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we can rewrite the statement
[tex](x-5)(A)=x^2-x-20[/tex]where A is the missing factor, A must be of the form
[tex](x+a)[/tex]where a is a constant, to obtain "a" we must bear in mind that the multiplication of the two constants must give us the third term and the sum of these must give us the second term
so
[tex]\begin{gathered} -5\times a=-20 \\ -5+a=-1 \end{gathered}[/tex]if we solve any equation, the value of a is 4
so a is 4 and the factor is
[tex](x+4)[/tex][tex](x-5)(x+4)=x^2-x-20[/tex]