Question

write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. write the polynomial in standard form. 2, 5 - i f(x) = \square

Explanation:

Step1: Identify conjugate zero

Since coefficients are rational, $5+i$ is also a zero.

Step2: Write factor form

$f(x)=(x-2)(x-(5-i))(x-(5+i))$

Step3: Simplify complex factors

First expand $(x-(5-i))(x-(5+i))$:

$$\begin{align*} &(x-5+i)(x-5-i)\\ =&(x-5)^2 - i^2\\ =&x^2-10x+25 - (-1)\\ =&x^2-10x+26 \end{align*}$$

Step4: Multiply with linear factor

Multiply $(x-2)$ with $x^2-10x+26$:

$$\begin{align*} &(x-2)(x^2-10x+26)\\ =&x(x^2-10x+26)-2(x^2-10x+26)\\ =&x^3-10x^2+26x-2x^2+20x-52\\ =&x^3-12x^2+46x-52 \end{align*}$$

Answer:

$x^3-12x^2+46x-52$