Question

homework assignment 5.5 zeros of polynomials
due friday by 11:59pm points 12 submitting an external tool
homework assignment 5.5 zeros of polynomials
score: 3/12 answered: 3/12
question 4
find a degree 3 polynomial with real coefficients having zeros 2 and 3 - 3i and a lead coefficient of 1. write p in expanded form.
p(x) =

Explanation:

Step1: Recall complex - conjugate root theorem

If a polynomial with real coefficients has a complex zero \(a + bi\), then its complex - conjugate \(a - bi\) is also a zero. Given the zero \(3-3i\), its complex - conjugate \(3 + 3i\) is also a zero of the polynomial \(P(x)\).

Step2: Write the polynomial in factored form

Since the zeros of the polynomial are \(x = 2\), \(x=3 - 3i\), and \(x = 3 + 3i\) and the leading coefficient is \(1\), we can write the polynomial as \(P(x)=(x - 2)[x-(3 - 3i)][x-(3 + 3i)]\).

Step3: Simplify \([x-(3 - 3i)][x-(3 + 3i)]\)

\[

$$\begin{align*} [x-(3 - 3i)][x-(3 + 3i)]&=(x - 3+3i)(x - 3 - 3i)\\ &=(x - 3)^2-(3i)^2\\ &=x^{2}-6x + 9+9\\ &=x^{2}-6x+18 \end{align*}$$

\]

Step4: Multiply by \((x - 2)\)

\[

$$\begin{align*} P(x)&=(x - 2)(x^{2}-6x + 18)\\ &=x(x^{2}-6x + 18)-2(x^{2}-6x + 18)\\ &=x^{3}-6x^{2}+18x-2x^{2}+12x - 36\\ &=x^{3}-8x^{2}+30x - 36 \end{align*}$$

\]

Answer:

\(P(x)=x^{3}-8x^{2}+30x - 36\)