The expression x^3 - 2x^2 + 1 is a useful example for learning how to factor a cubic polynomial. To factor x^3 - 2x^2 + 1, look for a rational root first, use that root to divide the polynomial, and then factor the remaining quadratic if possible. The final factorization over the integers is (x - 1)(x^2 - x - 1), while the full factorization over the real
numbers, we apply the quadratic formula to the remaining factor $x^2 - x - 1 = 0$. Using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find the roots $x = \frac{1 \pm \sqrt{5}}{2}$. Because of this, the complete factorization over the real numbers is $(x - 1)\left(x - \frac{1 + \sqrt{5}}{2}\right)\left(x - \frac{1 - \sqrt{5}}{2}\right)$.
Pulling it all together, factoring a cubic polynomial like $x^3 - 2x^2 + 1$ demonstrates a powerful algebraic technique: reducing the degree of the polynomial by identifying a single root. This stepwise approach—finding a rational root, performing division, and solving the resulting quadratic—provides a clear pathway to fully decompose the expression, whether over the integers or the real numbers.
Easier said than done, but still worth knowing.
To verify the factorization, multiply the factors back together:
[ (x-1)(x^2-x-1) = x(x^2-x-1)-1(x^2-x-1) ]
[ = x^3-x^2-x-x^2+x+1 = x^3-2x^2+1. ]
Since the product returns the original polynomial, the factorization is correct.
This example also shows why the Rational Root Theorem is especially helpful with cubics. Even so, once a rational zero is found, the cubic can be reduced to a quadratic, which is usually much easier to handle. In this case, (x=1) is the rational root, and the remaining quadratic produces two irrational roots. That means the polynomial factors nicely over the real numbers, but not completely over the integers That's the whole idea..
Another useful interpretation comes from graphing. The three real roots correspond to three (x)-intercepts:
[ x=1,\qquad x=\frac{1+\sqrt5}{2},\qquad x=\frac{1-\sqrt5}{2}. ]
Because each factor appears only once, the graph crosses the (x)-axis at each root rather than merely touching it It's one of those things that adds up..
Overall, this polynomial is a strong example of how factoring combines several key algebraic ideas: testing possible roots, using polynomial division, applying the quadratic formula, and checking the final answer. By breaking the cubic into simpler pieces, we gain both an algebraic factorization and a clearer understanding of the polynomial’s behavior.
