Answer :
To find the zeros of the polynomial [tex]\( n(x) = 3x^3 - x^2 - 39x + 13 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that satisfy the equation:
[tex]\[ 3x^3 - x^2 - 39x + 13 = 0 \][/tex]
This requires solving the cubic equation. After solving for [tex]\( x \)[/tex], we obtain the exact zeros of the polynomial.
The solutions to the polynomial equation are:
[tex]\[ x = \frac{1}{3}, \quad x = -\sqrt{13}, \quad x = \sqrt{13} \][/tex]
Therefore, the zeros of the polynomial [tex]\( n(x) = 3x^3 - x^2 - 39x + 13 \)[/tex] in exact form are:
[tex]\[ x = \frac{1}{3}, \][/tex]
[tex]\[ x = -\sqrt{13}, \][/tex]
[tex]\[ x = \sqrt{13}. \][/tex]
These values of [tex]\( x \)[/tex] are the points where the polynomial [tex]\( n(x) \)[/tex] intersects the x-axis.
[tex]\[ 3x^3 - x^2 - 39x + 13 = 0 \][/tex]
This requires solving the cubic equation. After solving for [tex]\( x \)[/tex], we obtain the exact zeros of the polynomial.
The solutions to the polynomial equation are:
[tex]\[ x = \frac{1}{3}, \quad x = -\sqrt{13}, \quad x = \sqrt{13} \][/tex]
Therefore, the zeros of the polynomial [tex]\( n(x) = 3x^3 - x^2 - 39x + 13 \)[/tex] in exact form are:
[tex]\[ x = \frac{1}{3}, \][/tex]
[tex]\[ x = -\sqrt{13}, \][/tex]
[tex]\[ x = \sqrt{13}. \][/tex]
These values of [tex]\( x \)[/tex] are the points where the polynomial [tex]\( n(x) \)[/tex] intersects the x-axis.
