Question

1. determine the third degree polynomial with roots -3 and 6 - 2i.

\\( p(x)=x^3 - 9x^2 + 4x + 120 \\)
\\( p(x)=x^3 - x^2 + 4x + 240 \\)
\\( p(x)=x^3 - 18x^2 - 4x + 120 \\)
\\( p(x)=x^3 + 9x^2 - 4x + 120 \\)

Explanation:

Step1: Recall Complex Conjugate Root Theorem

For a polynomial with real coefficients, if \( a+bi \) is a root, then \( a - bi \) is also a root. Given root \( 6 - 2i \), its conjugate \( 6+2i \) is also a root. So the three roots of the third - degree polynomial are \( r_1=-3 \), \( r_2 = 6 - 2i \), \( r_3=6 + 2i \).

Step2: Use the Factor Theorem

If \( r \) is a root of a polynomial \( P(x) \), then \( (x - r) \) is a factor of \( P(x) \). So the polynomial can be written as:
\( P(x)=(x + 3)[x-(6 - 2i)][x-(6 + 2i)] \)

Step3: Simplify the product of the complex - factor terms

First, simplify \( [x-(6 - 2i)][x-(6 + 2i)] \).
Using the formula \( (a - b)(a + b)=a^{2}-b^{2} \), where \( a=(x - 6) \) and \( b = 2i \).
\( [x-(6 - 2i)][x-(6 + 2i)]=(x - 6)^{2}-(2i)^{2} \)
Expand \( (x - 6)^{2}=x^{2}-12x + 36 \) and \( (2i)^{2}=4i^{2}=-4 \) (since \( i^{2}=-1 \))
So \( (x - 6)^{2}-(2i)^{2}=x^{2}-12x + 36+4=x^{2}-12x + 40 \)

Step4: Multiply by the remaining factor

Now, multiply \( (x + 3) \) with \( x^{2}-12x + 40 \):
\( P(x)=(x + 3)(x^{2}-12x + 40) \)
Using the distributive property \( a(b + c + d)=ab+ac + ad \), where \( a = x+3 \), \( b=x^{2} \), \( c=-12x \), \( d = 40 \)
\( (x + 3)(x^{2}-12x + 40)=x(x^{2}-12x + 40)+3(x^{2}-12x + 40) \)
\(=x^{3}-12x^{2}+40x+3x^{2}-36x + 120 \)
Combine like terms:
\( x^{3}+(-12x^{2}+3x^{2})+(40x-36x)+120=x^{3}-9x^{2}+4x + 120 \)

Answer:

\( P(x)=x^{3}-9x^{2}+4x + 120 \) (First Option)