Question

find all other zeros of $p(x)=x^{3}-5x^{2}+16x - 30$, given that $1 + 3i$ is a zero.
(if there is more than one zero, separate them with commas.)

Explanation:

Step1: Use Conjugate Root Theorem

For a polynomial with real coefficients, if \(a + bi\) is a zero, then its conjugate \(a - bi\) is also a zero. Given \(1 + 3i\) is a zero, so \(1 - 3i\) is also a zero.

Step2: Find the quadratic factor

The quadratic factor corresponding to zeros \(1 + 3i\) and \(1 - 3i\) is \((x-(1 + 3i))(x-(1 - 3i))\).
Simplify this:
\[

$$\begin{align*} &(x - 1 - 3i)(x - 1 + 3i)\\ =&[(x - 1)-3i][(x - 1)+3i]\\ =&(x - 1)^2-(3i)^2\\ =&x^2 - 2x + 1+9\\ =&x^2 - 2x + 10 \end{align*}$$

\]

Step3: Divide the polynomial by the quadratic factor

Divide \(P(x)=x^3 - 5x^2 + 16x - 30\) by \(x^2 - 2x + 10\) using polynomial long division or synthetic division.
Using polynomial long division:
Divide \(x^3 - 5x^2 + 16x - 30\) by \(x^2 - 2x + 10\).
\(x^3\div x^2=x\), multiply \(x^2 - 2x + 10\) by \(x\) to get \(x^3 - 2x^2 + 10x\).
Subtract from \(x^3 - 5x^2 + 16x - 30\):
\((x^3 - 5x^2 + 16x - 30)-(x^3 - 2x^2 + 10x)=-3x^2 + 6x - 30\)
Now, \(-3x^2\div x^2=-3\), multiply \(x^2 - 2x + 10\) by \(-3\) to get \(-3x^2 + 6x - 30\).
Subtract: \((-3x^2 + 6x - 30)-(-3x^2 + 6x - 30)=0\)
So, \(P(x)=(x^2 - 2x + 10)(x - 3)\)

Step4: Find the remaining zero

Set \(x - 3 = 0\), we get \(x = 3\)

Answer:

\(1 - 3i, 3\)