factors of constant term) / (factors of leading coefficient). For $x^3 - 6x^2 + 11x - 6$: try $\pm 1, \pm 2, \pm 3, \pm 6$. Test $f(1) = 1 - 6 + 11 - 6 = 0$ ✓. Start with $x = 1$.

Try it in the visualization

Test different values of $x$. When $f(x) = 0$, the point lights up green on the graph. Synthetic division shows the factoring step. All roots are marked.