To find the zeroes, we set f(x) = 0. Once we do that we can use the Zero Product Property (if a product is zero, so are its pieces) and factoring.
0 = 2x² + 32
0 = 2(x² + 16) Divide 2 out of both terms
Since 2 ≠ 0, we can move onto the other piece.
0 = x² + 16
-16 = x² Subtract 16 on both sides
√-16 or -√-16 = x Take the square root of both sides
Square roots can be broken into pieces, and i = √-1. We use that to help finish.
x = √-16
x = √-1 * √16
x = √-1 * 4
x = i * 4 = 4i
Repeating the root simplification with x = -√-16 yields x = -4i.
Thus the zeroes are x = 4i and x = -4i
The zeros of the function given are 4i and -4i
Given the function f(x) = 2x^2 + 32
The zeros occurs at the point where f(x) = 0
Substitute this into the function to have:
0 = 2x^2 + 32
-2x^2 = 32
x^2 = -32/2
x^2 = -16
x = √-16
x = ±4i
Hence the zeros of the function given are 4i and -4i
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