How do you factor a perfect square trinomial $x^{2} - 6 x + 9$ ?

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Solution

Factorise the expression:
Given expression is $x^{2} - 6 x + 9$ .
$\begin{array}{rcl} x^{2} - 6 x + 9 & = & x^{2} - 3 x - 3 x + 9 (Splitting the middle term) \\ \Rightarrow & x^{2} - 6 x + 9 = x (x - 3) - 3 (x - 3) (Taking common term out) \\ ∴ x^{2} - 6 x + 9 & = & (x - 3) (x - 3) \end{array}$
Therefore, factor of $x^{2} - 6 x + 9$ are $(x - 3) (x - 3)$ .

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