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

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Answer 1 · 2016-01-19T21:32:28

$(x-3)^2$

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

$x^2 +2ax+ a^2$ , which gives the answer automatically:
$(x+a)^2$ , since it is a notable case.

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

second degree polynomials:

$alpha=(-b +-sqrt(b^2-4ac))/(2a)$

so, it will be:

$alpha=(6 +-sqrt(6^2-4*1*9))/(2*1)$

$=(6 +-sqrt(36-36))/(2)$

$=(6+-0)/2$

which can be devided in

$alpha1=(6+0)/2$ and $alpha2=(6-0)/2$

$alpha1=3$ or $alpha2=3$

So the factorization is

$(x-alpha1)(x-alpha2)$

$(x-3)(x-3)$ or $(x-3)^2$

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