Question

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 0,1, and 1+i

Answer 1

Answer:

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

Step-by-step explanation:

The standard polynomial is given by:

$f(x)=a(x-p)(x-q)...$

Where a is the leading coefficient and p and q are the zeros.

We are given that the zeros are x = 0, 1, and 1 + i.

First, by the Complex Root Theorem, if 1 + i is a zero, then 1 - i must also be a zero.

Therefore:

$f(x)=1(x-(0))(x-(1))(x-(1+i))(x-(1-i))$

Simplify:

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

Expand the third and fourth factors:

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

Distribute:

$=x^2-x-ix-(x(1-i)-(1+i)(1-i))$

Expand:

$=x^2-x-ix-x+ix+(1-i+i-i^2)$

Therefore:

$=x^2-2x+(2)=x^2-2x+2$

Hence, our polynomial function is:

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