# Is it possible to factor y=x^2-6x+9 ? If so, what are the factors?

Jan 19, 2016

${\left(x - 3\right)}^{2}$

#### Explanation:

Imagine you haven't realized that expression is of the type:

${x}^{2} + 2 a x + {a}^{2}$, which gives the answer automatically:
${\left(x + a\right)}^{2}$, since it is a notable case.

In these cases you have to use the general method for solving

second degree polynomials:

$\alpha = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

so, it will be:

$\alpha = \frac{6 \pm \sqrt{{6}^{2} - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

$= \frac{6 \pm \sqrt{36 - 36}}{2}$

$= \frac{6 \pm 0}{2}$

which can be devided in

$\alpha 1 = \frac{6 + 0}{2}$ and $\alpha 2 = \frac{6 - 0}{2}$

$\alpha 1 = 3$ or $\alpha 2 = 3$

So the factorization is

$\left(x - \alpha 1\right) \left(x - \alpha 2\right)$

$\left(x - 3\right) \left(x - 3\right)$ or ${\left(x - 3\right)}^{2}$