## Algebra 2 (1st Edition)

$-3, -\frac{5}{3},\frac{1}{2}$
We are given the polynomial function: $$f(x)=6x^3+25x^2+16x-15.$$ $\bf{Step\text{ }1}$ First we will list the possible rational zeros. The leading coefficient is $6$ and the constant term is $-15$. So the possible rational zeros are: $$\pm 1,\pm3,\pm 5,\pm 15,\pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{5}{2},\pm \dfrac{15}{2},\pm \dfrac{1}{3},\pm \dfrac{5}{3},\pm \dfrac{1}{6},\pm \dfrac{5}{6}.$$ $\bf{Step\text{ }2}$ Choose reasonable values from the list using the graph of the function. We find the reasonable values are: $$x=-3, x=-\dfrac{3}{2}, x=-\dfrac{5}{3},x=\dfrac{1}{2}, x=\dfrac{1}{3}$$ $\bf{Step\text{ }3}$ Check the values using synthetic division until a zero is found: Test $x=-3$: Because we got that the remainder is $0$, $-3$ is a zero of $f$. $\bf{Step\text{ }4}$ Write the result of the synthetic division: \begin{align*} f(x)&=(x+3)(6x^2+7x-5).\\ \end{align*} $\bf{Step\text{ }5}$ Find the remaining zeros of $f$ by solving $6x^2+7x-5=0$ using the quadratic formula: \begin{align*} x&=\dfrac{-7\pm\sqrt{7^2-4(6)(-5)}}{2(6)}\\ &=\dfrac{-7\pm 13}{12}\\ x_1&=\dfrac{-7-13}{12}=-\dfrac{5}{3}\\ x_2&=\dfrac{-7+13}{12}=\dfrac{1}{2}. \end{align*} The real zeros of $f$ are $-3, -\frac{5}{3},\frac{1}{2}$.