Question

Find all zeros of the function f(x)=6x^3-29x^2+36x-9 . Enter the zeros separated by commas.

Answer 1

To find the zeros of the function \( f(x) = 6x^3 - 29x^2 + 36x - 9 \), we'll start by using the Rational Root Theorem to look for potential rational roots.

The possible rational roots are the factors of the constant term (-9) divided by the factors of the leading coefficient (6).

The factors of -9 are: \( \pm 1, \pm 3, \pm 9 \).

The factors of 6 are: \( \pm 1, \pm 2, \pm 3, \pm 6 \).

So the possible rational roots are:

\[
\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{3} (which is 1), \pm 9, \pm \frac{9}{2}, \pm \frac{9}{3} (which is 3), \pm \frac{9}{6} (which is \frac{3}{2})
\]

We can test some of these possible rational roots using synthetic division or direct substitution.

Let's test \( x = 1 \):

\[
f(1) = 6(1)^3 - 29(1)^2 + 36(1) - 9 = 6 - 29 + 36 - 9 = 4 \quad (\text{not zero})
\]

Now, testing \( x = 3 \):

\[
f(3) = 6(3)^3 - 29(3)^2 + 36(3) - 9
= 6(27) - 29(9) + 108 - 9
= 162 - 261 + 108 - 9
= 0 \quad (\text{this is a zero})
\]

Now that we have found one zero, \( x = 3 \), we will use synthetic division to divide \( f(x) \) by \( x - 3 \):

\[
\begin{array}{r|rrrr}
3 & 6 & -29 & 36 & -9 \\
& & 18 & -33 & 9 \\
\hline
& 6 & -11 & 3 & 0 \\
\end{array}
\]

This gives us the quotient \( 6x^2 - 11x + 3 \). We can now find the zeros of \( 6x^2 - 11x + 3 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 6 \cdot 3}}{2 \cdot 6}
\]
\[
= \frac{11 \pm \sqrt{121 - 72}}{12} = \frac{11 \pm \sqrt{49}}{12} = \frac{11 \pm 7}{12}
\]
Calculating the two possible values:

1. \( x = \frac{18}{12} = \frac{3}{2} \)
2. \( x = \frac{4}{12} = \frac{1}{3} \)

The zeros of the function \( f(x) = 6x^3 - 29x^2 + 36x - 9 \) are therefore:

\[
\boxed{3, \frac{3}{2}, \frac{1}{3}}
\]