Question

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

Answer 1

Answer:

$f(x)=3x^3+x^2+3x+1$

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$

If the polynomial function must be with integer coefficients, then it has a form

$f(x)=3x^3+x^2+3x+1$