Question

Use distributive property to factor method 1. Use distributive property to expand method 2.

Answer 1

Step-by-step explanation:

I guess method 1 means to deal with whole factors.

x + 5 = (x - 2)(x + 5)

for (x + 5) <> 0 we can divide both sides by this factor :

1 = x - 2

x = 3

for the second solution we deal with

x + 5 = 0

x = -5

so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.

method 2 : we multiply the expression out and solve it then

x + 5 = (x - 2)(x + 5)

x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10

0 = x² + 2x - 15

the general solution to such a square equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -15

x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =

= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =

= -1 ± 4

x1 = -1 + 4 = 3

x2 = -1 - 4 = -5

and you get the 2 solutions again. as expected, they are the same as with method 1, of course.