The zeros of the function f (x) = (x+2) ^2 - 25 are?

Zeros of the function

f (x) = (x + 2) ² - 25

f (x) = (x + 2) (x + 2) - 25

f (x) = x (x + 2) + 2 (x + 2) - 25

f (x) = x (x) + x (2) + 2 (x) + 2 (2) - 25

f (x) = x² + 2x + 2x + 4 - 25

f (x) = x² + 4x + 4 - 25

f (x) = x² + 4x - 21

x² + 4x - 21 = 0

x = - (4) + / - √ ((4) ² - 4 (1) (-21))

2 (1)

x = - 4 + / - √ (16 + 84)

2

x = - 4 + / - √ (100)

2

x = - 4 + / - 10

2

x = - 2 + 5

x = - 2 + 5 x = - 2 - 5

x = 3 x = - 7

f (x) = x² + 4x - 21

f (3) = (3) ² + 4 (3) - 21

f (3) = 9 + 12 - 21

f (3) = 21 - 21

f (3) = 0

(x, f (x)) = (3, 0)

or

f (x) = x² + 4x - 21

f (-7) = (-7) ² + 4 (-7) - 21

f (-7) = 49 - 28 - 21

f (-7) = 21 - 21

f (-7) = 0

(x, f (x)) = (-7, 0)

Vertex

X - Intercept

-b/2a = - (4) / 2 (1) = - 4/2 = - 2

Y - Intercept

y = x² + 4x - 21

y = (-2) ² + 4 (-2) - 21

y = 4 - 8 - 21

y = - 4 - 21

y = - 25

(x, y) = (-2, - 25)

Y=x^2+4x+4-25

y=x^2+4x-21

y = (x+7) (x-3)

zeros are - 7 and 3