Question

Finding a polynomial of a given degree with given zeros

Answer 1

Since f(x) is a polynomial with 3rd degree, then it will have 3 roots (zeroes)

One of them is real and the other two are complex conjugate roots

Since the real root is 4, then

x = 4

Since the complex root is (1 - i), then

The other root will be the conjugate of it (1 + i)

x = (1 - i)

x = (1 + i)

To find f(x) we will multiply the three factors of it

We can get the factors from the zeroes

$x=4$

Subtract 4 from both sides

$\begin{gathered} x-4=4-4 \\ x-4=0 \end{gathered}$

The first factor is (x - 4)

$\begin{gathered} x=1-i \\ x-(1-i)=(1-i)-(1-i) \\ x-1+i=0 \end{gathered}$

The second factor is (x - 1 + i)

The third factor is (x - 1 - i)

$f(x)=(x-4)(x-1+i)(x-1-i)$

We will multiply them to find f(x)

$\begin{gathered} (x-1+i)(x-1-i)= \\ x^2-x-ix-x+1+i+ix-i-(i^2)= \\ x^2-2x+1-(-1)= \\ x^2-2x+1+1= \\ x^2-2x+2 \end{gathered}$

Multiply it by (x - 4)

$\begin{gathered} f(x)=(x-4)(x^2-2x+2) \\ f(x)=x^3-2x^2+2x-4x^2+8x-8 \\ f(x)=x^3-6x^2+10x-8 \end{gathered}$

The answer is

$f(x)=x^3-6x^2+10x-8$