Question

Write a possible third degree polynomial with integer coefficient that have zeros: 1 2i, -4. Assume the leading coefficient to be 1

Answer 1

Answer:

The polynomial is:

$p(x) = x^3 + 2x^2 - 3x + 20$

Step-by-step explanation:

A third degree polynomial can be written in function of it's zeros $x_1, x_2, x_3$ the following way:

$p(x) = a(x - x_1)(x - x_2)(x - x_3)$

In which a is the leading coefficient.

Integer coefficient that have zeros: 1+2i, 1-2i, -4

Leading coefficient: 1

So

$p(x) = 1(x - (1+2i))(x - (1-2i))(x - (-4))$

$p(x) = (x - 1 -2i)(x - 1 + 2i)(x + 4)$

$p(x) = ((x-1)^2 - (2i)^2)(x + 4)$

$p(x) = (x^2 - 2x + 1 - 4i^2)(x + 4)$

Since $i^2 = -1$

$p(x) = (x^2 - 2x + 1 + 4)(x + 4)$

$p(x) = (x^2 - 2x + 5)(x + 4)$

$p(x) = x^3 + 4x^2 - 2x^2 - 8x + 5x + 20$

$p(x) = x^3 + 2x^2 - 3x + 20$