## College Algebra (10th Edition)

The zeros of the given function are: $\color{blue}{\left\{i, -i, -1, 1\right\}}$ The completely factored form of the given function is: $\color{blue}{P(x) = (x-1)(x+1)(x-i)(x+i)}$
RECALL: $a^2-b^2=(a-b)(a+b)$ The given polynomial function can be written as $P(x) = (x^2)^2-1^2$. Factor the difference of two squares using the formula above with $a=x^2$ and $b=1$ to obtain: $P(x) = (x^2+1)(x^2-1)$ Factor the second binomial using the same formula above to obtain: $P(x) = (x^2+1)(x-1)(x+1)$ Equate each factor to zero then solve each equation to obtain: \begin{array}{ccccc} &x^2+1=0 &\text{or} &x-1=0 &\text{or} &x+1=0 \\&x^2=-1 &\text{or} &x=1 &\text{or} &x=-1 \\&x=\pm\sqrt{-1} &\text{or} &x=1 &\text{or} &x=-1 \\&x=\pm i &\text{or} &x=1 &\text{or} &x=-1 \end{array} Thus, the zeros of the given function are: $\color{blue}{\left\{i, -i, -1, 1\right\}}$ The completely factored form of the given function is: $\color{blue}{P(x) = (x-1)(x+1)(x-i)(x+i)}$
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