If \(f(a)=0\), then \((x-a)\) is a factor of \(f(x)\).

This gives a quick way to test whether a linear factor divides exactly into a polynomial.

To test whether \((x-a)\) is a factor of \(f(x)\):

1. Substitute \(x=a\) into \(f(x)\).
2. Simplify to find \(f(a)\).
3. If \(f(a)=0\), then \((x-a)\) is a factor.
4. If \(f(a)\ne 0\), then \((x-a)\) is not a factor.

Show that \((x-2)\) is a factor of

\[ f(x)=x^3-3x^2-4x+12 \]

Substitute \(x=2\):

\[ f(2)=2^3-3(2^2)-4(2)+12 \] \[ =8-12-8+12 \] \[ =0 \]

Determine whether \((x+1)\) is a factor of

\[ f(x)=2x^3+x^2-5x+4 \]

Since \((x+1)=(x-(-1))\), substitute \(x=-1\):

\[ f(-1)=2(-1)^3+(-1)^2-5(-1)+4 \] \[ =-2+1+5+4 \] \[ =8 \]

The factor theorem is often used to find unknown constants.

Example: Find \(k\) if \((x-3)\) is a factor of

\[ f(x)=x^3+kx^2-5x+6 \]

Since \((x-3)\) is a factor, \(f(3)=0\).

\[ f(3)=3^3+k(3^2)-5(3)+6 \] \[ 27+9k-15+6=0 \] \[ 18+9k=0 \] \[ 9k=-18 \] \[ k=-2 \]

The factor theorem is a special case of the remainder theorem.

When \(f(x)\) is divided by \((x-a)\), the remainder is \(f(a)\).

If the remainder is 0, then \((x-a)\) is a factor.

Once a factor is found, divide the polynomial by that factor to get the quotient.

Example: If \((x-2)\) is a factor of

\[ x^3-3x^2-4x+12 \]

then divide by \((x-2)\) to get:

\[ x^3-3x^2-4x+12=(x-2)(x^2-x-6) \]

and then factor further if possible.