Question

My teacher put this on our take home test but we haven’t done any problems like this can anyone explain?

Answer 1

Zeros of a polynomial can be defined as the points where the polynomial becomes zero as a whole.

But the zeros of the polynomial are at x=1 and x=3i.

But one of the zeros above is a complex number. So, we have;

`\begin{gathered} x=1 \\ x^2=-9 \\ x=\sqrt[]{-9} \\ x=\pm3i \\ x=3i \\ x=-3i \end{gathered}`

Thus, the missing zero of the polynomial is negative 3i (-3i).

(b) Thus, the polynomial in a factored form is;

`\begin{gathered} x-1=0,x-3i=0,x+3i=0 \\ (x-1)(x-3i)(x+3i)=0 \end{gathered}`

(c) In standard form, we have;

`\begin{gathered} (x-1)(x-3i)(x+3i)=0 \\ (x-1)(x(x+3i)-3i(x+3i))=0 \\ (x-1)(x^2+3ix-3ix-9(i^2))=0 \\ (x-1)(x^2-9(-1))=0 \\ (x-1)(x^2+9)=0 \\ \end{gathered}`

We expand further, we have;

`\begin{gathered} x(x^2+9)-1(x^2+9)=0 \\ x^3+9x-x^2-9=0 \\ x^3-x^2+9x-9=0 \end{gathered}`