Example: Consider the third degree polynomial
$$
f(x)=x^3-9x^2+25x-25
$$
- Use a graphing calculator to find a real zero of \(f(x)\).
From the figure: \(x=-2\) is a zero of \(f(x)\) is equivalent to \((x+2)\) is a factor of \(f(x)\).
- 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
$$
x_{2,3}=\frac{4\pm\sqrt{16-4\cdot 5}}{2}=\frac{4\pm\sqrt{-4}}{2}=\frac{-4\pm 2i}{2}=2\pm i
$$
- Write \(f(x)\) as a product of linear factors.
$$ f(x) = (x-5)(x-[2+i])(x-[2-i]) $$
-
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}