## Algebra 2 (1st Edition)

$1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$
We are given the polynomial function: $$f(x)=4x^3-20x+16.$$ $\bf{Step\text{ }1}$ First we will list the possible rational zeros. The leading coefficient is $4$ and the constant term is $16$. So the possible rational zeros are: $$\pm 1,\pm2,\pm 4,\pm 8,\pm 16,\pm \dfrac{1}{2},\pm \dfrac{1}{4}.$$ $\bf{Step\text{ }2}$ Choose reasonable values from the list using the graph of the function. We find the only reasonable value is: $$x=1.$$ $\bf{Step\text{ }3}$ Check this value using synthetic division: Test $x=1$. Because we got that the remainder is $0$, $1$ is a zero of $f$. $\bf{Step\text{ }4}$ Factor out the binomial using the result of the synthetic division: \begin{align*} f(x)&=(x-1)(4x^2+4x-16)\\ &=4(x-1)(x^2+x-4).\\ \end{align*} $\bf{Step\text{ }5}$ Find the remaining zeros of $f$ by solving $x^2+x-4=0$ using the quadratic formula: \begin{align*} x&=\dfrac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}\\ &=\dfrac{-1\pm\sqrt{17}}{2}. \end{align*} The real zeros of $f$ are $1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$. 