Given:) \(-1, 4 - 2i\)

Used Formula:

\((a+b)(a-b) = a^2 - b^2\)

Calculation:

If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.

Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.

Hence, below mentioned equation is written as.

\([x - (-1)] [x - (4 - 2i)][x-(4 + 2i)] = 0\)

\((x + 1) (x - 4 + 2i)(x - 4 - 2i) = 0\)

Further use arithmetic rule.

\((a + b)(a - b) = a^2 - b^2\)

Here, \(a = x - 4, b = 2i\)

Now, the polynomial equation is,

\((x + 1)[(x - 4)^2 - (2i)^2] = 0\)

Use arithmetic rule.

\((a - b)^2 = a^2 - 2ab + b^2\)

And \(i^2 = -1\)

Now, the polynomial equation is,

\((x + 1) (x^2 - 8x + 16 + 4) = 0\)

\((x + 1) (x^2 - 8x + 20) = 0\)

\((x^3 + x^2 - 8x^2 - 8x + 20x + 20) = 0\)

\(x^3 - 7x^2 + 12x + 20 = 0\)

Hence, the polynomial equation of giveb roots \(-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0\)

Used Formula:

\((a+b)(a-b) = a^2 - b^2\)

Calculation:

If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.

Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.

Hence, below mentioned equation is written as.

\([x - (-1)] [x - (4 - 2i)][x-(4 + 2i)] = 0\)

\((x + 1) (x - 4 + 2i)(x - 4 - 2i) = 0\)

Further use arithmetic rule.

\((a + b)(a - b) = a^2 - b^2\)

Here, \(a = x - 4, b = 2i\)

Now, the polynomial equation is,

\((x + 1)[(x - 4)^2 - (2i)^2] = 0\)

Use arithmetic rule.

\((a - b)^2 = a^2 - 2ab + b^2\)

And \(i^2 = -1\)

Now, the polynomial equation is,

\((x + 1) (x^2 - 8x + 16 + 4) = 0\)

\((x + 1) (x^2 - 8x + 20) = 0\)

\((x^3 + x^2 - 8x^2 - 8x + 20x + 20) = 0\)

\(x^3 - 7x^2 + 12x + 20 = 0\)

Hence, the polynomial equation of giveb roots \(-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0\)