Write each quadratic function below in terms of linear factors.a)f(x)=x²-25b)f(x)=x²+25c)f(x)=4x²+25d)f(x)=x²-2x+1e)f(x)=x²-2x+4

a) (x + 5) (x - 5)

b) (x + 5i) (x - 5i)

c) (x + (5i/2)) (x - (5i/2))

d) (x-1)(x-1)

e) x +i√3 +1) (x -i√3+1)

Step-by-step explanation:

To solve this, we will need to factorize each quadratic function making it equal to zero first and then proceeding to find x  

a) f(x) = x²-25

x²-25 = 0

⇒(x + 5) (x - 5)

b) f(x)=x²+25

x² + 25 = 0

x²= -25

x = √-25

x = √25i

x = ±5i

⇒(x + 5i) (x - 5i)

c) f(x)=4x²+25

4x²+25 = 0

4x²= -25

x² = -25/4

x = ±√(-25/4)

x = ±(√25i)/2

x = ±5i /2

⇒(x + (5i/2)) (x - (5i/2))

d) f(x)=x²-2x+1

x²-2x+1 = 0

⇒(x - 1)²

e) f(x)=x²-2x+4

x²-2x+4 = 0

x²-2x = -4

x²-2x +1 = -4 +1

x²-2x + 1 = -3

(x-1)² +3 = 0

(x-1)²= -3

x-1 = √-3

x = ±√3i +1

⇒(x +i√3 +1) (x -i√3+1)

The answer is NOT C.The correct answer is A.They intersect at one point.