Solve the quadratic function.
 \[{k^2} - 11K - 102\].

Hint: We have a polynomial of degree 2 and it is called quadratic expression. We can solve the quadratic equation using the factorization method, or by using the quadratic formula, or by graphing method, or by completing the square method. The easy method is the factorization method, that is we split the middle term according to some condition.

Complete step-by-step solution:
Given, \[{k^2} - 11K - 102\].
Now consider the expression \[{k^2} - 11K - 102\]. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation \[a{k^2} + bk + c = 0\]. We have \[a = 1\], \[b = - 11\] and \[c = - 102\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that\[{b_1} \times {b_2} = ac\] and\[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = 6\] and \[{b_2} = - 17\]. Because \[{b_1} \times {b_2} = - 102\] \[(a \times c)\] and \[{b_1} + {b_2} = - 11(b)\].
Now we write \[{k^2} - 11K - 102\] as,
\[ = {k^2} + 6k - 17K - 102\]
Taking ‘k’ common in the first two terms and taking $-17$ common in the remaining two terms we have,
\[ = k(k + 6) - 17(K + 6)\]
Again taking \[(K + 6)\] common we have,
\[ = (K + 6)(k - 17)\].
The factors of \[{k^2} - 11K - 102\] are \[(K + 6)\] and \[(k - 17)\]. This is the required solution.

Note: Here instead of the variable ‘x’ we have variable ‘k’. The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. If we want to find the zeros or root of the given quadratic expression, we need to equate the obtained factors to zero.
\[(K + 6)(k - 17) = 0\]
Now using the zero-product principle we have
\[ \Rightarrow (K + 6) = 0\] and \[(k - 17) = 0\]
\[ \Rightarrow K = - 6\] and \[k = 17\]. These are the required roots.