Roots and Factor Theorem

What is a Root?

A root (or zero) of a polynomial is a value that makes the polynomial equal to zero.

For p(x)=x25x+6p(x) = x^2 - 5x + 6, the roots are the solutions to:

x25x+6=0x^2 - 5x + 6 = 0

We can factor this: (x2)(x3)=0(x - 2)(x - 3) = 0

So the roots are x=2x = 2 and x=3x = 3.


Check: Plug them back in:

xxp(x)=x25x+6p(x) = x^2 - 5x + 6
22410+6=04 - 10 + 6 = 0
33915+6=09 - 15 + 6 = 0

The Factor Theorem

This is the key insight connecting roots and factors:

If rr is a root of p(x)p(x), then (xr)(x - r) is a factor of p(x)p(x).

And vice versa:

If (xr)(x - r) is a factor of p(x)p(x), then rr is a root.


Example:

  • If x=4x = 4 is a root of p(x)p(x), then (x4)(x - 4) is a factor.
  • If (x+2)(x + 2) is a factor of p(x)p(x), then x=2x = -2 is a root.

Why is This Useful?

Finding factors from roots:

If you can find a root, you’ve found a factor.

Finding roots from factors:

Once factored, roots are obvious.

Reducing polynomials:

Once you find one root rr, divide out (xr)(x - r) and work with a smaller polynomial.


Example: Factoring a Cubic

Factor: x36x2+11x6x^3 - 6x^2 + 11x - 6


Step 1: Find a root

Try small integers: ±1,±2,±3,\pm 1, \pm 2, \pm 3, \ldots

Try x=1x = 1:

16+116=01 - 6 + 11 - 6 = 0

So x=1x = 1 is a root, which means (x1)(x - 1) is a factor.


Step 2: Divide out the factor

Divide x36x2+11x6x^3 - 6x^2 + 11x - 6 by (x1)(x - 1).

Result: x25x+6x^2 - 5x + 6

So: x36x2+11x6=(x1)(x25x+6)x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)


Step 3: Factor the remaining quadratic

x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3)


Final answer:

x36x2+11x6=(x1)(x2)(x3)x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)

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