We will first factor \(x^2-2x-35\) using the \(ac\)-method.

Here \(a=1,\, b=-2\text{,}\) and \(c=-35\text{.}\)

\begin{equation*}
\begin{aligned}
ac=1(-35)=-35
\end{aligned}
\end{equation*}

Now we factor \(-35\text{,}\) and search for factors whose sum is \(-2\text{.}\)

\begin{equation*}
\begin{aligned}
-35\amp = -1(35)\\
\amp = -5(7)\\
\amp = \alert{-7(5)}\\
\end{aligned}
\end{equation*}

In this case, the sum of the factors \(-7\) and \(5\) equals the middle coefficient, \(-2\text{.}\) Therefore, \(-2x=5x-7x\text{,}\) and we can write

\begin{equation*}
\begin{aligned}
x^2-2x-35+6=x^2+5x-7x-35.
\end{aligned}
\end{equation*}

We factor the equivalent expression by grouping.

\begin{equation*}
\begin{aligned}
x^2-2x-35+6\amp =x^2+5x-7x-35\\
\amp = x(x+5)-7(x+5)\\
\amp = (x+5)(x-7)
\end{aligned}
\end{equation*}

Our factored form is \((x+5)(x-7)\text{.}\) Now we can use the zero factor property to solve by setting each variable factor equal to zero.

\begin{equation*}
\begin{aligned}
x+5\amp= 0\\
x\amp =-5
\end{aligned}
\end{equation*}

OR

\begin{equation*}
\begin{aligned}
x-7\amp= 0\\
x\amp =7
\end{aligned}
\end{equation*}

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor.

The solutions are \(x=-5,7\text{.}\)