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: 2/12 answered: 2/12 question 3 find a degree 3 polynomial with real coefficients having zeros 2 and 4i and a lead coefficient of 1. write p in expanded form. p(x) =

Explanation:

Step1: Recall conjugate root theorem

If a polynomial with real - coefficients has a complex zero \(a + bi\), then its conjugate \(a - bi\) is also a zero. Given the zero \(4i\), its conjugate \(-4i\) is also a zero of the polynomial \(P(x)\) since the polynomial has real coefficients.

Step2: Write the polynomial in factored form

If \(r_1\), \(r_2\), and \(r_3\) are the zeros of a polynomial, then the polynomial can be written as \(P(x)=a(x - r_1)(x - r_2)(x - r_3)\). Here, \(a = 1\), \(r_1=2\), \(r_2 = 4i\), and \(r_3=-4i\). So \(P(x)=(x - 2)(x - 4i)(x + 4i)\).

Step3: Simplify \((x - 4i)(x + 4i)\)

Using the difference - of - squares formula \((a - b)(a + b)=a^{2}-b^{2}\), where \(a=x\) and \(b = 4i\), we have \((x - 4i)(x + 4i)=x^{2}-(4i)^{2}=x^{2}+16\).

Step4: Multiply \((x - 2)\) and \((x^{2}+16)\)

\(P(x)=(x - 2)(x^{2}+16)=x\cdot x^{2}+x\cdot16-2\cdot x^{2}-2\cdot16=x^{3}-2x^{2}+16x - 32\).

Answer:

\(P(x)=x^{3}-2x^{2}+16x - 32\)