Question 3(Multiple Choice Worth 1 points) (07.03 MC) Factor completely 3x^2-x+5 Prime (3x+1)(x-5) (3x-1)(x+5) (x+1)(3x-5)

To factor the quadratic expression \(3x^2 - x + 5\) completely, we need to check if it can be factored into the form \((ax + b)(cx + d)\).<br /><br />First, let's use the quadratic formula to find the roots of the quadratic equation \(3x^2 - x + 5 = 0\):<br /><br />The quadratic formula is given by:<br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />For the quadratic equation \(3x^2 - x + 5 = 0\), the coefficients are:<br />\[ a = 3, \quad b = -1, \quad c = 5 \]<br /><br />Substitute these values into the quadratic formula:<br />\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} \]<br />\[ x = \frac{1 \pm \sqrt{1 - 60}}{6} \]<br />\[ x = \frac{1 \pm \sqrt{-59}}{6} \]<br /><br />Since the discriminant (\(\sqrt{-59}\)) is negative, the quadratic equation has no real roots. This means that the quadratic expression \(3x^2 - x + 5\) cannot be factored over the real numbers.<br /><br />Therefore, the correct answer is:<br />\[ \text{Prime} \]<br /><br />So, the correct choice is:<br />\[ \text{Prime} \]