How can I factor this polynomial: k^2 - 17k + 16?

(k-1)(k-16)

\({k}^{2} -17k+16=0 \)

delta b²-4ac = (-17)²-4*1*16 = 225

\({k}_{1}= \frac{17+15}{2} = 16\)

\({k}_{2}=\frac{17-15}{2} = 1\)

\({k}^{2} -17k+16= (k-16)(k-1) \)