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caylus
Hello,

f(x)=(x-3)(x-7)=x²-10x+21

Answer C

Answer:

Function which has real zeros at x = 3 and x = 7 is:

f(x)= x²-10x+21

Step-by-step explanation:

function has real zeros at x = 3 and x = 7

⇒ (x-3)(x-7)=0

⇒ x(x-7)-3(x-7)=0

⇒ x²-7x-3x+21=0

⇒ x²-10x+21=0

f(x)= x²-10x+21 has real zeros at x=3 and x=7

Hence, function which has real zeros at x = 3 and x = 7 is:

f(x)= x²-10x+21