Questions: Find the polynomial function in standard form that has the zeros listed. 4i f(x)=x^2+8x+16 f(x)=x^2-16 f(x)=x^2+16 f(x)=x^2-8x+16

Transcript text: Find the polynomial function in standard form that has the zeros listed. $4 i$ $f(x)=x^{2}+8 x+16$ $f(x)=x^{2}-16$ $f(x)=x^{2}+16$ $f(x)=x^{2}-8 x+16$

Solution

Solution Steps

Step 1: Identify the given zeros

The given zero is $ 4i $. Since complex zeros come in conjugate pairs for polynomials with real coefficients, the other zero is $ -4i $.

Step 2: Form the polynomial from the zeros

The polynomial can be formed by multiplying the factors corresponding to the zeros: \[ (x - 4i)(x + 4i) \]

Step 3: Expand the polynomial

Expand the product: \[ (x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 - 16i^2 \] Since $ i^2 = -1 $, this simplifies to: \[ x^2 - 16(-1) = x^2 + 16 \]

Step 4: Compare with the given options

The polynomial $ x^2 + 16 $ matches one of the given options: \[ f(x) = x^2 + 16 \]

Final Answer

$\boxed{x^2 + 16}$

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