Respuesta :
Answer:
(x + 1 + [tex]\sqrt{7}[/tex] )(x + 1 - [tex]\sqrt{7}[/tex] )
Step-by-step explanation:
There are no integer values with product - 6 and sum + 2
Calculate the zeros then obtain the factors , that is
if x = a is a zero then (x - a) is a factor
Given
x² + 2x - 6 ( equate to zero )
x² + 2x - 6 = 0 ( add 6 to both sides )
x² + 2x = 6
To complete the square
add ( half the coefficient of the x- term)² to both sides
x² + 2(1)x + 1 = 6 + 1
(x + 1)² = 7 ( take the square root of both sides )
x + 1 = ± [tex]\sqrt{7}[/tex] ( subtract 1 from both sides )
x = - 1 ± [tex]\sqrt{7}[/tex]
zeros are x = - 1 - [tex]\sqrt{7}[/tex] and x = - 1 + [tex]\sqrt{7}[/tex]
Then factors are
(x - (- 1 - [tex]\sqrt{7}[/tex]) ) , (x - (- 1 + [tex]\sqrt{7}[/tex] ) ), that is
(x + 1 + [tex]\sqrt{7}[/tex]) and (x + 1 - [tex]\sqrt{7}[/tex] )
Thus
x² + 2x - 6 = (x + 1 + [tex]\sqrt{7}[/tex] )(x + 1 - [tex]\sqrt{7}[/tex] ) ← in factored form
