To find the polynomial function whose roots include $4$ and an imaginary sign, we interpret this as a root of the form $a \pm bi$, where $a$ is the real part and $b$ is the imaginary part. However, roots of polynomials with real coefficients must occur in conjugate pairs. If one root is $4$ and there is an imaginary root, let the roots be:

1. $4$ (real root)
2. $a + bi$ (imaginary root)
3. $a - bi$ (conjugate of the imaginary root)

Step-by-Step Derivation

Assuming the imaginary roots are $i$ (simplest case where $a = 0, b = 1$):

• The roots are: $4$, $i$, and $-i$.

The polynomial is formed as: $P(x) = (x - 4)(x - i)(x + i)$

Now simplify:

1. Expand the imaginary part: $(x - i)(x + i) = x^2 + 1$

2. Combine with the real root: $P(x) = (x - 4)(x^2 + 1)$

Expand fully: $P(x) = x(x^2 + 1) - 4(x^2 + 1)$ $P(x) = x^3 + x - 4x^2 - 4$

Rearranging: [ P(x) = x^3 - 4x^2 + x - 4