Factor. \[ c^{2}+6 c+9 \]

Step 1 ：\(c^2\) and \(9\) are perfect squares, since \(c^2=(c)^2\) and \(9=(3)^2\).

Step 2 ：\(6c\) is twice the product of the roots of \(c^2\) and \(9\), since \(6c=2(c)(3)\).

Step 3 ：So, \(c^2+6c+9 = (c)^2+2(c)(3)+(3)^2\).

Step 4 ：We can use the square of a sum pattern to factor: \(a^2 +2ab+ b^2 =(a+b)^2\). In this case, \(a=c\) and \(b=3\).

Step 5 ：Therefore, \((c)^2+2(c)(3)+(3)^2 =(c+3)^2\).

Step 6 ：In conclusion, \(c^2+6c+9=(c +3)^2\).

Step 7 ：You can always check your factorization by expanding it.

Step 8 ：So, the final answer is \(\boxed{c^2+6c+9=(c +3)^2}\).