Respuesta :
[tex]x^2-4x-5 =\\\\x^2+x-5x-5=\\\\x(x+1)-5(x+1)=\\\\(x-5)(x+1)[/tex]
Easy. Given expression in current equation.
[tex]\mathbf{x^2 - 4x - 5}[/tex]
First break those following expressions in the current equation into grouped form, as to, relate the value as completely equal and not altering the actual expression. So;
[tex]\mathbf{(x^2 + x) + (- 5x - 5)}[/tex]
Factor out the variable of 'x' from the expression of [tex]\mathbf{x^2 + x}[/tex]. We are getting by factoring 'x'; [tex]\mathbf{x (x + 1)}[/tex].
Factor out the numbered negative value of '5' from the expression of [tex]\mathbf{- 5x - 5}[/tex]. We are getting by factoring '5'; [tex]\mathbf{- 5 (x + 1)}[/tex].
[tex]\mathbf{\therefore \quad x (x + 1) - 5 (x + 1)}[/tex]
Factor out the expression as a common term on both side, to obtain the final answer, that is, [tex]\mathbf{(x + 1)}[/tex]
[tex]\boxed{\mathbf{\underline{(x + 1)(x - 5)}}}[/tex]
Hope it helps.
[tex]\mathbf{x^2 - 4x - 5}[/tex]
First break those following expressions in the current equation into grouped form, as to, relate the value as completely equal and not altering the actual expression. So;
[tex]\mathbf{(x^2 + x) + (- 5x - 5)}[/tex]
Factor out the variable of 'x' from the expression of [tex]\mathbf{x^2 + x}[/tex]. We are getting by factoring 'x'; [tex]\mathbf{x (x + 1)}[/tex].
Factor out the numbered negative value of '5' from the expression of [tex]\mathbf{- 5x - 5}[/tex]. We are getting by factoring '5'; [tex]\mathbf{- 5 (x + 1)}[/tex].
[tex]\mathbf{\therefore \quad x (x + 1) - 5 (x + 1)}[/tex]
Factor out the expression as a common term on both side, to obtain the final answer, that is, [tex]\mathbf{(x + 1)}[/tex]
[tex]\boxed{\mathbf{\underline{(x + 1)(x - 5)}}}[/tex]
Hope it helps.