Find all other zeros of $\newline$ $P(x)=x^{3}+5x^{2}+8x+6$ , given that $\newline$ $-1-i$ is a zero. $\newline$ (If there is more than one zero, separate them with commas.)

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Q. Find all other zeros of

\newline

P(x)=x^{3}+5x^{2}+8x+6

, given that

\newline

-1-i

is a zero.

\newline

(If there is more than one zero, separate them with commas.)

Complex Conjugate Root Theorem: Since $-1-i$ is a zero of the polynomial $P(x)$ , and the coefficients of $P(x)$ are real numbers, the complex conjugate of $-1-i$ , which is $-1+i$ , must also be a zero of $P(x)$ due to the Complex Conjugate Root Theorem.
Factorization of Polynomial: To find the remaining zeros, we can use the known zeros to factor the polynomial. The factors corresponding to the zeros $-1-i$ and $-1+i$ are $(x - (-1 - i)) = (x + 1 + i)$ and $(x - (-1 + i)) = (x + 1 - i)$ , respectively.
Quadratic Factor Calculation: We can multiply these two factors to find the quadratic factor of $P(x)$ :
$(x + 1 + i)(x + 1 - i) = x^2 + x - ix + x + 1 - i - ix + i - i^2$
Since $i^2 = -1$ , this simplifies to:
$x^2 + 2x + (1 - (-1)) = x^2 + 2x + 2$
Remaining Zero Determination: Now we can divide the original polynomial $P(x)$ by the quadratic factor we found to determine the remaining zero: $\frac{P(x)}{x^2 + 2x + 2} = \frac{x^3 + 5x^2 + 8x + 6}{x^2 + 2x + 2}$ We can perform polynomial long division or synthetic division to find the quotient.
Polynomial Long Division: Performing the division, we set up the long division as follows: $\newline$ $x + 3$ $\newline$ $\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$ $\newline$ $x^2+2x+2 | x^3 + 5x^2 + 8x + 6$ $\newline$ $\quad\quad - (x^3 + 2x^2 + 2x)$ $\newline$ $\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$ $\newline$ $\quad\quad\quad\quad\quad 3x^2 + 6x$ $\newline$ $\quad\quad\quad\quad\quad - (3x^2 + 6x + 6)$ $\newline$ $\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$ $\newline$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 0$ $\newline$ The quotient is $x + 3$ , and the remainder is $\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$ $0$ , which means the division is exact.
Final Zero Determination: The quotient $x + 3$ gives us the remaining linear factor of $P(x)$ , which corresponds to the zero $-3$ . Therefore, the zeros of $P(x)$ are $-1-i$ , $-1+i$ , and $-3$ .