Find all zeros of the function f(x) = 8x^3 - 18x^2 - 71x + 60. Enter the zeros separated by commas.

Steps to Solve

1. Rational Root Theorem

The Rational Root Theorem states that if a polynomial has integer coefficients, then every rational root of the polynomial has the form $p/q$ where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. For the given polynomial $f(x) = 8x^3 - 18x^2 - 71x + 60$, the constant term is 60 and the leading coefficient is 8.

2. List Possible Rational Roots

List the factors of 60: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60$. List the factors of 8: $\pm 1, \pm 2, \pm 4, \pm 8$. Possible rational roots are of the form $\frac{\text{factor of 60}}{\text{factor of 8}}$. Therefore, possible roots include: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{5}{4}, \pm \frac{15}{4}, \pm \frac{1}{8}, \pm \frac{3}{8}, \pm \frac{5}{8}, \pm \frac{15}{8}$.

3. Test Possible Roots

We test these possible roots by plugging them into the original equation $f(x)$. We start with integers that are easy to calculate. Let's test $x = 1/2$:

$f(\frac{1}{2}) = 8(\frac{1}{2})^3 - 18(\frac{1}{2})^2 - 71(\frac{1}{2}) + 60 = 8(\frac{1}{8}) - 18(\frac{1}{4}) - \frac{71}{2} + 60 = 1 - \frac{9}{2} - \frac{71}{2} + 60 = 61 - \frac{80}{2} = 61 - 40 = 21 \neq 0$

Let's test $x = -3$:

$f(-3) = 8(-3)^3 - 18(-3)^2 - 71(-3) + 60 = 8(-27) - 18(9) + 213 + 60 = -216 - 162 + 213 + 60 = -378 + 273 = -105 \neq 0$

Let's test $x = 4$:

$f(4) = 8(4)^3 - 18(4)^2 - 71(4) + 60 = 8(64) - 18(16) - 284 + 60 = 512 - 288 - 284 + 60 = 572 - 572 = 0$.

So, $x = 4$ is a root.

4. Perform Polynomial Division

Since $x = 4$ is a root, $(x - 4)$ is a factor of $f(x)$. We can perform polynomial division to find the other factor. Dividing $8x^3 - 18x^2 - 71x + 60$ by $(x-4)$ gives us $8x^2 + 14x - 15$.

5. Solve the Quadratic Equation

Now we need to find the roots of the quadratic equation $8x^2 + 14x - 15 = 0$. We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 8$, $b = 14$, and $c = -15$.

$x = \frac{-14 \pm \sqrt{14^2 - 4(8)(-15)}}{2(8)} = \frac{-14 \pm \sqrt{196 + 480}}{16} = \frac{-14 \pm \sqrt{676}}{16} = \frac{-14 \pm 26}{16}$.

So, $x = \frac{-14 + 26}{16} = \frac{12}{16} = \frac{3}{4}$ and $x = \frac{-14 - 26}{16} = \frac{-40}{16} = -\frac{5}{2}$.

6. List All Roots The three roots are $4, 3/4, -5/2$.