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Let ​f(x)=x2+12x+32.​

What are the zeros of the function?

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User BoltKey
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Explanation:

To find the zeros of the function f(x), we need to solve the equation f(x) = 0.

f(x) = x^2 + 12x + 32

Setting f(x) equal to zero and factoring, we get:

0 = x^2 + 12x + 32

0 = (x + 4)(x + 8)

Using the zero product property, we set each factor equal to zero and solve for x:

x + 4 = 0 or x + 8 = 0

x = -4 or x = -8

Therefore, the zeros of the function f(x) are -4 and -8.

User Kleinohad
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