Using factoring and reducing divide (x2+5x-6) by (x+6)

Hello,

Let's find the roots of equation "x^2 +5x-6''

We have :

a = 1
b = 5
c = -6

Delta = b^2 - 4ac

Delta = 5^2 - 4.1.(-6)

Delta = 25 + 24

Delta = 49

Then,

x = [ -b +/- √( Delta) ]/2a

Replacing the values:

x = [ -5 +/- √(49) ] /2.1

x = [ -5 +/- 7]/2

Then,

x' = (-5-7)/2 <=> -6

And

x" = (-5+7)/2 <=> 1

As ax^2+bx+c = a(x-x')(x-x")

Then,

x^2 +5x-6 = 1.(x-(-6))(x-1)

= (x+6).(x-1)

Now replacing this equation:

(x^2+5x-6)/(x+6) = (x+6).(x-1) / (x+6)

Cuting, (x+6) we stay with:

= (x+1)

= x+1

I hope this has helped!