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.
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Testing \( x = 1 \):
\[ V(1) = 1^3 - 7(1^2) + 13(1) - 3 = 1 - 7 + 13 - 3 = 4 \quad (\text{not a root}) \]
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Testing \( x = -1 \):
\[ V(-1) = (-1)^3 - 7(-1)^2 + 13(-1) - 3 = -1 - 7 - 13 - 3 = -24 \quad (\text{not a root}) \]
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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:
\[ \boxed{3, 2 + \sqrt{3}, 2 - \sqrt{3}} \]