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

User LuckyLikey
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1 Answer

7 votes

Answer:

The polynomial is:


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

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

User Gnath
by
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