Question

Find a polynomial of least degree with real coefficients in standard form that has 0 and the square root of 7i as zeros.

A. x² - 7
B. x³ - 7
C. x² + 7
D. x³ + 7

Answer 1

To find the polynomial with 0 and \(\sqrt{7i}\) as zeros, we also include the complex conjugate \(-\sqrt{7i}\). Multiplying the factors associated with these zeros yields the polynomial x^3 + 7x, and eliminating the linear term, as it's not given, results in the polynomial x^3 + 7.

Step-by-step explanation:

To find a polynomial of least degree with real coefficients that has the given zeros, namely 0 and \(\sqrt{7i}\), we must first recognize that the nonreal zero \(\sqrt{7i}\) implies that its complex conjugate \(-\sqrt{7i}\) must also be a zero of the polynomial, because coefficients are real. Hence, our polynomial must have at least three zeros: 0, \(\sqrt{7i}\), and \(-\sqrt{7i}\).

Using these zeros, we build the polynomial by multiplying the factors associated with each:

x (the zero of 0)

(x - \sqrt{7i}) (the zero of \(\sqrt{7i}\))

(x + \sqrt{7i}) (the zero of \(-\sqrt{7i}\))

The process of multiplying these factors results in the polynomial:

\(x(x - \sqrt{7i})(x + \sqrt{7i}) = x(x^2 + 7) = x^3 + 7x\)

To get a polynomial in standard form, we then set the linear term to zero since it is not part of the problem's given zeros. Therefore, our answer is D. x^3 + 7.