College Algebra 7th Edition

$P(x)=(x-2)^2(x^2+2x+4)$ Zero: $2 \text{ (multiplicity 2)}$ Refer to the graph below.
Factor the polynomial completely to obtain: \begin{align*} P(x)&=(x^4-2x^3)+(-8x+16)\\ &=x^3(x-2)+(-8)(x-2)\\ &=(x-2)(x^3-8)\\ &=(x-2)(x-2)(x^2+2x+4)\\ &=(x-2)^2(x^2+2x+4)\\ \end{align*} To find the zeros, use the Zero-Product Property by equating each factor to $0$, then solve each equation to obtain: \begin{align*} (x-2)^2&=0 &\text{or}& &x^2+2x+4=0\\ x-2&=0 &\text{or}& &\text{(no real solution)} \\ x&=2 \text{ (multiplicity 2)}\\ \end{align*} The zero of the function is $2$. Use a graphing utility to graph $P(x)$. Refer to the graph above.