Answer:
[tex](x-5)(x+1)[/tex]
Step-by-step explanation:
we have
[tex]x^{2} -4x-5[/tex]
To find the factors ------> solve the quadratic equation
equate the function to zero
[tex]x^{2} -4x-5=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} -4x-5=0[/tex]
so
[tex]a=1\\b=-4\\c=-5[/tex]
substitute
[tex]x=\frac{-(-4)(+/-)\sqrt{-4^{2}-4(1)(-5)}} {2(1)}[/tex]
[tex]x=\frac{4(+/-)\sqrt{36}} {2}[/tex]
[tex]x=\frac{4(+/-)6} {2}[/tex]
[tex]x=\frac{4(+)6} {2}=5[/tex]
[tex]x=\frac{4(-)6} {2}=-1[/tex]
therefore
the factors are
[tex](x-5)(x+1)[/tex]
