Question

# Polynomial equation with real coefficients that has roots -1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.

Polynomial arithmetic
Polynomial equation with real coefficients that has roots $$-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.$$

2021-01-20
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 $$-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0$$