College Algebra 7th Edition

(a) $\left\{0, 1-i, 1+i\right\}$. (b) $P(x) = x(x^2-2x+2)$
$\bf{(a) \text{ Zeros}}$ Factor the polynomial completely to obtain: $P(x) = x(x^2-2x+2$) Equate each factor to zero then solve each equation to obtain: \begin{array}{ccccc} &x=0 &\text{or} &x^2-2x+2=0\end{array} The second equation can be solved using the quadratic formula $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. The trinomial has $a=1, b=-2, c=2$. Substitute these values into the quadratic formula above to obtain: $x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(2)}}{2(1)} \\x=\dfrac{2\pm \sqrt{4-8}}{2} \\x=\dfrac{2\pm \sqrt{-4}}{2} \\x=\dfrac{2\pm \sqrt{4(-1)}}{2} \\x=\dfrac{2\pm 2\sqrt{-1}}{2} \\x=\dfrac{2\pm 2i}{2} \\x=1\pm i$ Thus, the zeros of the function are: $\left\{0, 1-i, 1+i\right\}$. $\bf{(b) \text{ Completely Factored Form}}$ From part (a) above, the completely factored form of $P(x)$ is: $P(x) = x(x^2-2x+2)$