Example: Consider the third degree polynomial $$ f(x)=x^3-9x^2+25x-25 $$
  1. Use a graphing calculator to find a real zero of \(f(x)\).
    Cubic
    From the figure: \(x=-2\) is a zero of \(f(x)\) is equivalent to \((x+2)\) is a factor of \(f(x)\).
  2. Find all the zeros (aka roots) of \(f(x)\).
    Since \((x+2)\) is a factor of \(f(x)\), \((x+2)\) divides \(f(x)\) with a zero remainder. $$ \require{enclose} \begin{array}{r} x^2-4x+5\hspace{19mm} \\[-3pt] x-5 \enclose{longdiv}{x^3-9x^2+25x-25}\kern-.2ex \\[-3pt] \underline{x^3-5x^2\phantom{-25x-25..}} \\[-3pt] -4x^2+25x-25 \\[-3pt] \underline{-4x^2+20x\phantom{.+50}} \\[-3pt] \phantom{x^4-4x^3-25x^2+}5x-25 \\[-3pt] \underline{5x-25} \\[-3pt] \phantom{00}0 \end{array} $$ Hence $$ f(x)=x^3-9x^2+25x-25=(x-5)(x^2-4x+5) $$ Roots of the \((x^2-4x+5)\) factor are $$ x_{2,3}=\frac{4\pm\sqrt{16-4\cdot 5}}{2}=\frac{4\pm\sqrt{-4}}{2}=\frac{-4\pm 2i}{2}=2\pm i $$
  3. Write \(f(x)\) as a product of linear factors.
    $$ f(x) = (x-5)(x-[2+i])(x-[2-i]) $$
  4. Verify the correctness of your factorization above by direct multiplication.
    \begin{align} f(x) &= (x-5)(x-[2+i])(x-[2-i]) \\ &= (x-5)([x-2]-i])([x-2]+i]) \\ &= (x-5)\left([x-2]^2-i^2 \right) \\ &= (x-5)\left( x^2-4x+4+1 \right) \\ &= (x-5)\left( x^2-4x+5 \right) \\ &= x^3-9x^2+25x-25 \end{align}