Question

Find all the zeros. Write the answer in exact form.
V(x)=x^3-7x^2+13x-3

Answer 1

To find the zeros of the polynomial \( V(x) = x^3 - 7x^2 + 13x - 3 \), we can use various methods such as factoring, synthetic division, or the Rational Root Theorem. Since this is a cubic polynomial, let's start by trying to find rational roots using the Rational Root Theorem.

The potential rational roots can be the factors of the constant term (\(-3\)) divided by the factors of the leading coefficient (\(1\)). The potential rational roots are:

\[
\pm 1, \pm 3
\]

We will test these possible roots in the polynomial to find at least one zero.

1. **Testing \( x = 1 \)**:

\[
V(1) = 1^3 - 7(1^2) + 13(1) - 3 = 1 - 7 + 13 - 3 = 4 \quad (\text{not a root})
\]

2. **Testing \( x = -1 \)**:

\[
V(-1) = (-1)^3 - 7(-1)^2 + 13(-1) - 3 = -1 - 7 - 13 - 3 = -24 \quad (\text{not a root})
\]

3. **Testing \( x = 3 \)**:

\[
V(3) = 3^3 - 7(3^2) + 13(3) - 3 = 27 - 63 + 39 - 3 = 0 \quad (\text{is a root})
\]

Since \( x = 3 \) is a root, we can use synthetic division to factor the polynomial.

Using synthetic division to divide \( V(x) \) by \( x - 3 \):

\[
\begin{array}{r|rrrr}
3 & 1 & -7 & 13 & -3 \\
& & 3 & -12 & 3 \\
\hline
& 1 & -4 & 1 & 0 \\
\end{array}
\]

The result is \( x^2 - 4x + 1 \).

Now, we have \( V(x) = (x - 3)(x^2 - 4x + 1) \).

Next, we find the zeros of the quadratic \( x^2 - 4x + 1 \). We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1, b = -4, c = 1 \):

\[
x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}
\]

So, the zeros of the polynomial \( V(x) \) are:

\[
x = 3, \quad x = 2 + \sqrt{3}, \quad x = 2 - \sqrt{3}
\]

In exact form, the zeros of \( V(x) \) are:

\[
\boxed{3, 2 + \sqrt{3}, 2 - \sqrt{3}}
\]