How do you write a polynomial function of least degree that has real coefficients, the following given zeros 3-i,5i and a leading coefficient of 1?

Question

Caleb Alexander · Accepted Answer

Any polynomial with real coefficients and imaginary or complex zeros, must have zeros that are conjugate pairs, therefore, the polynomial must have the following factors:

#y = (x - 3 + i)(x - 3 - i)(x - 5i)(x + 5i)#

Multiplying the last two factors is easy; it is the sum of two squares:

#y = (x - 3 + i)(x - 3 - i)(x^2 + 25)#

Multiplying the first two factors is a bit more difficult:

#y = (x^2 - 3x - ix -3x + 9 + 3i + ix -3i - i^2)(x^2 + 25)#

Combine like terms:

#y = (x^2 - 6x + 9 - i^2)(x^2 + 25)#

Use the identity #i^2 = -1#:

#y = (x^2 - 6x +10)(x^2 + 25)#

Multiply the remaining factors:

#y = x^4 - 6x^3 +10x^2 + 25x^2 -150x + 250#

Combine like terms:

#y = x^4 - 6x^3 + 35x^2 -150x + 250#

Benjamin Achtenberg · Answer

undefined To write a polynomial function with real coefficients and the given zeros, you can use the complex conjugate theorem. Since the zeros given are 3-i, 5i, and 3+i (since complex zeros always come in conjugate pairs), the factors of the polynomial are (x - 3 + i), (x - 3 - i), and (x - 5i). Multiplying these factors together gives the polynomial function.