Question

Asking again... Please help! 11/20, will give brainliest. I do not tolerate spam answers!

Answer 1

Answer:

(x + 1 + $\sqrt{7}$ )(x + 1 - $\sqrt{7}$ )

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 = ± $\sqrt{7}$ ( subtract 1 from both sides )

x = - 1 ± $\sqrt{7}$

zeros are x = - 1 - $\sqrt{7}$ and x = - 1 + $\sqrt{7}$

Then factors are

(x - (- 1 - $\sqrt{7}$) ) , (x - (- 1 + $\sqrt{7}$ ) ), that is

(x + 1 + $\sqrt{7}$) and (x + 1 - $\sqrt{7}$ )

Thus

x² + 2x - 6 = (x + 1 + $\sqrt{7}$ )(x + 1 - $\sqrt{7}$ ) ← in factored form