The polynomial equation with zeroes 2i, -2i, 2 is [tex]x^3 -2x^2 + 4x - 8 = 0[/tex]
Given that zeros of polynomial are 2i, -2i, 2
To find: polynomial equation in standard form
zeros of polynomial are 2i, -2i, 2. So we can say,
x = 2i
x = -2i
x = 2
Or x - 2i = 0 and x + 2i = 0 and x - 2 = 0
Multiplying the above factors, we get the polynomial equation
[tex](x - 2i)(x + 2i)(x - 2) = 0\\[/tex] ------- eqn 1
Using a algebraic identity,
[tex](a - b)(a + b) = a^2 - b^2[/tex]
Thus [tex](x - 2i)(x + 2i) = x^2 - (2i)^2[/tex]
We know that [tex]i^2 = -1[/tex]
[tex]Thus (x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 -4(-1) = x^2 + 4[/tex]
Substitute the above value in eqn 1
[tex](x^2 + 4)(x - 2) = 0[/tex]
Multiply each term in first bracket with each term in second bracket
[tex]x^3 -2x^2 + 4x - 8 = 0[/tex]
Thus the required equation of polynomial is found