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Write a polynomial function in standard form with real coefficients whose zeros include 1,4i, and -4i

User Jawsware
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2 Answers

3 votes

Answer:

f(x) = x^3 - x^2 + 16x - 16

Explanation:

If a polynomial has the zeros 1, 4i, and -4i, then it must have the factors (x - 1), (x - 4i), and (x + 4i). This is because a factor of (x - a) produces a root of x = a.

To find the polynomial, we can multiply these factors together:

(x - 1)(x - 4i)(x + 4i)

= (x - 1)(x^2 - (4i)^2)

= (x - 1)(x^2 + 16)

= x^3 + 16x - x^2 - 16

So the polynomial function in standard form with real coefficients whose zeros include 1, 4i, and -4i is:

f(x) = x^3 - x^2 + 16x - 16

User Atlanto
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8.7k points
3 votes
If the zeros of a polynomial are 1, 4i, and -4i, then the polynomial can be factored as follows:

(x - 1)(x - 4i)(x + 4i)

To write this polynomial in standard form, we need to multiply out the factors and simplify:

(x - 1)(x - 4i)(x + 4i)
= (x - 1)(x^2 - (4i)^2)
= (x - 1)(x^2 - 16i^2)
= (x - 1)(x^2 + 16)

Expanding the product, we get:

x^3 + 16x - x^2 - 16

So the polynomial function with real coefficients whose zeros include 1, 4i, and -4i is:

f(x) = x^3 - x^2 + 16x - 16

:))
User Guillaume Gaujac
by
8.1k points

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