Is it possible to factor y=x^3 - 3x^2 + 2 ? If so, what are the factors?

Mar 7, 2016

Factors of y=x^3−3x^2+2 are $y = \left(x - 1\right) \left({x}^{2} - 2 x + 2\right)$

Explanation:

One of the easiest way to factorize a simple polynomial (here I mean those with single variable (here $x$) and integers as coefficients) is identify zeros of the polynomials.

Zeros of the polynomials are those numbers, which when put in place of variable, result in polynomial's value becoming zero. These are generally factors of the independent term (without $x$), which is $2$ here. Factors of $2$ are $\left\{1 , - 1 , 2 , - 2\right\}$.

Clearly $1$ is one such zero of y=x^3−3x^2+2, then $\left(x - 1\right)$ is one such factor. Dividing x^3−3x^2+2 by $\left(x - 1\right)$ we get ${x}^{2} - 2 x + 2$ and hence $y = \left(x - 1\right) \left({x}^{2} - 2 x + 2\right)$.

As the polynomial $\left({x}^{2} - 2 x + 2\right)$ is quadratic, one can check discriminant to find if rational factors are possible or not. Discriminant of quadratic polynomial $\left(a {x}^{2} + b x + c\right)$ is $\left({b}^{2} - 4 a c\right)$. If it is negative no real factors exist and if it is square of a number, rational roots are possible.

Here, $\left({b}^{2} - 4 a c\right) = \left({\left(- 2\right)}^{2} - 4 \times 1 \times 2\right) = \left(4 - 8\right) = - 4$ and as such no further real roots exist.

So factors of y=x^3−3x^2+2 are $y = \left(x - 1\right) \left({x}^{2} - 2 x + 2\right)$